Pretorsion theories, stable category and preordered sets

Abstract

We show that in the category of preordered sets there is a natural notion of pretorsion theory, in which the partially ordered sets are the torsion-free objects and the sets endowed with an equivalence relation are the torsion objects. Correspondingly, it is possible to construct a stable category factoring out the objects that are both torsion and torsion-free.

1 Introduction

Every preorder can be obtained from an equivalence relation and a partial order. In this paper, we show that this can be interpreted in the language of a suitably defined (pre)torsion theory. More precisely, we consider the category $\mathbf {Preord}$ of all preordered sets $(A,\rho )$ and its full subcategories ParOrd and Equiv, whose objects are all partially ordered sets $(A,\le )$ and all sets $(A,\sim )$, where $\sim$ is an equivalence relation on A, respectively. Call trivial objects of Preord the objects of $\mathbf {Preord}$ that are in both ParOrd and Equiv. Trivial objects are the objects of $\mathbf {Preord}$ of the form $(A,=)$, where $=$ is the equality relation on A. Call trivial morphisms the morphisms in $\mathbf {Preord}$ that factor through a trivial object. Factoring out the category $\mathbf {Preord}$ modulo the trivial objects and the trivial morphisms, we get a category Preord, called the stable category of Preord. The category Preord minus the empty object, denoted by Preord$^*$, turns out to be a pointed category, that is, a category with a zero object. In the pointed category Preord$^*$, we have the notions of kernel and cokernel, which have their natural counterpart in $\mathbf {Preord}$. We call these natural counterparts prekernels and precokernels. This allows us to define short preexact sequences: if $f:A\rightarrow B$ and $g:B\rightarrow C$ are two morphisms in $\mathbf {Preord}$, we say that is a short preexact sequence in $\mathbf {Preord}$ if f is a prekernel of g and g is a precokernel of f. For example, for every object $(A,\rho )$ in $\mathbf {Preord}$, let $\sim$ be an equivalence relation on A contained in $\rho$. Then

is a short preexact sequence in $\mathbf {Preord}$. Here k is the identity map** and $\pi$ is the canonical projection. For any partially ordered set $(A,\rho )$, let $\simeq _\rho$ denote the equivalence relation on A defined, for every $a,b\in A$, by $a\simeq _\rho b$ if $a\rho b$ and $b\rho a$. Then

is a short preexact sequence in $\mathbf {Preord}$, $(A,\simeq _\rho )\in {}$Equiv and $(A/\simeq _\rho ,\le _\rho )\in {}$ParOrd. It is now easy to generalize what we have seen above and define pretorsion theories for arbitrary categories ${{\mathcal C}}$. Our definition is as follows. Let ${{\mathcal C}}$ be an arbitrary category. A pretorsion theory$({{\mathcal T}},{{\mathcal F}})$ for ${{\mathcal C}}$ consists of two classes ${{\mathcal T}},{{\mathcal F}}$ of objects of ${{\mathcal C}}$, closed under isomorphism, satisfying the following two conditions. Set ${{\mathcal Z}}{:}{=}{{\mathcal T}}\cap {{\mathcal F}}$, let ${\text {Triv}}_{{{\mathcal Z}}}(X,Y)$ be the set of all morphisms $X\rightarrow Y$ in ${{\mathcal C}}$ that factor though an object of ${{\mathcal Z}}$, and call ${{\mathcal Z}}$-preexact any sequence preexact relatively to ${{\mathcal Z}}$. The two conditions sufficient to define our pretorsion theories $({{\mathcal T}},{{\mathcal F}})$ are:

(1) For every object B of ${{\mathcal C}}$ there is a short ${{\mathcal Z}}$-preexact sequence

with $A\in {{\mathcal T}}$ and $C\in {{\mathcal F}}$.

(2) ${\text {Hom}}_{{{\mathcal C}}}(T,F)={\text {Triv}}_{{{\mathcal Z}}}(T, F)$ for every object $T\in {{\mathcal T}}$, $F\in {{\mathcal F}}$.

The pair (Equiv, PreOrd) turns out to be a pretorsion theory in $\mathbf {Preord}$.

The category Preord$^*$ provides an example of a category that arises as a quotient category of a topological category (the category of Alexandroff-discrete spaces, i.e. of preordered sets), but nevertheless has features typical of categories of commutative algebra (short exact sequences).

2 Preorders, partial orders and equivalence relations

A preorder on a set A is a relation on A that is reflexive and transitive. We will denote by Preord the category of all preordered sets. Its objects are the pairs $(A,\rho )$, where A is a set and $\rho$ is a preorder on A. The morphisms $f:(A,\rho )\rightarrow (A',\rho ')$ in Preord are the map**s f of A into $A'$ such that $a\rho b$ implies $f(a)\rho ' f(b)$ for all $a,b\in A$. As usual, when there is no danger of confusion or when the preorder is clear from the context, we will denote the preordered set $(A,\rho )$ simply by A.

Remark 2.1

We now fix some trivial notations useful in the sequel. The category Preord has arbitrary coproducts and arbitrary products. Given any family $\{\,(A_i,\rho _i)\mid i\in I\,\}$ of preordered sets, the coproduct of the family is the disjoint union $A=\coprod _{i\in I}A_i ({:}{=}\{(a,i)\mid i\in I,a\in A_i \})$, endowed with the coproduct preorder$\rho$ defined, for every $(a,i), (b,j)\in A$, by $(a,i)\rho (b,j)$ if $i=j$ and $a\rho _i b.$ The product in Preord of the family is its Cartesian product with the componentwise preorder.

The main examples of preordered sets $(A,\rho )$ are those in which the preorder $\rho$ is a partial order (i.e. $\rho$ is antisymmetric) or an equivalence relation (i.e. $\rho$ is symmetric). The full subcategories of Preord whose objects are all preordered sets $(A,\rho )$ with $\rho$ a partial order (an equivalence relation) will be denoted by ParOrd (Equiv, respectively).

In our first proposition, we describe preorders on a set A. They are obtained from an equivalence relation $\sim$ on A and a partial order on the quotient set $A/\sim$.

Proposition 2.2

Let A be a set. There is a one-to-one correspondence between the set of all preorders $\rho$ on A and the set of all pairs $(\sim ,\le )$, where $\sim$ is an equivalence relation on A and $\le$ is a partial order on the quotient set $A/\sim$. The correspondence associates with every preorder $\rho$ on A the pair $(\simeq _\rho ,\le _\rho )$, where $\simeq _\rho$ is the equivalence relation defined, for every $a,b\in A$, by $a\simeq _\rho b$ if $a\rho b$ and $b\rho a$, and $\le _\rho$ is the partial order on $A/\simeq _\rho$ defined, for every $a,b\in A$, by $[a]_{\simeq _\rho }\le [ b]_{\simeq _\rho }$ if $a\rho b$. Conversely, for any pair $(\sim ,\le )$ with $\sim$ an equivalence relation on A and $\le$ a partial order on $A/\sim$, the corresponding preorder $\rho _{(\sim ,\le )}$ on A is defined, for every $a,b\in A$, by $a\rho _{(\sim ,\le )} b$ if $[a]_{\sim }\le [b]_{\sim }$.

The objects of Preord that are objects in both the full subcategories ParOrd and Equiv are the objects of the form $(A,=)$, where $=$ denotes the equality relation on A. We will call them the trivial objects of Preord. The full subcategory of Preord whose objects are all trivial objects $(A,=)$ is isomorphic to the category of sets.

A morphism $f:(A,\rho )\rightarrow (A',\rho ')$ in Preord is trivial if it factors through a trivial object, that is, if there exist a trivial object $(B,=)$ and morphisms $g:(A,\rho )\rightarrow (B,=)$ and $h:(B,=)\rightarrow (A',\rho ')$ in Preord with $f=hg$. The following lemma has an easy proof:

Lemma 2.3

Let $(A,\rho )$ and $(A',\rho ')$ be objects in the category Preord and $f:A\rightarrow A'$ be a morphism. Then $f:(A,\rho )\rightarrow (A',\rho ')$ is a trivial morphism in Preord if and only if $a\rho b$ implies $f(a)=f(b)$ for all $a,b\in A$.

We will denote by ${\text {Hom}}(A,A')$ the set of all morphisms $A\rightarrow A'$ in the category Preord and by ${\text {Triv}}(A,A')$ the set of all trivial morphisms $A\rightarrow A'$.

In order to justify the terminology that follows, it is now convenient to introduce some notions of topology. Recall that a topological space is Alexandroff-discrete provided that the intersection of any family of open subsets is an open subset. The full subcategory of Top whose objects are all Alexandroff-discrete spaces is isomorphic to the category Preord. The category isomorphism associates with any preordered set $(X,\rho )$, the topological space $(X,\tau _\rho )$, whose open sets are the subsets A of X such that, if $a\in A$, $x\in X$ and $x\rho a$, then $x\in A$. It is immediately seen that $(X,\tau _\rho )$ is an Alexandroff-discrete space. Moreover, for any $x,y\in X$, $x\rho y$ if and only if y belongs to the closure $\overline{\{x\}}$ of x with respect to the topology $\tau _\rho$. A map** $X\rightarrow Y$ is a morphism $(X,\rho )\rightarrow (Y,\sigma )$ in Preord if and only if it is a continuous map** $(X,\tau _\rho )\rightarrow (Y,\tau _\sigma )$. Conversely, let $(X,\tau )$ be any Alexandroff-discrete space. The associated preorder $\le _{\tau }$ on X is defined, for every, $x,y\in X$, by $x\le _{\tau }y$ if $\overline{\{y\}}\subseteq \overline{\{x\}}$.

It is easy to see that:

Proposition 2.4

A morphism $f:(A,\rho )\rightarrow (A',\rho ')$ in Preord is a monomorphism in Preord if and only if it is an injective map**, and is an epimorphism if and only if it is a surjective map**.

Remark 2.5

Trivial objects are exactly the projective objects in Preord. Also, all trivial objects $(X,=)$ with $X\ne \emptyset$ are (projective) generators in Preord. These two facts can be easily checked directly. Also see [1].

Now we define a suitable quotient category of the category Preord, which we will call the stable category and denote by Preord. As far as quotient categories are concerned, we follow [9, pp. 51–52]. For every object $(A,\rho )$ of Preord, we will say that a subset B of A is a clopen subset if, for every $a\in A{\setminus } B$ and $b\in B$, and .

Remarks 2.6

(1) Let $(A,\rho )$ be a preordered set, and let $B\subseteq A$. By definition, B is a clopen subset, according to the definition just given, if and only if B is a clopen subset in the Alexandroff-discrete space $(A,\tau _\rho )$.

(2) Let $(A_i,\rho _i),\ i \in I$, be preordered sets, and let $(\coprod _{i\in I}A_i, \rho )$ be their coproduct in $\mathbf {Preord}$. Fix an index $i\in I$. By definition, if $j\in I{\setminus }\{i\}$, $a\in A_i$ and $b\in A_j$, then and . Hence the canonical image $A_i\times \{i\}$ of $A_i$ in the coproduct is a clopen subset of $\coprod _{i\in I}A_i$.

Definition 2.7

Let A be a preordered set. We say that A is indecomposable in $\mathbf {Preord}$ if A is non-empty, and for any preordered sets B, C, we have that $A\cong B\coprod C$ in $\mathbf {Preord}$ implies $B=\emptyset$ or $C=\emptyset$. (An equivalent definition of indecomposable, or connected, object which makes sense in any category ${{\mathcal C}}$ is the following: an object A is indecomposable: whenever $A\cong \coprod _{i\in I}B_i$, then there is an index $i\in I$ for which the coproduct injection $B_i\rightarrow A$ is an isomorphism. Notice that the case $I=\emptyset$ shows that such A cannot be initial in the category ${{\mathcal C}}$.)

From Remark 2.6(2), we get that:

Lemma 2.8

The following conditions are equivalent for a non-empty preordered set $(A,\rho )$:

• 1. $(A,\rho )$ is indecomposable.

• 2. The Alexandroff-discrete space $(A,\tau _\rho )$ is connected, that is, its clopen subsets are only $\emptyset$ and A.

For any preordered set $(A,\rho )$, let $\equiv _\rho$ be the equivalence relation generated by $\rho$, that is, the transitive closure of the relation $\rho \rho ^o$ (i.e. $x\equiv _\rho y$ if and only if there exist elements $x_0=x,x_1,\ldots , x_n=y\in A$ such that, for every $i=1,2,\dots n$, $x_{i-1}\rho x_{i}$ or $x_{i}\rho x_{i-1}$).

The proof of the following lemma is elementary:

Lemma 2.9

Let $(A,\rho )$ be a preordered set.

• 1. If $B\subseteq A$ is clopen and $a\in B$, then $[a]_{\equiv _\rho }\subseteq B$.

• 2. For any $a\in A$, the equivalence class $[a]_{\equiv _\rho }$ is clopen and connected.

• 3. Every clopen subset of A is a union of equivalence classes modulo $\equiv _\rho$.

• 4. The connected components of A are exactly the equivalence classes of A modulo $\equiv _\rho$.

Proof

(1) Suppose $x\in [a]_{\equiv _\rho }$. By definition, there are elements $x_0,x_1,\ldots ,x_n\in A$ such that $a=x_0$, $x=x_n$, and for every $i=1,2,3,\dots ,n$, $x_{i-1}\rho x_{i}$ or $x_{i}\rho x_{i-1}$. Since $a\in B$ and either $a\rho x_1$ or $x_1\rho a$, the fact that B is clopen implies that $x_1\in B$. Then $x\in B$ by induction.

(2) Assume $x\in [a]_{\equiv _\rho }$ and $y\in A$. If $x\rho y$ or $y\rho x$, then $x\equiv _\rho y$, since $\equiv _\rho$ is generated by $\rho$. It follows $y\in [a]_{\equiv _\rho }$, and this proves that $[a]_{\equiv _\rho }$ is clopen. Suppose there are non-empty clopen sets U, V of $(A,\tau _\rho )$ such that $[a]_{\equiv _\rho }=U\cup V$. If $u\in U,\ v\in V$, then $u\equiv _\rho v \equiv _\rho a$. By (1), it follows that $[a]_{\equiv _\rho }\subseteq U\cap V$, proving that $[a]_{\equiv _\rho }$ is connected.

(3) Let B be a clopen subset of A. By (1), it is clear that $B=\bigcup _{a\in B}[a]_{\equiv _\rho }$.

(4) By (2), every equivalence class modulo $\equiv _\rho$ is a connected component of A, since it is connected and clopen. Conversely, let Y be a connected component of A. Note that, for every $a\in Y$, $[a]_{\equiv _\rho }\cap Y$ is clopen in Y relatively to the subspace topology. Since $Y=\bigcup _{a\in Y}([a]_{\equiv _\rho }\cap Y)$ and every union of clopen subsets is clopen because $(A,\tau _\rho )$ is an Alexandroff-discrete space, the fact that Y is connected implies that there exists an element $a_0\in Y$ such that $Y=[a_0]_{\equiv _\rho }\cap Y$. In other words, $Y\subseteq [a_0]_{\equiv _\rho }$, and since Y is a connected component and $[a_0]_{\equiv _\rho }$ is connected, we infer that $[a_0]_{\equiv _\rho }=Y$. $\square$

An immediate consequence of Lemmas 2.8 and 2.9 is that

$$\begin{aligned} A=\mathop {\dot{\bigcup }}\limits _{[a]_{\equiv _\rho }\in A/\equiv _\rho }[a]_{\equiv _\rho } \end{aligned}$$

is the unique coproduct decomposition of a preordered set $(A,\rho )$ into a disjoint union of indecomposable clopen subsets. That is, every Alexandroff-discrete space is the coproduct of its connected components.

3 The stable category

We will now construct from Preord a pointed quotient category in which one may consider the key instruments of homological algebra, like short exact sequences (Sect. 3). For every pair of objects $(A,\rho ),(A',\rho ')$ in Preord, let $R_{A,A'}$ be the relation on the set ${\text {Hom}}(A,A')$ defined, for every $f,g:(A,\rho )\rightarrow (A',\rho ')$, by $fR_{A,A'}g$ if there exists a clopen subset B of A such that $f|_B$ and $g|_B$ are two trivial morphisms and $f|_{A{\setminus } B}=g|_{A{\setminus } B}$. By definition, given arbitrary trivial morphisms $f,g:A\rightarrow A'$, we have that $fR_{A,A'}g$. In the next lemma, we make use of the terminology of [9, p. 51].

Lemma 3.1

The assignment $(A,A')\mapsto R_{A,A'}$ is a congruence on the category Preord.

Proof

The relations $R_{A,A'}$ are clearly reflexive and symmetric. They are also transitive, because if $f,g,h:(A,\rho )\rightarrow (A',\rho ')$, $fR_{A,A'}g$ and $gR_{A,A'}h$, then there exist clopen subsets B, C of A such that $f|_B$, $g|_B$, $g|_C$ and $h|_C$ are four trivial morphisms, $f|_{A{\setminus } B}=g|_{A{\setminus } B}$ and $g|_{A{\setminus } C}=h|_{A{\setminus } C}$. Then $B\cup C$ is a clopen subset of A. Let us prove that $f|_{B\cup C}$ is a trivial morphism. Suppose $x,y\in B\cup C$ and $x\rho y$. Then we have two cases: either $x,y\in B$ or $x,y\in C{\setminus } B$. If $x,y\in B$, then $f(x)=f(y)$ because $f|_B$ is a trivial morphism. If $x,y\in C{\setminus } B$, then $f(x)=g(x)$ and $f(y)=g(y)$ because $f|_{A{\setminus } B}=g|_{A{\setminus } B}$. As $g|_C$ is a trivial morphism, it follows that $g(x)=g(y)$; hence, $f(x)=f(y)$. This proves that $f|_{B\cup C}$ is a trivial morphism. Similarly, $h|_{B\cup C}$ is a trivial morphism. Finally, if $x\in A{\setminus }(B\cup C)$, then $f(x)=g(x)$ because $f|_{A{\setminus } B}=g|_{A{\setminus } B}$ and $g(x)=h(x)$ because $g|_{A{\setminus } C}=h|_{A{\setminus } C}$. Thus $f(x)=h(x)$. This proves that $f|_{A{\setminus } (B\cup C)}=h|_{A{\setminus } (B\cup C)}$. Therefore, $R_{A,A'}$ is an equivalence relation on ${\text {Hom}}(A,A')$ for each pair $(A,A')$ of objects of Preord.

Finally, suppose that $f,g:(A,\rho )\rightarrow (A',\rho ')$ with $fR_{A,A'}g$, $h:(B,\sigma )\rightarrow (A,\rho )$ and $\ell :(A',\rho ')\rightarrow (C,\tau )$. Then there exists a clopen subset X of A such that $f|_X$ and $g|_X$ are two trivial morphisms and $f|_{A{\setminus } X}=g|_{A{\setminus } X}$. Then $h^{-1}(X)$ is a clopen subset of $(B,\sigma )$. The map** $\ell f h|_{h^{-1}(X)}$ is a trivial morphism, because if $b,b'\in h^{-1}(X)$ and $b\sigma b'$, then $h(b),h(b')\in X$ and $h(b)\rho h(b')$. Since $f|_X$ is a trivial morphism, we get that $fh(b)=fh(b')$. Therefore, $\ell fh(b)=\ell fh(b')$. This proves that $\ell f h|_{h^{-1}(X)}$ is a trivial morphism. Similarly, $\ell g h|_{h^{-1}(X)}$ is a trivial morphism. It remains to prove that $\ell f h|_{B{\setminus } h^{-1}(X)}=\ell g h|_{B{\setminus } h^{-1}(X)}$. Suppose $b\in B{\setminus } h^{-1}(X)$. Then $h(b)\in A{\setminus } X$, so that $f(h(b))=g(h(b))$ because $f|_{A{\setminus } X}=g|_{A{\setminus } X}$. Therefore, $\ell (f(h(b)))=\ell (g(h(b)))$. $\square$

It is therefore possible to construct the quotient category Preord${}{:}{=}{}$Preord / R. We will call it the stable category. Its objects are all preordered sets $(A,\rho )$, like in Preord. The morphisms $(A,\rho )\rightarrow (A',\rho ')$ are the equivalence classes of ${\text {Hom}}(A,A')$ modulo $R_{A,A'}$, that is,

$$\begin{aligned} {\text {Hom}}_{{\underline{\mathbf{Preord}}}}(A,A'){:}{=}{\text {Hom}}_{\text{ Preord }}(A,A')/{ R_{A,A'}}. \end{aligned}$$

There is a canonical functor $F:{}$Preord${}\rightarrow {}$ Preord. Our notation will be $F(A)=\underline{A}$ for every object A and $F(f)=\underline{f}$ for every morphism $f:A\rightarrow A'$.

Remark 3.2

The functor F preserves coproducts. In fact, let $(A_i,\rho _i)$, $i\in I$, be preordered sets and let $C{:}{=}(\coprod _{i\in I}A_i, \rho )$ be their coproduct in $\mathbf {Preord}$, so C is the disjoint union and $\rho$ is the coproduct preorder. For any $i\in I$, let $\varepsilon _i:A_i\rightarrow C$ be the canonical embedding (for all $x\in A_i, \varepsilon _i(x){:}{=}(x,i)$). Fix any preordered set Y, and for all $i\in I$, consider morphisms $\underline{f_i}:\underline{A_i}\rightarrow \underline{Y}$ in $\underline{\mathbf {Preord}}$. By the universal property of C in $\mathbf {Preord}$, there exists a unique morphism $f:C\rightarrow Y$ (in $\mathbf {Preord}$) such that $f_i=f\circ \varepsilon _i$ for all $i\in I$. In particular, $\underline{f_i}=\underline{f}\circ \underline{\varepsilon _i}$. Suppose there is a morphism $\underline{g}:\underline{C}\rightarrow \underline{Y}$ in $\underline{\mathbf {Preord}}$ such that $\underline{f_i}=\underline{g}\circ \underline{\varepsilon _i}$, for any $i\in I$. According to the last condition, for every $i\in I$ there exists a clopen subset $B_i$ of $A_i$ such that $f_i|_{B_i}$ and $(g\circ \varepsilon _i)|_{B_i}$ are trivial morphisms and $f_i|_{A_i{\setminus } B_i} =(g\circ \varepsilon _i)|_{A_i{\setminus } B_i}$. By Remark 2.6(2), the canonical image $B_i\times \{i\}$ of $B_i$ in C is clopen in C, and since $(C,\tau _\rho )$ is Alexandroff-discrete, $B{:}{=}\bigcup _{i\in I}B_i\times \{i\}$ is clopen in C. Now it is straightforward to check that $f|_B, g|_B$ are trivial morphisms and that $f|_{C{\setminus } B}=g|_{C{\setminus } B}$. Hence $\underline{ f}=\underline{g}$. This concludes the proof.

For every preordered set $(A,\rho )$, let $\pi :A\rightarrow A/\equiv _\rho$, $\pi :a\mapsto [a]_{\equiv _\rho }$, be the canonical projection. Let $A^*$ be the clopen subset of A defined by

$$\begin{aligned} A^*{:}{=} \bigcup _{\begin{array}{c} a\in A \\ |\pi (a)|> 1 \end{array}}\pi (a). \end{aligned}$$

We will say that $(A,\rho )$ is reduced if $A=A^*$. Therefore, every preordered set A is the disjoint union, in a unique way, of two clopen sets: the reduced object $A^*$ and the trivial object $A{\setminus } A^*$.

Theorem 3.3

The following conditions are equivalent for two non-empty preordered sets $(A,\rho ),(A',\rho ')$:

• 1. $(A,\rho ),(A',\rho ')$ are isomorphic objects in $\underline{\mathbf {Preord}}$.

• 2. $(A^*,\rho ),(A'^*,\rho ')$ are isomorphic objects in $\mathbf {Preord}$.

• 3. There exist trivial objects $X,X'$ in $\mathbf {Preord}$ such that $A\coprod X\cong A'\coprod X'$ in $\mathbf {Preord}$.

Proof

First of all, we prove some preliminary steps.

Step 1. For every non-trivial preordered set $(A,\rho )$, the sets Aand $A^*$with the induced preorder are isomorphic objects in Preord.

From A non-trivial, we have $A^*\ne \emptyset$. Fix an element $a_0$ in $A^*$. Let $f:A\rightarrow A^*$ be defined by $f(x)=x$ for every $x\in A^*$ and $f(x)=a_0$ for every $x\in A{\setminus } A^*$. Let $\varepsilon :A^*\rightarrow A$ be the inclusion. Then $f\varepsilon$ is the identity $1_{A^*}$ on $A^*$, and $\varepsilon f\,R_{A,A}1_{A}$, because (1) $\varepsilon f|_{A^*}=1_{A}|_{A^*}$, (2) $\varepsilon f|_{A^*}$ is the constant map** equal to $a_0$, hence a trivial morphism, and (3) $1_{A}|_{A{\setminus } A^*}$ is the trivial morphism because ${A{\setminus } A^*}$ is a trivial object. Hence A and $A^*$ are isomorphic in $\underline{\mathbf {Preord}}$.

Step 2. If $(A^*,\rho )$is a reduced preordered set and fis an endomorphism of $(A^*,\rho )$such that $f\,R_{A^*,A^*}1_{A^*}$, then $f=1_{A^*}$.

Suppose $f\,R_{A^*,A^*}1_{A^*}$. Then there exists a clopen subset B of $A^*$ such that $f|_B$ and $1_{A^*}|_B$ are two trivial morphisms and $f|_{A^*{\setminus } B}=1_{A^*}|_{A^*{\setminus } B}$. But $1_{A^*}|_B$ trivial morphism and $A^*$ reduced implies $B=\emptyset$. Thus $f=1_{A^*}$ follows from $f|_{A^*{\setminus } B}=1_{A^*}|_{A^*{\setminus } B}$.

Step 3. Two reduced preordered sets $(A^*,\rho ),({A'}^*,\rho ')$are isomorphic objects in Preordif and only if they are isomorphic objects inPreord.

Assume $(A^*,\rho )$ and $({A'}^*,\rho ')$ isomorphic in Preord. Then there exist morphisms $f:A^*\rightarrow {A'}^*$ and $g:{A'}^*\rightarrow A^*$ in Preord such that $gf\,R_{A^*,A^*}1_{A^*}$ and $fg\,R_{{A'}^*,{A'}^*}1_{{A'}^*}$. By Step 2, we have $gf=1_{A^*}$ and $fg=1_{{A'}^*}$. Thus $(A^*,\rho )$ and $({A'}^*,\rho ')$ are isomorphic in Preord. The converse is clear.

Step 4. If $(A,\rho ),(A',\rho ')$ are non-empty preordered sets that are isomorphic in Preord , then $(A,\rho )$ is trivial if and only if $(A',\rho ')$ is trivial.

Suppose $(A,\rho )$ trivial. By definition, there are morphisms $f:(A',\rho ')\rightarrow (A,\rho )$, $g:(A,\rho )\rightarrow (A',\rho ')$ such that $\underline{gf}=\underline{1_{A'}}$. Hence there is a clopen subset $B\subseteq A'$ with $1_{A'}|_B$ trivial and $gf|_{A'{\setminus } B}=1_{A'}|_{A'{\setminus } B}$. Take $x,y\in A'$ such that $x\rho 'y$. If $x,y\in B$, the fact that $1_{A'}|_B$ is trivial implies $x=y$. Otherwise, we must have $x,y\in A'{\setminus } B$, since B is clopen. Since $(A,\rho )$ is trivial and $gf|_{A'{\setminus } B}=1_{A'}|_{A'{\setminus } B}$, it immediately follows $x=y$.

Step 5. If Ais a non-empty preordered set and Xis a trivial object in $\mathbf {Preord}$, then Aand $A\coprod X$(the coproduct in $\mathbf {Preord}$) are isomorphic objects in $\underline{\mathbf {Preord}}$.

Two mutually inverse isomorphisms in $\underline{\mathbf {Preord}}$ are induced: (1) by the coproduct embedding $A\rightarrow A\coprod X$ and (2) by the coproduct $A\coprod X\rightarrow A$ of the identity map** $1_A$ of A and any constant map** $X\rightarrow A$.

We are now ready to prove the equivalence of the three conditions in the statement of the theorem.

(1)$\Rightarrow$(2). If $A,A'$ are isomorphic in $\underline{\mathbf {Preord}}$, they are either both trivial or both non-trivial by Step 4. In the first case, $A^*=A'^*=\emptyset$. In the second case, the conclusion follows from Steps 1 and 3.

(2)$\Rightarrow$(3). If $(A^*,\rho ),(A'^*,\rho ')$ are isomorphic in $\mathbf {Preord}$, it suffices to take $X{:}{=}A'{\setminus } A'^*$ and $X'{:}{=}A{\setminus } A^*$, and (3) holds.

(3)$\Rightarrow$(1). Assume $A\coprod X\cong A'\coprod X'$ in $\mathbf {Preord}$ with $X,X'$ trivial objects. If A is trivial, then $A'$ is also trivial, and trivial non-empty objects are isomorphic in $\underline{\mathbf {Preord}}$. If A is non-trivial, then $A'$ is also non-trivial, so $A\cong A\coprod X\cong A'\coprod X'\cong A'$ in $\underline{\mathbf {Preord}}$ by Step 5. $\square$

Remark 3.4

It is worth noting that the assumption that $A,A'$ are non-empty in the statement of Theorem 3.3 is necessary. Indeed, if $(A,=)$ is a non-empty trivial object $\mathbf {Preord}$ and $A'=\emptyset$, then $(A,=),A'$ are obviously not isomorphic in $\underline{\mathbf {Preord}}$, but $A^*=A'^*=\emptyset$.

Definition 3.5

Let $f:A\rightarrow A'$ be a morphism in Preord. We say that a morphism $k:X\rightarrow A$ in $\mathbf {Preord}$ is a prekernel of f if the following properties are satisfied:

• 1. fk is a trivial morphism.

• 2. Whenever $\lambda :Y\rightarrow A$ is a morphism in $\mathbf {Preord}$ and $f\lambda$ is trivial, then there exists a unique morphism $\lambda ':Y\rightarrow X$ in $\mathbf {Preord}$ such that $\lambda =k\lambda '$.

Recall that, for every map** $f:A\rightarrow A'$, the equivalence relation $\sim _f$ on A, associated with f, is defined, for every $a,b\in A$, by $a\sim _f b$ if $f(a)=f(b)$. It is easy to show that every morphism in $\mathbf {Preord}$ has a prekernel. More precisely:

Proposition 3.6

Let $f:(A,\rho )\rightarrow (A',\rho ')$ be a morphism in $\mathbf {Preord}$. Then a prekernel of f is the morphism $k:(A,\rho {\,\,}\cap \sim _f)\rightarrow (A,\rho )$, where k is the identity map** and $\sim _f$ is the equivalence relation on A associated with f.

We do not give a proof of the next proposition here, because it will be proved in a more general form in Proposition 5.1.

Proposition 3.7

Let $f:A\rightarrow A'$ be a morphism in $\mathbf {Preord}$, and let $\mu :X\rightarrow A$ be a prekernel of f. The following properties hold.

• 1. $\mu$ is a monomorphism.

• 2. If $\lambda :Y\rightarrow A$ is any other prekernel of f, then there exists a unique isomorphism $\lambda ':Y\rightarrow X$ such that $\lambda =\mu \lambda '$.

The category Preord has an initial object (the empty set) and terminal objects (the preordered sets with one element). Thus initial objects are different from terminal objects in Preord. Hence the same is true in its quotient category Preord${}{:}{=}{}$Preord / R. In order to define kernels, cokernels and exact sequences, we need a pointed category, and we have just seen that Preord is not a pointed category. But to get a pointed category, it is sufficient to eliminate the empty set, i.e. consider the full subcategory Preord$^*$ of Preord whose objects are all $(A,\rho )$ with $A\ne \emptyset$ and the full subcategory Preord$^*$ of the category Preord whose objects are all objects of Preord except for the empty set. More precisely, the category Preord$^*$ is a pointed category because its zero object is $\underline{0}=\underline{T}$ for every non-empty trivial object $(T,=)$. For any pair of objects $\underline{A},\underline{A'}$ of Preord$^*$, it is possible therefore to define the zero morphism$\underline{0}_{\underline{A},\underline{A'}}:\underline{A}\rightarrow \underline{A'}$ as the composite morphism of the unique morphism of $\underline{A}$ into $\underline{0}$ and the unique morphism of $\underline{0}$ into $\underline{A'}$. For instance, for any constant map** $c:A\rightarrow A'$, we find that $\underline{c}:\underline{A}\rightarrow \underline{A'}$ is clearly $\underline{0}_{\underline{A},\underline{A'}}:\underline{A}\rightarrow \underline{A'}$. More generally, for a morphism $g:A\rightarrow A'$, one has that $\underline{g}=\underline{0}_{\underline{A},\underline{A'}}$ if and only if g is a trivial morphism. By definition, the kernel of any morphism $\underline{f}:\underline{A}\rightarrow \underline{A'}$ in Preord$^*$ is the equalizer of $\underline{f}$ and the zero morphism $\underline{0}_{\underline{A},\underline{A'}}$.

Remark 3.8

By Theorem 3.3, the situation for the category $\mathbf {Preord}^*$ is extremely similar to the case of the category $\text{ mod- }R$ of finitely presented right modules over a semiperfect ring R [10, Introduction and Theorem 1.4]. In both cases, every object is, in a unique way, a coproduct of a projective object and an object with no nonzero projective direct summands, we factor out modulo projective objects, two objects are isomorphic in the stable category if and only if they are stably isomorphic (recall that A and B are stably isomorphic if there are projective P and Q such that $A\oplus P\cong B\oplus Q$), every stable isomorphism class of finitely generated modules over a semiperfect ring contains a unique minimal element (up to isomorphism) and so on.

Proposition 3.9

If $f:(A,\rho )\rightarrow (A',\rho ')$ is a morphism in Preord with $A,A'\ne \emptyset$ and $k:(A,\rho {\,\,}\cap \sim _f)\rightarrow (A,\rho )$ is a prekernel of f, then $\underline{k}$ is a kernel of $\underline{f}$ in the pointed category Preord$^*$.

Proof

We must show that $\underline{k}$ satisfies the universal property of kernels. Since fk is trivial (see Proposition 3.6), the morphism $\underline{fk}$ is the zero morphism. Take any morphism $k':(K',\tau )\rightarrow (A,\rho )$ in $\mathbf {Preord}^*$ such that $\underline{ fk'}$ is the zero morphism, i.e. $fk'$ is trivial. By Proposition 3.6, there exists a unique morphism $u:(K',\tau )\rightarrow (A,\rho {\,\,}\cap \sim _f)$ such that $k'=ku$. In particular, $\underline{k'}=\underline{ku}$. For the uniqueness of $\underline{u}$ in $\underline{\mathbf {Preord}}^*$, we have that the identity $A\rightarrow A$ is another prekernel by Proposition 3.6, so that k is an isomorphism by Proposition 3.7(2). $\square$

Example 3.10

(Quotient preordered set) Let $(A,\rho )$ be a preordered set and $\sim$ an equivalence relation on A. Then $\rho$ induces a well-defined preorder $\rho '$ on the quotient set $A/\sim$ (via the position $[a_1]_\sim \,\,\rho '\,\,[a_2]_\sim$ if $a_1\,\rho \, a_2$ for all $a_1,a_2\in A$) if and only if $\sim {}\subseteq \rho$. In fact, if the relation $\rho '$ on the quotient set $A/\sim$ is a well-defined reflexive relation, then $a_1\sim a_2$ implies $[a_1]_\sim =[a_2]_\sim$, so $[a_1]_\sim \rho '[a_2]_\sim$, i.e. $a_1\,\rho \, a_2$. This proves that $\sim {}\subseteq \rho$. Conversely, if $\sim {}\subseteq \rho$, then $[a_1]_\sim =[a_2]_\sim$ and $[a_3]_\sim =[a_4]_\sim$ imply $a_1\,\rho \,a_3$ if and only if $a_2\,\rho \,a_4$.

If these two equivalent conditions hold, then the canonical projection $\pi :A\rightarrow A/\sim$ is a morphism in Preord and its prekernel is the identity $k:(A,\sim )\rightarrow (A,\rho )$.

For simplicity of notation, we will indicate by the same symbol $\rho$ the preorder induced by $\rho$ on the quotient set $A/\sim$, provided $\sim {}\subseteq \rho$.

Remark 3.11

Let A be a set and let $\sim ,\ \rho$ be, respectively, an equivalence relation and a preorder on A such that $\sim {}\subseteq \rho$. As we have seen, the canonical projection $\pi :(A,\rho )\rightarrow (A/\sim ,\rho )$ is a morphism in $\mathbf {Preord}$. Then:

• 1. If $B\subseteq A$ is a clopen set, then $\pi (B)$ is clopen in $Q{:}{=}A/\sim$.

• 2. $\underline{\pi }$ is an epimorphism in $\underline{\mathbf {Preord}}$.

Proof

(1) Take elements $p\in \pi (B)$ and $q \in Q{\setminus }\pi (B)$, thus $p=\pi (b), q=\pi (x)$ for some $b\in B,x\in A{\setminus } B$. Since B is clopen in A, we have and and, by definition, and .

(2) Let $g,h:(Q,\rho )\rightarrow (T,\tau )$ be morphisms in $\mathbf {Preord}$ with $\underline{g\pi }=\underline{h\pi }$ in $\underline{\mathbf {Preord}}$. By definition, there is a clopen set $B\subseteq A$ such that $(g\pi )|_B, (h\pi )|_B$ are trivial morphisms and $(g\pi )|_{A{\setminus } B}=(h\pi )|_{A{\setminus } B}$. By part (1), $B'{:}{=}\pi (B)$ is clopen in Q and it is easily seen that $\pi (A{\setminus } B)=Q\setminus B'$. Then $g|_{B'},h|_{B'}$ are trivial morphisms and $g|_{Q{\setminus } B'}=g|_{Q{\setminus } B'}$. Hence $\underline{g}=\underline{h}$, and the conclusion follows. $\square$

Definition 3.12

Let $f:A\rightarrow A'$ be a morphism in $\mathbf {Preord}$. A precokernel of f is a morphism $p:A'\rightarrow X$ such that:

• 1. pf is a trivial map.

• 2. Whenever $\lambda :A'\rightarrow Y$ is a morphism such that $\lambda f$ is trivial, then there exists a unique morphism $\lambda _1:X\rightarrow Y$ with $\lambda =\lambda _1 p$.

If $f:(A,\rho )\rightarrow (A',\rho ')$ is a morphism in Preord, consider the canonical projection $c:(A',\rho ')\rightarrow (A'/\zeta _f,\rho '\vee \zeta _f)$, where $\zeta _f$ is the equivalence relation on $A'$ generated by the set $\{\,(f(a_1),f(a_2))\mid a_1,a_2\in A,\ a_1\rho a_2\,\}$ and $\vee$ is the least upper bound in the complete lattice of all preorders on the set $A'$ (that is, $\rho '\vee \zeta _f$ is the preorder on $A'$ generated by $\rho '$ and $\zeta _f$). Thus $\zeta _f$ is the transitive closure of the relation $\{\,(a',a')\mid a'\in A'\,\}\cup \{\,(f(a_1),f(a_2))\mid a_1,a_2\in A,\ a_1\sim _{\rho } a_2\,\}$. Since the equivalence relation $\zeta _f$ is contained in the preorder $\rho '\vee \zeta _f$, it is possible to construct the quotient preordered set (Example 3.10), and thus, $\rho '\vee \zeta _f$ induces a preorder on the quotient set $A'/\zeta _f$ (which will be still denoted by $\rho '\vee \zeta _f$, with a small abuse of notation).

Proposition 3.13

Let $f:(A,\rho )\rightarrow (A',\rho ')$ be a morphism in $\mathbf {Preord}$. Then the canonical projection $c:(A',\rho ')\rightarrow (A'/\zeta _f,\rho '\vee \zeta _f)$ is a precokernel of f.

Proof

We must prove that cf is a trivial morphism and that, for every morphism $g:(A',\rho ')\rightarrow (B,\sigma )$ with gf a trivial morphism, there exists a unique morphism $g':(A'/\zeta _f,\rho '\vee \zeta _f)\rightarrow (B,\sigma )$ such that $g=g'c$. If $x,y\in A$ and $x\rho y$, then $f(x)\zeta _f f(y)$ by the definition of $\zeta _f$. So $cf(x)=cf(y)$. This shows that cf is a trivial morphism. Now let $g:(A',\rho ')\rightarrow (B,\sigma )$ be a morphism in Preord with gf a trivial morphism. If $x,y\in A$ and $x\rho y$, we get that $gf(x)=gf(y)$. It follows that $\zeta _f\subseteq \,\sim _g$. This shows that the position $[a']_{\zeta _f}\mapsto g(a)$ defines a map** $g':A'/\zeta _f\rightarrow B$, and clearly g factors uniquely through the canonical projection c and the map** $g'$. The map** $g'$ is a morphism $(A'/\zeta _f,\rho '\vee \zeta _f)\rightarrow (B,\sigma )$ in Preord. To see this, consider elements $[a]_{\zeta _f},[\alpha ]_{\zeta _f}$ such that $a(\zeta _f\vee \rho ')\alpha$. By definition, there are elements $x_0{:}{=}a,x_1,\ldots ,x_n{:}{=}\alpha \in A'$ such that, for every $0\le i\le n-1$, either $x_i\zeta _f x_{i+1}$ or $x_i\rho ' x_{i+1}$. In the first case, since $\zeta _f\subseteq \sim _g$, it follows that $g'(x_i)=g'(x_{i+1})$. In the second case, since g is a morphism, $g(x_i)\sigma g(x_{i+1})$. Thus in both cases, we have $g'([x_i]_{\zeta _f})=g(x_i)\sigma g(x_{i+1})=g'([x_{i+1}]_{\zeta _f})$. Hence $g'([a]_{\zeta _f})\sigma g'([\alpha ]_{\zeta _f})$, because $\sigma$ is transitive. $\square$

It is not difficult to prove the next result directly, but it will also be an immediate consequence of our more general Proposition 5.2.

Proposition 3.14

Let $f:A\rightarrow A'$ be a morphism in $\mathbf {Preord}$.

• 1. Every precokernel of f is an epimorphism.

• 2. If $p:A'\rightarrow X$, $q:A'\rightarrow Y$ are cokernels of f, then there exists a unique isomorphism $\varphi :X\rightarrow Y$ such that $q=\varphi p$.

Proposition 3.15

Let $f:(A,\rho )\rightarrow (B,\sigma )$ be a morphism in $\mathbf {Preord}$ and let $\pi :(B,\sigma )\rightarrow (B/\zeta _f,\sigma \vee \zeta _f)$ be its precokernel. Then

• 1. $\zeta _f\subseteq {} \equiv _\sigma$, where $\equiv _\sigma$ denotes the equivalence relation generated by $\sigma$.

• 2. If $C\subseteq B$ is a clopen set, then $\pi (C)\subseteq B/\zeta _f$ is a clopen set.

• 3. $\underline{\pi }$ is an epimorphism in $\underline{\mathbf {Preord}}$.

Proof

(1) By definition, $\zeta _f$ is generated by

$$\begin{aligned} G{:}{=}\{(f(a_1),f(a_2))\mid a_1,a_2\in A, a_1\rho a_2 \}, \end{aligned}$$

and thus, $G\subseteq \sigma \subseteq \,\equiv _\sigma$, because f is a morphism in $\mathbf {Preord}$. This allows to conclude the proof.

(2) Set $Q{:}{=}B/\zeta _f$ and consider elements $\eta \in \pi (C)$ and $\lambda \in Q{\setminus } \pi (C)$. Then $\eta =\pi (c)$ and $\lambda =\pi (b)$ for some $c\in C$, $b\in B{\setminus } C$. If $\eta (\sigma \vee \zeta _f)\lambda$, by definition $c(\sigma \vee \zeta _f)b$. Thus there are elements $x_0{:}{=}c,x_1,\ldots , x_n=b\in B$ such that, for $0\le i< n$, either $x_i\sigma x_{i+1}$ or $x_i\zeta _f x_{i+1}$. For $i=0$, we have either $c \sigma x_1$ or $c\zeta _f x_1$. In the first case, we infer $x_1 \in C$, since C is clopen and $c\in C$. In the second case, there are elements $y_0{:}{=}c,y_1,\ldots , y_m{:}{=}x_1\in B$ such that, for $0\le j< m$, either $y_j\sigma y_{j+1}$ or $y_{j+1}\sigma y_j$. In particular, since $c\sigma y_1$ or $y_1\sigma c$ and C is clopen, it follows that $y_1\in C$. By induction on j, we have $y_m=x_1\in C$. This proves that in both cases $c\sigma x_1$ and $c\zeta _f x_1$, we have $x_1\in C$. By induction on i, we get $x_n=b\in C$, a contradiction.

(3) Argue as in Remark 3.11(2), noting that if C is a clopen subset of B, then $\pi (B{\setminus } C)=Q\setminus \pi (C)$. $\square$

Proposition 3.16

Let $f:(A,\rho )\rightarrow (B,\sigma )$ be a morphism in $\mathbf {Preord}^*$, and let

$$\begin{aligned} \pi :(B,\sigma )\rightarrow (B/\zeta _f, \sigma \vee \zeta _f) \end{aligned}$$

be a precokernel of f. Then $\underline{\pi }$ is a cokernel of $\underline{ f}$ in the pointed category $\underline{\mathbf {Preord}}^*$.

Proof

By Proposition 3.13, $\pi f$ is trivial, that is, $\underline{\pi f}=0$ in $\underline{\mathbf {Preord}}^*$. Now, let $g:(B,\sigma )\rightarrow (C,\tau )$ be a morphism in $\mathbf {Preord}$ such that $\underline{gf}=0$ in $\underline{\mathbf {Preord}}^*$, that is, gf is trivial. By Proposition 3.13 again, there exists a unique morphism $g':(B/\zeta _f, \sigma \vee \zeta _f)\rightarrow (C,\tau )$ in $\mathbf {Preord}$ such that $g=g'\pi$, so $\underline{g}=\underline{g'\pi }$. The uniqueness of such a $\underline{g'}$ in $\underline{\mathbf {Preord}}^*$ follows from the fact that $\underline{\pi }$ is an epimorphism in $\underline{\mathbf {Preord}}$, in view of Propositions 3.13, 3.14(1) and 3.15(3). $\square$

Definition 3.17

Let $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ be morphisms in $\mathbf {Preord}$. We say that is a short preexact sequence in $\mathbf {Preord}$ if f is a prekernel of g and g is a precokernel of f.

Example 3.18

Let A be any set, let $\rho$ be a preorder on A and let $\sim$ be an equivalence relation on A such that $\sim {}\subseteq {} \rho$. By Example 3.10, $\rho$ induces a preorder on the quotient set $A/\sim$, which we also denote by $\rho$. Then

is a short preexact sequence in $\mathbf {Preord}$, where k is the identity map and $\pi$ is the canonical projection. The equalities $\sim _\pi \cap {\,} \rho ={}\sim \cap {\,} \rho ={}\sim$ and Proposition 3.6 imply that k is a prekernel of $\pi$. Moreover, by definition $\zeta _k={}\sim$ and $\rho \vee \zeta _k=\rho {\,}\vee \sim {}=\rho$; hence, $\pi$ is a precokernel of k by Proposition 3.13.

In particular, if $\simeq _\rho$ is the equivalence relation on A defined by $a\simeq _\rho b$ if $a\rho b$ and $b\rho a$ (so that in particular we have $\simeq _\rho \subseteq \rho$) and $\le _\rho$ is the partial order on $A/\simeq _\rho$ induced by $\rho$ (see Proposition 2.2), then

is a short preexact sequence in $\mathbf {Preord}$.

Example 3.19

Let $f:(A,\rho )\rightarrow B$ be a morphism in $\mathbf {Preord}$. Consider the canonical prekernel $k:(A,\rho {\,\,} \cap \sim _f)\rightarrow (A,\rho )$ of f (Proposition 3.6). Let $\pi :(A,\rho )\rightarrow (A/\zeta _k, \rho \vee \zeta _k)$ be the canonical precokernel of k, according to Proposition 3.13. Then

is a short preexact sequence in $\mathbf {Preord}$. To prove it, we only need to show that k is a prekernel of $\pi$. This follows from the equalities $\rho \, \cap \sim _\pi =\rho \,\cap \zeta _k=\rho \,\cap \sim _f$ and Proposition 3.6.

In the next proposition, we determine all short preexact sequences in $\mathbf {Preord}$, up to isomorphism.

Proposition 3.20

Let be any short preexact sequence in $\mathbf {Preord}$. Then there exists a commutative diagram

where k is the identity map, $\pi$ is the canonical projection and

is a short preexact sequence.

Proof

By Proposition 3.6, k is a prekernel of g. Since, by assumption, f is also a prekernel for g, we infer from Proposition 3.7(2) the existence of an isomorphism $(X,\rho )\rightarrow (Y,\sigma \cap \sim _g)$ which makes the square on the left in diagram (*) commute. Now g is a precokernel of f, so that the existence of an isomorphism $(Z,\tau )\rightarrow (Y/\zeta _k,\zeta _k\vee \sigma )$ making the right square of the diagram commute will follow by showing that $\pi$ is a precokernel of f (Proposition 3.14(2)). First, we prove that $\pi f$ is trivial. Fix elements $x,y\in X$ with $x\,\rho \, y$. Since g is a precokernel of f, gf is trivial, and thus, $g(f(x))=g(f(y))$, that is, $f(x)\sim _g f(y)$. Moreover $f(x)\sigma f(y)$, because f is a morphism. Since, by definition, $\zeta _k$ is generated by $\{(y_1,y_2)\in Y\times Y\mid y_1\,(\sigma \cap \sim _g)\,y_2 \}$, it follows that $f(x)\,\zeta _k\, f(y)$, that is, $\pi f(x)=\pi f(y)$. This proves that $\pi f$ is trivial. Now, let $\lambda :(Y,\sigma )\rightarrow (T,\eta )$ be a morphism with $\lambda f$ trivial. We have to show that there exists a unique morphism ${\widetilde{\lambda }}:(Y/\zeta _k,\zeta _k\vee \sigma )\rightarrow (T,\eta )$ such that $\lambda ={\widetilde{\lambda }}\pi$. It is enough to show the existence of such a ${\widetilde{\lambda }}$, the uniqueness being a trivial consequence of the fact that $\pi$ is surjective. But, since g is a precokernel of f and $\lambda f$ is trivial, there exists a unique morphism $\lambda _1:(Z,\tau )\rightarrow (T,\eta )$ such that $\lambda =\lambda _1g$. Now $\zeta _k$ is the equivalence relation generated by $\sigma \,\cap \sim _g$ and $\sim _g$ is an equivalence relation, so $\zeta _k\subseteq \sim _g$. Thus $\lambda _1$ induces a well-defined map** ${\widetilde{\lambda }}:(Y/\zeta _k,\zeta _k\vee \sigma )\rightarrow (T,\eta )$, $[y]_{\zeta _k}\mapsto \lambda _1(g(y))$ and $\lambda _1$ clearly satisfies $\lambda ={\widetilde{\lambda }}\pi$. We claim that $\lambda _1$ is a morphism. In order to see this, take $[y]_{\zeta _k},[z]_{\zeta _k}\in Y/\zeta _k$ with $y(\sigma \vee \zeta _k) z$, and let $\beta _1{:}{=}y,\beta _2,\ldots , \beta _n{:}{=}z\in Y$ be such that, for every $1\le i<n$, either $\beta _i\sigma \beta _{i+1}$ or $\beta _i\zeta _k\beta _{i+1}$. In the first case, we infer $g(\beta _i)\tau g(\beta _{i+1})$, and thus, $\widetilde{\lambda }([\beta _i]_{\zeta _k})\eta {\widetilde{\lambda }}([\beta _{i+1}]_{\zeta _k})$, because $g,\lambda _1$ are morphisms. In the second case, we have, by definition, $[\beta _i]_{\zeta _k}=[\beta _{i+1}]_{\zeta _k}$. Thus in both cases we get $\widetilde{\lambda }([\beta _i]_{\zeta _k})\eta {\widetilde{\lambda }}([\beta _{i+1}]_{\zeta _k})$ for any $1\le i<n$. Since $\eta$ is transitive, it follows that ${\widetilde{\lambda }}([y]_{\zeta _k})\eta {\widetilde{\lambda }}([z]_{\zeta _k})$, proving that ${\widetilde{\lambda }}$ is a morphism.

Finally, is a short preexact sequence as we saw in Example 3.19. $\square$

4 Short exact sequences

In a pointed category, a short exact sequence is a pair of morphisms $f:A\rightarrow B$, $g:B\rightarrow C$ such that f is a kernel of g and g is a cokernel of f. As usual, the notation will be . We will now describe all short exact sequences in the pointed category Preord$^*$ up to isomorphism (similarly to the description of any short exact sequence in the category ${\text {Mod-}}R$ of right modules over a ring R, which is isomorphic to a short exact sequence of the form for suitable modules $A_R\le B_R$).

Proposition 4.1

For every short exact sequence

in Preord$^*$, there is a commutative diagram

where $\sim$ is an equivalence relation on B, k is the identity and $\pi$ is the canonical projection.

Proof

Let $f:A\rightarrow (B,\rho ),\ g:(B,\rho )\rightarrow C$ be representatives of $\underline{f},\ \underline{g}$. Consider the sequence of morphisms

in $\mathbf {Preord}^*$, where k is the identity and $\pi$ is the canonical projection. By Propositions 3.6 and 3.13, k is a prekernel of g and $\pi$ is a precokernel of k. Moreover the pair of morphisms $k,\pi$ is a short preexact sequence in $\mathbf {Preord}$ by Example 3.19. By Propositions 3.9 and 3.16, the canonical image

is an exact sequence in the stable category $\underline{\mathbf {Preord}}^*$. Since $\underline{f}, \underline{k}$ are kernels of $\underline{g}$, there is a unique isomorphism $\underline{A}\rightarrow \underline{(B,\rho \cap \sim _g)}$ in $\underline{\mathbf {Preord}}^*$ which makes the square on the left of diagram (**) commute. In order to get an isomorphism making the square on the right commute, it will suffice to show that $\underline{\pi }$ is a cokernel of $\underline{f}$. Let $\tau$ be the preorder on A and let $a,b\in A$ be such that $a\tau b$. Since $\underline{f},\underline{g}$ form a short exact sequence, we have, by definition, $\underline{g}\underline{f}=0$, i.e. gf is trivial. Thus $gf(a)=gf(b)$, and since f is a morphism, $f(a)\rho f(b)$, i.e. $f(a)(\rho \, \cap \sim _g)f(b)$. Since, by definition, $\zeta _k$ is the equivalence relation generated by $\rho \,\cap \sim _g$, it follows $[f(a)]_{\zeta _k}=[f(b)]_{\zeta _k}$, that is, $\pi f(a)=\pi f(b)$, and this proves that $\pi f$ is trivial, i.e. $\underline{\pi f}=0$. Now take any morphism $\underline{\lambda }:\underline{B}\rightarrow \underline{T}$ in $\underline{\mathbf {Preord}}^*$ satisfying $\underline{\lambda }\underline{f}=0$. Let $\lambda :B\rightarrow T$ be a representative of $\underline{ \lambda }$. By assumption, $\underline{g}$ is a cokernel of $\underline{f}$, so there exists a unique morphism $\underline{\lambda _1}:\underline{C}\rightarrow \underline{ T}$ in $\underline{\mathbf {Preord}}^*$ with $\underline{\lambda }=\underline{\lambda _1 g}$. Let $\lambda _1:C\rightarrow T$ be a representative of $\underline{\lambda _1}$. By definition, $\zeta _k\subseteq \sim _g$, and thus, there is a well-defined map** ${\widetilde{\lambda }}:(B/\zeta _k,\zeta _k\vee \rho )\rightarrow T$ such that ${\widetilde{\lambda }}([y]_{\zeta _k})=\lambda _1g(y)$, for every $y\in B$. By the same argument given in the proof of Proposition 3.20, it is easily seen that ${\widetilde{\lambda }}$ is a morphism in $\mathbf {Preord}^*$. We claim that the canonical image $\underline{{\widetilde{\lambda }}}$ of ${\widetilde{\lambda }}$ in $\underline{\mathbf {Preord}}^*$ is such that $\underline{\lambda }=\underline{{\widetilde{\lambda }}}\underline{\pi }$ (and it is clearly the unique one with this property, because $\underline{\pi }$ is an epimorphism by Proposition 3.15(3)). Since $\underline{\lambda }=\underline{\lambda _1}\underline{g}$, there is a clopen subset $\Lambda$ of $(B,\rho )$ such that $\lambda =\lambda _1g$ on $\Lambda$ and both $\lambda$ and $\lambda _1g$ are trivial on $B{\setminus }\Lambda$. It easily follows that $\lambda ={\widetilde{\lambda }}\pi$ on $\Lambda$ and that $\lambda , {\widetilde{\lambda }}\pi$ are trivial on $B{\setminus }\Lambda$. This concludes the proof. $\square$

Proposition 4.2

An identity morphism $k:(A,\sigma )\rightarrow (A,\rho )$ is a prekernel of a morphism in Preord$^*$ if and only if $\sigma =\rho {\,\,}\cap \equiv _\sigma$. Moreover, if these equivalent conditions hold, then k is the prekernel of the canonical projection $\pi :(A,\rho )\rightarrow (A/\equiv _\sigma , \rho \vee \equiv _\sigma )$.

Proof

If $k:(A,\sigma )\rightarrow (A,\rho )$ is a kernel of a morphism $f:(A,\rho )\rightarrow (A',\rho ')$ in Preord, then k is the morphism $k:(A,\rho {\,\,}\cap \sim _f)\rightarrow (A,\rho )$, where $\sim _f$ is the equivalence relation on A defined, for every $x,y\in A$, by $x\sim _fy$ if $f(x)=f(y)$ and k is the identity map**. Thus $\sigma =\rho {\,\,}\cap \sim _f$. We must prove that $\sigma =\rho \,\cap \equiv _\sigma$. Now $\sigma =\rho {\,\,}\cap \sim _f$ implies that $\sigma \subseteq \rho$ and that $\sigma$ is contained in the equivalence relation $\sim _f$. Hence the equivalence relation $\equiv _\sigma$ generated by $\sigma$ is contained in $\sim _f$. Therefore, $\rho \,\cap \equiv _\sigma {}\subseteq \rho \,\cap \sim _f{}=\sigma$. Conversely, $\sigma \subseteq \rho$ because the identity $k:(A,\sigma )\rightarrow (A,\rho )$ is a morphism in Preord$^*$, and $\sigma \subseteq \,\equiv _\sigma$ trivially. This concludes the proof of one of the implications of the first part of the statement. The rest follows trivially considering the canonical projection $\pi :(A,\rho )\rightarrow (A/\equiv _\sigma , \rho \vee \equiv _\sigma )$ and noticing that $\sim _\pi \,=\,\equiv _\sigma$. $\square$

The stable category Preord$^*$ has an unexpected behaviour, because it has features typical of abelian categories and semiabelian categories (short exact sequences). A study of its properties (how far is it from being abelian?) is a promising starting point for further investigations. We are grateful to the referee for this and other remarks. We do not study the structure of the quotient category Preord$^*$ in detail here, nor a possible more general setting where to define it, because it would lead us far from the subject dealt with in this paper.

Notice that similar questions and a similar construction of stable quotient category are given in [5], but the stable category presented in that article is not pointed.

Appendix: pretorsion theories

We now briefly present a general setting for the results proved in the previous sections. We have called this section “Appendix” because the notion of pretorsion $({{\mathcal T}},{{\mathcal F}})$ in general categories we need here will be presented in further detail in the paper [4]. We have decided to call our pairs $({{\mathcal T}},{{\mathcal F}})$ “pretorsion” theories, because our concept properly generalizes a number of torsion theories existing in the mathematical literature. See, for instance, [6,7,8].

Our definition of pretorsion theory is based on the familiar idea (propagated by Marco Grandis, Sandra Mantovani and others) to fix a class ${{\mathcal Z}}$ of objects in a category ${{\mathcal C}}$ and declare a morphism to be ${{\mathcal Z}}$-trivial when it factors through an object in ${{\mathcal Z}}$. With these morphisms replacing the zero morphisms of a pointed category, one then arrives to our notions instantaneously. Letting ${{\mathcal Z}}$ be the class of discretely preordered sets, our example of pretorsion theory in the category Preord is easily checked and hardly surprising.

Torsion theory in general categories are studied in [6,7,8]. These papers are rather technical, and we will only use a tiny part of the results presented there. Hence, in this section, we give an elementary presentation of the results we need. There is a large body in the topologically oriented literature, which pursues the (torsion, torsion-free) paradigm in the form of general types of (connectedness, total disconnectedness). See, in particular, the papers [2, 3] as well as the extensive literature cited there. The focus of these papers is on obtaining results in categories of topology that mirror, or even provide a common generalization of, the algebraic theory. Notice that, in our paper, the category of preordered sets is a topological category, a concept well studied in the literature [1] .

Fix an arbitrary category ${{\mathcal C}}$ and a class ${{\mathcal Z}}$ of objects of ${{\mathcal C}}$. For every pair $A,A'$ of objects of ${{\mathcal C}}$, we will write ${\text {Triv}}_{{{\mathcal Z}}}(A, B)$ for the set of all morphisms in ${{\mathcal C}}$ that factor through an object of ${{\mathcal Z}}$. We will call these morphisms ${{\mathcal Z}}$-trivial.

Let $f:A\rightarrow A'$ be a morphism in ${{\mathcal C}}$. We say that a morphism $k:X\rightarrow A$ in ${{\mathcal C}}$ is a ${{\mathcal Z}}$-prekernel of f if the following properties are satisfied:

• 1. fk is a ${{\mathcal Z}}$-trivial morphism.

• 2. Whenever $\lambda :Y\rightarrow A$ is a morphism in ${{\mathcal C}}$ and $f\lambda$ is ${{\mathcal Z}}$-trivial, then there exists a unique morphism $\lambda ':Y\rightarrow X$ in ${{\mathcal C}}$ such that $\lambda =k\lambda '$.

Proposition 5.1

Let $f:A\rightarrow A'$ be a morphism in ${{\mathcal C}}$ and let $\mu :X\rightarrow A$ be a ${{\mathcal Z}}$-prekernel for f. The following properties hold.

• 1. $\mu$ is a monomorphism.

• 2. If $\lambda :Y\rightarrow A$ is any other ${{\mathcal Z}}$-prekernel of f, then there exists a unique isomorphism $\lambda ':Y\rightarrow X$ such that $\lambda =\mu \lambda '$.

Proof

(1) Let T be an object of ${{\mathcal C}}$ and $g,h:T\rightarrow X$ be morphisms in ${{\mathcal C}}$ such that $\mu g=\mu h$. Since $\mu$ is a ${{\mathcal Z}}$-prekernel of f, the morphism $f\mu$ is ${{\mathcal Z}}$-trivial, so that, a fortiori, the morphism $f\mu g$ is also ${{\mathcal Z}}$-trivial. By applying part (2) of the definition of ${{\mathcal Z}}$-prekernel to $\mu g$, it follows that there exists a unique morphism $u:T\rightarrow X$ such that $\mu u=\mu g=\mu h$. From the uniqueness of u, we get that $u=g=h$.

(2) By part (2) of the definition of ${{\mathcal Z}}$-prekernel, there is a unique morphism $\lambda ':Y\rightarrow X$ such that $\lambda =\mu \lambda '$ and a unique morphism $\mu ':X\rightarrow Y$ such that $\mu =\lambda \mu '$. It follows that $\lambda =\lambda \mu '\lambda '$ and $\mu =\mu \lambda '\mu '$. By part (1) of this proposition, we obtain that $\mu '\lambda '=1_Y, \lambda '\mu '=1_X$. The conclusion is now clear. $\square$

Dually, a ${{\mathcal Z}}$-precokernel of f is a morphism $p:A'\rightarrow X$ such that:

• 1. pf is a ${{\mathcal Z}}$-trivial morphism.

• 2. Whenever $\lambda :A'\rightarrow Y$ is a morphism and $\lambda f$ is ${{\mathcal Z}}$-trivial, then there exists a unique morphism $\lambda _1:X\rightarrow Y$ with $\lambda =\lambda _1 p$.

If ${{\mathcal C}}^{{{\,\mathrm{op}\,}}}$ is the opposite category of ${{\mathcal C}}$, the ${{\mathcal Z}}$-precokernel of a morphism $f:A\rightarrow A'$ in ${{\mathcal C}}$ is the ${{\mathcal Z}}$-prekernel of the morphism $f:A'\rightarrow A$ in ${{\mathcal C}}^{{{\,\mathrm{op}\,}}}$. Hence, from Proposition 5.1, we get:

Proposition 5.2

Let $f:A\rightarrow A'$ be a morphism in a category ${\mathcal {C}}$ and let $p:A'\rightarrow X$ be a ${\mathcal {Z}}$-precokernel of f. Then the following properties hold.

• 1. p is an epimorphism.

• 2. If $q:A'\rightarrow Y$ is another ${\mathcal {Z}}$-precokernel of f, then there exists a unique isomorphism $\varphi :X\rightarrow Y$ satisfying $q=\varphi p$.

Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be morphisms in ${{\mathcal C}}$. We say that

is a short ${{\mathcal Z}}$-preexact sequence in ${{\mathcal C}}$ if f is a ${{\mathcal Z}}$-prekernel of g and g is a ${{\mathcal Z}}$-precokernel of f.

Clearly, if is a short ${{\mathcal Z}}$-preexact sequence in ${{\mathcal C}}$, then is a short ${{\mathcal Z}}$-preexact sequence in ${{\mathcal C}}^{{{\,\mathrm{op}\,}}}$.

It is now clear that if ${\mathcal {C}}{:}{=}\mathbf {Preord}$ and ${\mathcal {Z}}$ is the class of all objects of type $(A,=)$ (where A is any set and $=$ is the equality relation on A), then the ${\mathcal {Z}}$-prekernel and the ${\mathcal {Z}}$-precokernel of any morphism f in $\mathbf {Preord}$ exist and coincide with the prekernel (resp., cokernel) of f. In particular, a short ${\mathcal {Z}}$-preexact sequence in $\mathbf {Preord}$ is a short preexact sequence, as in Definition 3.17.

Definition 5.3

Let ${{\mathcal C}}$ be an arbitrary category. A pretorsion theory$({{\mathcal T}},{{\mathcal F}})$ for ${{\mathcal C}}$ consists of two classes ${{\mathcal T}},{{\mathcal F}}$ of objects of ${{\mathcal C}}$, closed under isomorphism, satisfying the following two conditions. Set ${{\mathcal Z}}{:}{=}{{\mathcal T}}\cap {{\mathcal F}}$.

(1) For every object B of ${{\mathcal C}}$ there is a short ${{\mathcal Z}}$-preexact sequence

with $A\in {{\mathcal T}}$ and $C\in {{\mathcal F}}$.

(2) ${\text {Hom}}_{{{\mathcal C}}}(T,F)={\text {Triv}}_{{{\mathcal Z}}}(T, F)$ for every object $T\in {{\mathcal T}}$, $F\in {{\mathcal F}}$.

Lemma 5.4

Let ${{\mathcal C}}$ be a category, ${{\mathcal Z}}$ a class of objects of ${{\mathcal C}}$ and

a short ${{\mathcal Z}}$-preexact sequence. Then:

(a) f is ${{\mathcal Z}}$-trivial if and only if g is an isomorphism.

(b) g is ${{\mathcal Z}}$-trivial if and only if f is an isomorphism.

Proof

(a) Suppose f${{\mathcal Z}}$-trivial. By definition, the identity morphism $1_B:B\rightarrow B$ is clearly a ${{\mathcal Z}}$-precokernel of f. By the uniqueness up to isomorphism of ${{\mathcal Z}}$-precokernels (see Proposition 5.2), there is a unique isomorphism $h:C\rightarrow B$ such that $hg=1_B$. Hence g is an isomorphism.

Conversely, assume that g is an isomorphism. Since gf is ${{\mathcal Z}}$-trivial, i.e. gf factors though an object in ${{\mathcal Z}}$, the same holds for f, that is, f is ${{\mathcal Z}}$-trivial.

(b) follows from (a) passing to the opposite category ${{\mathcal C}}^{{{\,\mathrm{op}\,}}}$. $\square$

Proposition 5.5

Let $({{\mathcal T}},{{\mathcal F}})$ be a pretorsion theory in a category ${{\mathcal C}}$, let ${{\mathcal Z}}={{\mathcal T}}\cap {{\mathcal F}}$, and let X be any object in ${{\mathcal C}}$.

• 1. If ${\text {Hom}}_{{{\mathcal C}}}(X,F)={\text {Triv}}_{{{\mathcal Z}}}(X,F)$ for every $F\in \mathcal F$, then $X\in {{\mathcal T}}$.

• 2. If ${\text {Hom}}_{{{\mathcal C}}}(T,X)={\text {Triv}}_{{{\mathcal Z}}}(T,X)$ for every $T\in {{\mathcal T}}$, then $X\in \mathcal F$.

Proof

By definition, there is a short ${{\mathcal Z}}$-preexact sequence with $A\in {{\mathcal T}}$ and $C\in {{\mathcal F}}$. If ${\text {Hom}}_{{{\mathcal C}}}(X,F)={\text {Triv}}_{{{\mathcal Z}}}(X,F)$ for every $F\in \mathcal F$, then g is ${{\mathcal Z}}$-trivial, so f is an isomorphism by the previous lemma. As ${{\mathcal T}}$ is closed under isomorphism, we get that $X\in {{\mathcal T}}$. This proves (1). Similarly for (2). $\square$

Notice that if $({{\mathcal T}},{{\mathcal F}})$ is a pretorsion theory in ${{\mathcal C}}$, then $({{\mathcal F}},{{\mathcal T}})$ is a pretorsion theory in ${{\mathcal C}}^{{{\,\mathrm{op}\,}}}$. Clearly, what we have proved in the previous sections shows that (Equiv, ParOrd) is a pretorsion theory in $\mathbf {Preord}$.

Remark 5.6

It would be nice to try to characterize all pretorsion theories in the category $\mathbf {Preord}$, but this seems to be hopeless. An example of pretorsion theory in $\mathbf {Preord}$ is given by the pair $(\mathbf {ParOrd},\mathbf {Equiv})$. Notice that here we have inverted the order of the two subcategories with respect to what we have done until now in the article. Now the torsion objects are partially ordered sets, and the torsion-free objects are sets with an equivalence relation. In this case, the torsion-free quotient of an object $(A,\rho )$ of $\mathbf {Preord}$ is $(A,\equiv _\rho )$, where $\equiv _\rho$ is the equivalence relation described between the statements of Lemmas 2.8 and 2.9. (It is the transitive closure of the relation $\rho \rho ^o$.) The torsion subobject of an object $(A,\rho )$ is the disjoint union of those clopen indecomposable subsets $[a]_{\equiv _\rho }$ of $(A,\rho )$ on which the preorder induced by $\rho$ is a partial order.

We consider this and other possible pretorsion theories on $\mathbf {Preord}$ much less interesting than that treated in the previous sections, which describes as any preordered set can be obtained from a partial order and an equivalence relation, in terms of a pretorsion theory.

Other easy natural examples of pretorsion theories in topological spaces, will be given in [4].

Remark 5.7

There is a notion of torsion theories for pointed categories given by Janelidze and Tholen in [8]. Their torsion theories are a special case of our notion of pretorsion theory. Namely, let ${{\mathcal C}}$ be a pointed category and let ${{\mathcal Z}}$ be the class of all zero objects on ${{\mathcal C}}$. Then, for any pair of fixed objects A, B of ${{\mathcal C}}$, the family of all ${{\mathcal Z}}$-trivial morphisms consists exactly of all zero morphisms. It immediately follows that, if $f:A\rightarrow B$, $k:X\rightarrow A$ are morphisms in ${{\mathcal C}}$, then k is a kernel of f if and only if k is a ${{\mathcal Z}}$-prekernel of f (where ${{\mathcal Z}}$ is as before) and similarly for cokernels. Thus short exact sequences, as defined in [8], are exactly our short ${{\mathcal Z}}$-preexact sequences. In [8], a torsion theory in ${{\mathcal C}}$ is a pair $({{\mathcal T}},{{\mathcal F}})$ of full subcategories of ${{\mathcal C}}$ that are closed under isomorphism, satisfying the following axioms:

• 1. ${\text {Hom}}(T,F)=0$, for every $T\in {{\mathcal T}},F\in {{\mathcal F}}$.

• 2. For every object X in ${{\mathcal C}}$, there are objects $T(X)\in {{\mathcal T}}, F(X)\in {{\mathcal F}}$ and a short exact sequence $T(X)\rightarrow X\rightarrow F(X)$.

The fact that every torsion theory $({{\mathcal T}},{{\mathcal F}})$ is an example of pretorsion theory will follow if we prove that ${{\mathcal T}}\cap {{\mathcal F}}$ is exactly the class ${{\mathcal Z}}$ of all zero objects of ${{\mathcal C}}$. Clearly, every zero object is in ${{\mathcal T}}\cap {{\mathcal F}}$. Conversely, let H be any object in ${{\mathcal T}}\cap {{\mathcal F}}$. By axiom (1), the identity $1_H:H\rightarrow H$ is the zero morphism, i.e. factors through a zero object. It immediately follows that H is (isomorphic to) a zero object.