3. Duality

3.1The Duality Principle

In the formal definition of a category there are objects $A,B,C,\dots$, arrows $f,g,h,\dots$ and four operations given by $\mathrm{dom}(f),\mathrm{cod}(f),1_A,g\circ f$ which satisfy the following seven axioms:

•	$\mathrm{dom}\left(1_A\right)=A$.
•	$\mathrm{cod}\left(1_A\right)=A$.
•	$f\circ 1_{\mathrm{dom}(f)}=f$.
•	$1_{\mathrm{cod}(f)}\circ f=f$.
•	$\mathrm{dom}(g\circ f)=\mathrm{dom}(f)$.
•	$\mathrm{cod}(g\circ f)=\mathrm{cod}(g)$.
•	$h\circ (g\circ f)=(h\circ g)\circ f$.

Of course, the operation “$g\circ f$” is only defined when $\mathrm{dom}(g)=\mathrm{cod}(f)$.

Given any sentence $\Sigma$ in the elementary language of category theory, we can form the “dual statement” ${\Sigma }^{\ast }$ by making the following replacements: $f\circ g$ for $g\circ f$, $\mathrm{cod}$ for $\mathrm{dom}$, $\mathrm{dom}$ for $\mathrm{cod}$. It is easy to see that then ${\Sigma }^{\ast }$ will also be a well-formed sentence. Next, suppose we have shown a sentence $\Sigma$ to entail one $\Delta$, that is,$$\Sigma ⟹\Delta ,$$without using any of the category axioms, then clearly$${\Sigma }^{\ast }⟹{\Delta }^{\ast }$$since the substituted terms are treated as mere undefined constants. But now observe that the axioms for category theory (CT) are themselves “self-dual” in the sense that we have $C{T}^{\ast }=CT$.

Therefore we have the following formal duality principle.

命题 3.1.1 (Formal duality). For any sentence $\Sigma$ in the language of category theory (CT) if $\Sigma$ follows from the axioms of categories, then its dual ${\Sigma }^{\ast }$ also follows, i.e.,$$(CT⟹\Sigma )\text{ implies }\left(CT⟹{\Sigma }^{\ast }\right)$$

Now assume that $\Sigma$ holds for any category $\mathcal{C}$. Then $\Sigma$ holds for any opposite category ${\mathcal{C}}^{\mathrm{op}}$. Hence ${\Sigma }^{\ast }$ holds in $\mathcal{C}=({\mathcal{C}}^{\mathrm{op}}{)}^{\mathrm{op}}$ for any category $\mathcal{C}$.

Therefore we have the following conceptual form of the duality principle.

命题 3.1.2 (Conceptual duality). For any statement $\Sigma$ about categories, if $\Sigma$ holds for all categories, then ${\Sigma }^{\ast }$ holds for all categories.

3.2Coproducts

定义 3.2.1. Let $\mathcal{C}$ be a category and let $A$ and $B$ be objects of $\mathcal{C}$. A coproduct of $A$ and $B$ in $\mathcal{C}$ is just the product of $A$ and $B$ in the opposite category ${\mathcal{C}}^{\mathrm{op}}$. This means an object $Q$ and arrows $q_1:A\to Q$ and $q_2:B\to Q$, called canonical injections, satisfying the following universal mapping property: given any diagram of the form$$A\stackrel{f_1}{\to }Z\stackrel{f_2}{\leftarrow }B$$there is a unique arrow $u:Q\to Z$ such that$$u\circ q_1=f_1\text{ and }u\circ q_2=f_2$$that is, the following diagram is commutative:

We denote the coproduct of $A$ and $B$ by $\left(Q,q_1,q_2\right)$. Sometimes $Q$ is also denoted by $A\coprod B$ or $A\oplus B$.

注 3.2.2. Sometimes (especially in the so-called additive categories), similarly to the case of products, the canonical injections $q_1:A\to A\oplus B$ and $q_2:B\to A\oplus B$ are also denoted by $\left[\begin{array}{c}1\\ 0\end{array}\right]:A\to A\oplus B$ and $\left[\begin{array}{c}0\\ 1\end{array}\right]:B\to A\oplus B$ respectively. The unique arrow $u:A\oplus B\to Z$ such that $u\circ q_1=f_1$ and $u\circ q_2=f_2$ is also denoted by$$\left[\begin{array}{cc}{f}_1& f_2\end{array}\right]:A\oplus B\to Z$$Then equalities involving compositions of arrows such as $u\circ q_1=f_1$ and $u\circ q_2=f_2$ may be rewritten in terms of matrix multiplications as $\left[\begin{array}{cc}{f}_1& f_2\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\end{array}\right]=f_1$ and $\left[\begin{array}{cc}{f}_1& f_2\end{array}\right]\cdot \left[\begin{array}{c}0\\ 1\end{array}\right]=f_2$.

注 3.2.3. Note that the canonical injections $q_1$ and $q_2$ need not be monomorphisms. Just consider the example with the canonical projections from products in the opposite category.

As usual, universal mapping property insures the following uniqueness result. It follows by the duality principle.

命题 3.2.4. The coproduct is unique up to an isomorphism.

More generally, one may define a coproduct of an arbitrary family of objects of a category, which is again unique up to an isomorphism.

定义 3.2.5. A coproduct of a family ${\left(A_i\right)}_{i\in I}$ of objects of a category $\mathcal{C}$ consists of an object $Q$, also denoted by $\coprod _{i\in I}{A}_i$ or $\underset{i\in I}{⨁}{A}_i$, and a family ${\left(q_i\right)}_{i\in I}$ of arrows, where $q_i:A_i\to Q$ for every $i\in I$, satisfying the following universal mapping property: given any object $Z$ of $\mathcal{C}$ and any family ${\left(f_i\right)}_{i\in I}$ of arrows, where $f_i:A_i\to Z$ for every $i\in I$, there is a unique arrow $u:Q\to Z$ such that $u\circ q_i=f_i$ for every $i\in I$.

Next we present some examples of coproducts in certain categories.

The Category $\mathbf{Set}$

The coproduct of two sets $A$ and $B$ is $\left(A\bigsqcup B,q_1,q_2\right)$, where$$A\bigsqcup B=\{(a,1)\mid a\in A\}\cup \{(b,2)\mid b\in B\}$$is the disjoint union of $A$ and $B,q_1:A\to A\bigsqcup B$ is the function defined by $q_1(a)=(a,1)$, and $q_2:B\to A\bigsqcup B$ is the function defined by $q_2(b)=(b,2)$.

Let $Z$ be a set and let $f_1:A\to Z$ and $f_2:B\to Z$ be functions. We look for a unique function $u:A\bigsqcup B\to Z$ such that the following diagram is commutative:

that is, $u\circ q_1=f_1$ and $u\circ q_2=f_2$. These equalities are equivalent to $u\left(q_1(a)\right)=f_1(a)$ and $u\left(q_2(b)\right)=f_2(b)$ for every $a\in A$ and $b\in B$, and furthermore, $u(a,1)=f_1(a)$ and $u(b,2)=f_2(b)$ for every $a\in A$ and $b\in B$. Note that $u$ is uniquely determined by this definition.

The construction of a coproduct may be easily generalized to an arbitrary family of sets.

The Category $\mathbf{Ab}$

The coproduct of two abelian groups $(A,+)$ and $(B,+)$ is $\left((A\times B,+),q_1,q_2\right)$, where $q_1:A\to$ $A\times B$ is the group homomorphism defined by $q_1(a)=(a,0)$, and $q_2:B\to A\times B$ is the group homomorphism defined by $q_2(b)=(0,b)$.

Note that $A\times B$ is a group with respect to the operation defined by$$\left(a_1,b_1\right)+\left(a_2,b_2\right)=\left(a_1+a_2,b_1+b_2\right),\forall \left(a_1,b_1\right),\left(a_2,b_2\right)\in A\times B.$$

Let $(Z,+)$ be an abelian group and let $f_1:A\to Z$ and $f_2:B\to Z$ be group homomorphisms. We look for a unique group homomorphism $u:A\times B\to Z$ such that the following diagram is commutative:

that is, $u\circ q_1=f_1$ and $u\circ q_2=f_2$. These equalities are equivalent to $u\left(q_1(a)\right)=f_1(a)$ and $u\left(q_2(b)\right)=f_2(b)$ for every $a\in A$ and $b\in B$, and furthermore, $u(a,0)=f_1(a)$ and $u(0,b)=f_2(b)$ for every $a\in A$ and $b\in B$. Note that a group homomorphism $u$ is uniquely determined by this definition, because we have$$u(a,b)=u((a,0)+(0,b))=u(a,0)+u(0,b)=f_1(a)+f_2(b)$$for every $a\in A$ and $b\in B$. One checks that the map $u$ defined as above is really a group homomorphism.

The construction of a coproduct may be generalized to an arbitrary family of abelian groups, but in a slightly different manner. For a family ${\left(A_i\right)}_{i\in I}$ of abelian groups, we denote$$\underset{i\in I}{⨁}{A}_i=\left\{{\left(a_i\right)}_{i\in I}\in \prod _{i\in I}{A}_i\mid {\left(a_i\right)}_{i\in I}\text{ has a finite number of non-zero elements }\right\}.$$Note that if $I$ is a finite set, then we have $\underset{i\in I}{⨁}{A}_i\cong {\prod }_{i\in I}{A}_i$.

The coproduct of the family ${\left(A_i\right)}_{i\in I}$ of abelian groups is $\left(\underset{i\in I}{⨁}{A}_i,{\left(q_i\right)}_{i\in I}\right)$, where for every $j\in I,q_j:A_j\to \underset{i\in I}{⨁}{A}_i$ is the group homomorphism defined by $q_j(a)={\left(a_i\right)}_{i\in I}$ with the properties that $a_j=a$ and $a_i=0$ for every $i\in I$ with $i\ne j$.

Poset Categories

Let $(L,⩽)$ be a lattice. Hence every two elements of $L$ have a supremum (i.e., smallest upper bound). Since $(L,⩽)$ is a poset, we may view it as a poset category. Recall that its objects are the elements of $L$, while an arrow $x\to y$ does exists if and only if $x⩽y$, where $x,y\in L$.

The coproduct of two elements $x,y\in L$ is $\left(\sup (x,y),q_1,q_2\right)$, where $q_1:x\to \sup (x,y)$ and $q_2:y\to \sup (x,y)$ are the unique arrows having the given domains and codomains.

Let $z\in L$ and let $x\to z$ and $y\to z$ be arrows. This means that $x⩽z$ and $y⩽z$. We look for a unique arrow $u:\sup (x,y)\to z$ such that the following diagram is commutative:

This means that $\sup (x,y)⩽z$. But this is true, because $z$ is an upper bound of $x$ and $y$, while $\sup (x,y)$ is the smallest upper bound of $x$ and $y$.

Note that if a poset $(L,⩽)$ is not a lattice, two elements of $L$ might not have a coproduct.

The construction of a coproduct may be easily generalized to an arbitrary family of elements, when $(L,⩽)$ is a complete lattice, that is, every family of elements of $L$ has an infimum and a supremum.

Categories with Coproducts

定义 3.2.6. A category $\mathcal{C}$ is said to have (binary) coproducts if any family of (two) objects of $\mathcal{C}$ has a coproduct.

例 3.2.7. We have seen that $\mathbf{Set}$ and $\mathbf{Ab}$ have (binary) coproducts, while poset categories may not have coproducts.

Let $\mathcal{C}$ be a category with binary coproducts, and let $f:A\to B$ and $f^{\prime }:A^{\prime }\to B^{\prime }$ be arrows in $\mathcal{C}$. Dually to the case of products, one may define a coproduct $f\oplus f^{\prime }:A\oplus A^{\prime }\to B\oplus B^{\prime }$ of $f$ and $f^{\prime }$.

Consider the coproducts $\left(A\oplus A^{\prime },p_1,p_2\right)$ and $\left(B\oplus B^{\prime },q_1,q_2\right)$. Let $f_1=q_1\circ f$ and $f_2=q_2\circ f^{\prime }$. By universal mapping property of the product $\left(A\oplus A^{\prime },p_1,p_2\right)$ there is a unique arrow $u:A\oplus A^{\prime }\to B\oplus B^{\prime }$ such that the following diagram is commutative:

that is, $u\circ p_1=f_1=q_1\circ f$ and $u\circ p_2=f_2=q_2\circ f^{\prime }$. We define $f\oplus f^{\prime }=u:A\oplus A^{\prime }\to B\oplus B^{\prime }$.

One may prove the following result.

命题 3.2.8. Let $\mathcal{C}$ be a category with binary coproducts. Then we have a covariant functor $L:\mathcal{C}\times \mathcal{C}\to \mathcal{C}$ defined by$$L\left(C,C^{\prime }\right)=C\oplus C^{\prime }$$for every object $\left(C,C^{\prime }\right)$ of $\mathcal{C}\times \mathcal{C}$, and$$L\left(f,f^{\prime }\right)=f\oplus f^{\prime }:A\oplus A^{\prime }\to B\oplus B^{\prime }$$for every arrow $\left(f,f^{\prime }\right)$ from $\mathcal{C}\times \mathcal{C}$ with $f:A\to B$ and $f^{\prime }:A^{\prime }\to B^{\prime }$.

For a category $\mathcal{C}$ with coproducts, one may generalize this construction to any finite family of arrows, and define a corresponding functor.

One may show the following associativity property by using universal mapping property of the coproduct.

命题 3.2.9.

Proposition 3.2.9 In any category $\mathcal{C}$ with binary coproducts, we have

$$A\oplus (B\oplus C)\cong (A\oplus B)\oplus C$$where $A,B,C$ are objects of $\mathcal{C}$.

One may show the following property.

命题 3.2.10. Let $\mathcal{C}$ be a locally small category with binary coproducts. Then for every object $A$ of $\mathcal{C}$, the contravariant functor ${\mathrm{Hom}}_{\mathcal{C}}(-,A):\mathcal{C}\to \mathbf{Set}$ preserves binary coproducts, that is, for every $B,C\in \mathcal{C}$, there is a bijection (i.e., isomorphism in $\mathbf{Set}$):$${\mathrm{Hom}}_{\mathcal{C}}(B\oplus C,A)\cong {\mathrm{Hom}}_{\mathcal{C}}(B,A)\oplus {\mathrm{Hom}}_{\mathcal{C}}(C,A)$$

3.3Equalizers

定义 3.3.1. Let $\mathcal{C}$ be a category and let $f,g:A\to B$ be arrows in $\mathcal{C}$. An equalizer of $f$ and $g$ consists of a pair $(E,e)$, where $E$ is an object of $\mathcal{C}$ and $e:E\to A$ is an arrow in $\mathcal{C}$ such that $f\circ e=g\circ e$ and it has the following universal mapping property: given any object $Z$ and any arrow $z:Z\to A$ in $\mathcal{C}$ such that $f\circ z=g\circ z$, there is a unique arrow $u:Z\to E$ in $\mathcal{C}$ such that $e\circ u=z$.

By universal mapping property of an equalizer one deduces the following result.

命题 3.3.2. An equalizer is uniquely determined up to an isomorphism.

命题 3.3.3. Any equalizer is a monomorphism.

证明. Let $f,g:A\to B$ be arrows in a category $\mathcal{C}$ having an equalizer ($E,e$). Let $\alpha ,\beta :Z\to E$ be arrows in $\mathcal{C}$ such that $e\circ \alpha =e\circ \beta$. Denote $z=e\circ \alpha =e\circ \beta :Z\to A$. Hence we have$$f\circ z=f\circ e\circ \alpha =g\circ e\circ \alpha =g\circ e\circ \beta =g\circ z.$$By universal mapping property of the equalizer, there is a unique arrow $u:Z\to E$ such that $e\circ u=z$. But we also have $e\circ \alpha =z$ and $e\circ \beta =z$. Hence we must have $u=\alpha =\beta$. This shows that $e$ is a monomorphism.

$\square$

The Category $\mathbf{Set}$

The equalizer of two functions $f,g:A\to B$ in $\mathbf{Set}$ is the pair $(E,e)$, where$$E=\{a\in A\mid f(a)=g(a)\}$$and $e:E\to A$ is the inclusion function.

For every $a\in E$ we have$$(f\circ e)(a)=f(e(a))=f(a)=g(a)=g(e(a))=(g\circ e)(a),$$hence $f\circ e=g\circ e$.

Now let $z:Z\to A$ be a function such that $f\circ z=g\circ z$. We look for a function $u:Z\to E$ such that $e\circ u=z$. This equality is equivalent to $e(u(x))=z(x)$ for every $x\in Z$, that is, $u(x)=z(x)$ for every $x\in Z$. Note that $z(x)\in E$, because $f(z(x))=g(z(x))$. Also, the equality $u(x)=z(x)$ uniquely determines $u$. Hence $(E,e)$ is an equalizer of $f,g$.

The Category $\mathbf{Vect}(K)$

The equalizer of two $K$-linear maps $f,g:A\to B$ in $\mathbf{Vect}(K)$ is the pair $(E,e)$, where$$E=\{a\in A\mid f(a)=g(a)\}$$and $e:E\to A$ is the inclusion $K$-linear map.

As in the category $\mathbf{Set}$, we have $f\circ e=g\circ e$. Also, for every $K$-linear map $z:Z\to A$ such that $f\circ z=g\circ z$, there is a unique function $u:Z\to E$ such that $e\circ u=z$. This equality is equivalent to $e(u(x))=z(x)$ for every $x\in Z$, that is, $u(x)=z(x)$ for every $x\in Z$.

Let us show that $u$ is a $K$-linear map. Let $k_1,k_2\in K$ and $x_1,x_2\in Z$. Then we have$$u\left(k_1{x}_1+k_2{x}_2\right)=z\left(k_1{x}_1+k_2{x}_2\right)=k_{1}z\left(x_1\right)+k_{2}z\left(x_2\right)=k_{1}u\left(x_1\right)+k_{2}u\left(x_2\right)$$Hence $u$ is a $K$-linear map.

In this category the equalizer of two $K$-linear maps is in fact a kernel of some $K$-linear map, namely$$E=\{a\in A\mid (f-g)(a)=0\}=\mathrm{Ker}(f-g)$$On the other hand, the kernel of a $K$-linear map is the equalizer of some $K$-linear maps, namely

$$\mathrm{Ker}(f)=\{a\in A\mid f(a)=0\}$$is the equalizer of the $K$-linear map $f:A\to B$ and the zero $K$-linear map $0:A\to B$.

Monoid Categories

Let $(M,\cdot )$ be a monoid. Recall that it may be viewed as a monoid category, where the single object is $M$, the arrows are the elements of $M$ and the composition is the multiplication of the elements of $M$. An equalizer of two elements $a,b\in M$ is an element $x\in M$ with $ax=bx$ and for every $z\in M$ such that $az=bz$ there is $u\in M$ such that $xu=z$. If $(M,\cdot )$ is a non-trivial group, then $a,b\in M$ with $a\ne b$ do not have an equalizer, because there is no $x\in M$ such that $ax=bx$.

定义 3.3.4. We say that a category has equalizers if every arrows $f,g:A\to B$ have an equalizer.

例 3.3.5. The categories $\mathbf{Set}$ and $\mathbf{Vect}(K)$ have equalizers, but monoid categories may not have equalizers.

3.4Coequalizer

定义 3.4.1. Let $\mathcal{C}$ be a category and let $f,g:A\to B$ be arrows in $\mathcal{C}$. A coequalizer of $f$ and $g$ consists of a pair $(Q,q)$, where $Q$ is an object of $\mathcal{C}$ and $q:B\to Q$ is an arrow in $\mathcal{C}$ such that $q\circ f=q\circ g$ and it has the following universal mapping property: given any object $Z$ and any arrow $z:B\to Z$ in $\mathcal{C}$ such that $z\circ f=z\circ g$, there is a unique arrow $u:Q\to Z$ in $\mathcal{C}$ such that $u\circ q=z$.

The following two results are dual to the corresponding ones for equalizers.

命题 3.4.2. A coequalizer is uniquely determined up to an isomorphism.

命题 3.4.3. Any coequalizer is an epimorphism.

The Category $\mathbf{Set}$

Let $f:A\to B$ be function. Then the homogeneous relation $\ker (f)=(A,A,\mathrm{Ker}(f))$ with the graph$$\mathrm{Ker}(f)=\left\{\left(x_1,x_2\right)\in A\times A\mid f\left(x_1\right)=f\left(x_2\right)\right\}$$is called the kernel of $f$. Note that the kernel of $f$ is an equivalence relation on $A$.

The following theorem will be useful.

定理 3.4.4 (Factorization theorem by a surjection). Let $f:A\to B$ be a function and let $g:A\to C$ be a surjective function such that $\mathrm{Ker}(g)\subseteq \mathrm{Ker}(f)$. Then there is a unique function $h:C\to B$ such that $f=h\circ g$.

Let $f,g:A\to B$ be functions. We define a relation $r=(B,B,R)$ by$$\left(b_1,b_2\right)\in R⟺\exists a\in A:b_1=f(a)\text{ and }{b}_2=g(a).$$Let $\overline{r}=(B,B,\overline{R})$ be the smallest equivalence relation on $B$ containing $R$. Then we may consider the partition$$B/\overline{R}=\{\overline{R}⟨b⟩\mid b\in B\}$$of $B$. Consider the function $\pi :B\to B/\overline{R}$ defined by $\pi (b)=\overline{R}⟨b⟩$. Note that the kernel of the function $\pi$ is$$\begin{array}{rl}\mathrm{Ker}(\pi )& =\left\{\left(b_1,b_2\right)\in B\times B\mid \pi \left(b_1\right)=\pi \left(b_2\right)\right\}\\ & =\left\{\left(b_1,b_2\right)\in B\times B\mid \overline{R}⟨b_1⟩=\overline{R}⟨b_2⟩\right\}=\overline{R}.\end{array}$$

We show that $(B/\overline{R},\pi )$ is a coequalizer of $f$ and $g$.

For every $a\in A$, we have $(f(a),g(a))\in \overline{R}$, which implies that $\overline{R}⟨f(a)⟩=\overline{R}⟨g(a)⟩$. It follows that$$(\pi \circ f)(a)=\overline{R}⟨f(a)⟩=\overline{R}⟨g(a)⟩=(\pi \circ g)(a),$$hence $\pi \circ f=\pi \circ g$.

Now let $z:B\to Z$ be a function such that $z\circ f=z\circ g$. We show that$$\overline{R}\subseteq \mathrm{Ker}(z)=\left\{\left(b_1,b_2\right)\in B\times B\mid z\left(b_1\right)=z\left(b_2\right)\right\}$$(the kernel of the function $z$ ). Let $\left(b_1,b_2\right)\in \overline{R}$. Then there is a finite number of elements $c_0,\dots ,c_n\in B$ such that $b_1=c_0,b_2=c_n$ and for every $i\in \{1,\dots ,n\}$, we have either $\left(c_{i-1},c_i\right)\in$ $R$ or $\left(c_i,c_{i-1}\right)\in R$. We may assume that $\left(b_1,b_2\right)\in R$, because otherwise we may proceed inductively. Since $\left(b_1,b_2\right)\in R$, there is $a\in A$ such that $b_1=f(a)$ and $b_2=g(a)$. Then we have$$z\left(b_1\right)=z(f(a))=z(g(a))=z\left(b_2\right),$$hence $\left(b_1,b_2\right)\in \mathrm{Ker}(z)$. Thus, we have $\mathrm{Ker}(\pi )=\overline{R}\subseteq \mathrm{Ker}(z)$. Using the factorization theorem of the function $z$ by the surjective function $\pi$, there is a unique function $u:B/\overline{R}\to Z$ such that $u\circ \pi =z$.

The Category $\mathbf{Ab}$

The following theorem will be useful.

定理 3.4.5 (Factorization Theorem by an Epimorphism). Let $f:A\to B$ be a homomorphism of abelian groups, and let $g:A\to A^{\prime }$ be an epimorphism of abelian groups with $\mathrm{Ker}(g)\subseteq \mathrm{Ker}(f)$. Then there exists a unique homomorphism of abelian groups $h:A^{\prime }\to B$ such that $f=h\circ g$, that is, the following diagram is commutative:

证明. Let $a^{\prime }\in A^{\prime }$. Since $g$ is an epimorphism, there exists $a\in A$ such that $g(a)=a^{\prime }$. If there exists $a_0\in A$ such that $g\left(a_0\right)=a^{\prime }$, then we have $g(a)=g\left(a_0\right)$. It follows that $g\left(a-a_0\right)=0$, hence $a-a_0\in \mathrm{Ker}(g)\subseteq \mathrm{Ker}(f)$. Then $f\left(a-a_0\right)=0$, hence $f(a)=f\left(a_0\right)$.

It follows that we can define the function $h:A^{\prime }\to B$ by $h\left(a^{\prime }\right)=f(a)$, where $f(a)$ is uniquely determined as above. We have $h(g(a))=a$ for every $a\in A$. Hence $f=h\circ g$.

We show that $h$ is a homomorphism of abelian groups. Let $a_1^{\prime },a_2^{\prime }\in A^{\prime }$. Then there exist $a_1,a_2\in A$ such that $g\left(a_1\right)=a_1^{\prime }$ and $g\left(a_2\right)=a_2^{\prime }$. Hence $h\left(a_1^{\prime }\right)=f\left(a_1\right)$ and $h\left(a_2^{\prime }\right)=f\left(a_2\right)$. We have$$g\left(a_1+a_2\right)=g\left(a_1\right)+g\left(a_2\right)=a_1^{\prime }+a_2^{\prime }.$$It follows that$$h\left(a_1^{\prime }+a_2^{\prime }\right)=h\left(g\left(a_1+a_2\right)\right)=f\left(a_1+a_2\right)=f\left(a_1\right)+f\left(a_2\right)=h\left(a_1^{\prime }\right)+h\left(a_2^{\prime }\right)$$Thus $h$ is a homomorphism of abelian groups.

For uniqueness, suppose that there exists a homomorphism $h^{\prime }:A^{\prime }\to B$ such that $f=h^{\prime }\circ g$. It follows that $h\circ g=h^{\prime }\circ g$. Since $g$ is an epimorphism, we have $h=h^{\prime }$.

$\square$

Let $(A,+)$ and $(B,+)$ be abelian groups and let $f,g:A\to B$ be group homomorphisms. Since $I=\mathrm{Im}(f-g)$ is a subgroup of $B$, we may consider the factor group$$Q=B/I=\{b+I\mid b\in B\}$$where the operation is defined by$$\left(b_1+I\right)+\left(b_2+I\right)=\left(b_1+b_2\right)+I,\forall b_1,b_2\in B$$The factor group $B/\mathrm{Im}(f-g)$ is also called a cokernel of the group homomorphism $f-g$, and we denote it by $\mathrm{Coker}(f-g)$.

Consider the group homomorphism $q:B\to Q=B/I$ defined by $q(b)=b+I$. We have $\mathrm{Ker}(q)=\{b\in B\mid q(b)=I\}=\{b\in B\mid b+I=I\}=I$.

We show that $(Q,q)$ is a coequalizer of $f$ and $g$.

For every $a\in A$ we have $f(a)-g(a)=(f-g)(a)\in I$, hence $f(a)+I=g(a)+I$. Then for every $a\in A$ we have$$(q\circ f)(a)=f(a)+I=g(a)+I=(q\circ g)(a),$$hence $q\circ f=q\circ g$.

Now let $(Z,+)$ be an abelian group and let $z:B\to Z$ be a group homomorphism such that $z\circ f=z\circ g$. We show that $I\subseteq \mathrm{Ker}(z)$. To this end, let $b\in I$. Then $b=(f-g)(a)$ for some $a\in A$. It follows that$$z(b)=z((f-g)(a))=z(f(a)-g(a))=z(f(a))-z(g(a))=0,$$hence $b\in \mathrm{Ker}(z)$. Thus we have $\mathrm{Ker}(q)=I\subseteq \mathrm{Ker}(z)$. Using the factorization theorem for the group homomorphism $z$ by the epimorphism $q$, there is a unique group homomorphism $u:B/I\to Z$ such that $u\circ q=z$.

In this category the coequalizer of two group homomorphisms is in fact the cokernel of some group homomorphism, namely$$Q=\mathrm{Coker}(f-g)$$On the other hand, the cokernel of a group homomorphism is the coequalizer of some group homomorphism, namely$$\mathrm{Coker}(f)=B/\mathrm{Im}(f)=B/\mathrm{Im}(f-0)$$is the coequalizer of the group homomorphism $f:A\to B$ and the zero group homomorphism $0:A\to B$.

Monoid Categories

Let $(M,\cdot )$ be a monoid, which may be viewed as a monoid category. A coequalizer of two elements $a,b\in M$ is an element $x\in M$ with $xa=xb$ and for every $z\in M$ such that $za=zb$ there is $u\in M$ such that $ux=z$. If $(M,\cdot )$ is a non-trivial group, then $a,b\in M$ with $a\ne b$ do not have a coequalizer, because there is no $x\in M$ such that $xa=xb$.

定义 3.4.6. We say that a category has coequalizers if every arrows $f,g:A\to B$ have a coequalizer.

例 3.4.7. The categories $\mathbf{Set}$ and $\mathbf{Ab}$ have coequalizers, while monoid categories may not have coequalizers.