2. Abstract Structure

2.1Epimorphisms and Monomorphisms

定义 2.1.1. Let $\mathcal{C}$ be a category. An arrow $f:A\to B$ in $\mathcal{C}$ is called a:

(1)	monomorphism (or briefly $mono$) if for every arrows $g,h:C\to A$ such that $f\circ g=f\circ h$, we have $g=h$.
(2)	epimorphism (or briefly $epi$) if for every arrows $g,h:B\to C$ such that $g\circ f=h\circ f$, we have $g=h$.
(3)	bimorphism if it is both a monomorphism and an epimorphism.

引理 2.1.2. Every isomorphism is a bimorphism.

证明. Let $f:A\to B$ be an isomorphism in a category $\mathcal{C}$. Let $g,h:C\to A$ be arrows in $\mathcal{C}$ such that $f\circ g=f\circ h$. Compose on the left by $f^{-1}$ in order to get $g=h$. Hence $f$ is a monomorphism. Similarly, one shows that $f$ is an epimorphism. Thus, $f$ is a bimorphism.

$\square$

One may also prove the following property.

命题 2.1.3. Let $f:A\to B$ and $g:B\to C$ be arrows in a category $\mathcal{C}$.

(1)	If $f,g$ are monomorphisms (epimorphisms), then so is $g\circ f$.
(2)	If $g\circ f$ is monomorphism, then so is $f$.
(3)	If $g\circ f$ is epimorphism, then so is $g$.

例 2.1.4.  

(1)	In $\mathbf{Set}$ monomorphisms coincide with injective functions, epimorphisms coincide with surjective functions, and bimorphisms coincide with isomorphisms and with bijective functions.
(2)	In many usual concrete categories, monomorphisms coincide with injective arrows. In not so many usual categories epimorphisms coincide with surjective arrows. For instance, in $\mathbf{Mon}$ monomorphisms coincide with injective monoid homomorphisms, but epimorphisms do not coincide with surjective monoid homomorphisms. Indeed, the inclusion $\mathrm{map}i:\mathbb{N}\to \mathbb{Z}$, defined by $i(x)=x$, is an epimorphism in $\mathbf{Mon}$, but it is not surjective. This is also an example of a bimorphism, which is not an isomorphism.
(3)	There is a large class of categories, called abelian categories (e.g., the categories $\mathbf{Ab}$ and $\mathrm{Vect}(K)$), in which bimorphisms coincide with isomorphisms.

定义 2.1.5. Let $\mathcal{C}$ be a category. An arrow $f:A\to B$ in $\mathcal{C}$ is called a:

(1)	section (or split monomorphism) if it has a left inverse arrow, that is, there is an arrow $g:B\to A$ such that $g\circ f=1_A$.
(2)	retraction (or split epimorphism) if it has a right inverse arrow, that is, there is an arrow $g:B\to A$ such that $f\circ g=1_B$.

注 2.1.6. An arrow is an isomorphism if and only if it is a section and a retraction.

引理 2.1.7.  

(1)	Every section is a monomorphism.
(2)	Every retraction is an epimorphism.

证明.  

(1)	Let $f:A\to B$ be a section in a category $\mathcal{C}$. Hence there is an arrow $r:B\to A$ such that $r\circ f=1_A$. Let $g,h:B\to C$ be arrows in $\mathcal{C}$ such that $f\circ g=f\circ h$. Then we have $$r\circ f\circ g=r\circ f\circ h,$$which implies that $g=h$. Hence $f$ is a monomorphism.
(2)	Similarly, one shows that every retraction is an epimorphism.

$\square$

One may show the following property.

命题 2.1.8. The following are equivalent for an arrow $f:A\to B$ in any category $\mathcal{C}$:

(1)	$f$ is an isomorphism.
(2)	$f$ is both a monomorphism and a retraction.
(3)	$f$ is both a section and an epimorphism.

例 2.1.9.  

(1)

In $\mathbf{Set}$ every monomorphism (i.e., injective function) is a section, except those of the form $\varnothing \to A$ with $A\ne \varnothing$. In $\mathbf{Set}$ the condition that every epimorphism is a retraction is equivalent to the axiom of choice.

(2)

In $\mathbf{Ab}$ the inclusion homomorphism $i:2\mathbb{Z}\to \mathbb{Z}$, defined by $i(2x)=2x$, is clearly a monomorphism. Suppose that it is a section. Then there is a group homomorphism $g:\mathbb{Z}\to 2\mathbb{Z}$ such that $g\circ i=1_{2\mathbb{Z}}$. We have $2g(1)=g(2)=g(i(2))=2$, hence $1=g(1)\in 2\mathbb{Z}$, a contradiction. Therefore, there are monomorphisms, which are not sections. In $\mathbf{Ab}$ the homomorphism $f:{\mathbb{Z}}_4\to {\mathbb{Z}}_2$ defined by $f(\bar{x})=\hat{x}$ is clearly an epimorphism. Suppose that it is a retraction. Then there is a group homomomorphism $g:{\mathbb{Z}}_2\to {\mathbb{Z}}_4$ such that $f\circ g=1_{{\mathbb{Z}}_2}$. The order of $g(\hat{1})$ divides the order of $\hat{1}$, which is $2$. Hence $g(\hat{1})\in \{\overline{0},\overline{2}\}$. But then $f(g(\hat{1}))=\hat{0}\ne \hat{1}=1_{{\mathbb{Z}}_2}(\hat{1})$, a contradiction. Therefore, there are epimorphisms which are not retractions.

命题 2.1.10. Let $F:\mathcal{C}\to \mathcal{D}$ be a (covariant) functor. Then:

(1)	$F$ preserves sections in the sense that if $f$ is a section, then $F(f)$ is also a section.
(2)	$F$ preserves retractions in the sense that if $f$ is a retraction, then $F(f)$ is also a retraction.
(3)	$F$ preserves isomorphisms in the sense that if $f$ is an isomorphism, then $F(f)$ is also an isomorphism.

证明. Let $f:A\to B$ be a section in $\mathcal{C}$. Then there is an arrow $g:B\to A$ such that $g\circ f=1_A$. This implies that$$F(g)\circ F(f)=F(g\circ f)=F\left(1_A\right)=1_{F(A)}$$hence $F(f)$ is a section in $\mathcal{D}$. Similarly, $F$ preserves retractions. The fact that $F$ preserves isomorphisms is a consequence of the first two properties.

$\square$

2.2Initial and Terminal Objects

定义 2.2.1. Let $\mathcal{C}$ be a category. An object $C^{\prime }$ of $\mathcal{C}$ is called:

(1)	initial if for every object $C$ of $\mathcal{C}$, there is a unique arrow $C^{\prime }\to C$.
(2)	terminal if for every object $C$ of $\mathcal{C}$, there is a unique arrow $C\to C^{\prime }$.

Sometimes an initial object is denoted by $0$, while a terminal object is denoted by $1$.

命题 2.2.2. Initial and terminal objects are unique up to an isomorphism.

证明. Assume that $C^{\prime },C^{\prime \prime }$ are initial objects of a category $\mathcal{C}$. Then there is a unique arrow $f:C^{\prime }\to C^{\prime \prime }$ and a unique arrow $g:C^{\prime \prime }\to C^{\prime }$. Note that we have the arrows $g\circ f:C^{\prime }\to C^{\prime }$ and $1_{C^{\prime }}:C^{\prime }\to C^{\prime }$. Since $C^{\prime }$ is an initial object, we must have $g\circ f=1_{C^{\prime }}$. Also, note that we have the arrows $f\circ g:C^{\prime \prime }\to C^{\prime \prime }$ and $1_{C^{\prime \prime }}:C^{\prime \prime }\to C^{\prime \prime }$. Since $C^{\prime \prime }$ is an initial object, we must have $f\circ g=1_{C^{\prime \prime }}$. Hence $f:C^{\prime }\to C^{\prime \prime }$ is an isomorphism. Hence initial objects are unique up to an isomorphism.

Similarly, one shows that terminal objects are unique up to an isomorphism.

$\square$

例 2.2.3.  

(1)	In $\mathbf{Set}$ the initial object is $\varnothing$, while the terminal object is any single-element set.
(2)	In $\mathbf{Grp}$ the initial object is the trivial group, while the terminal object is again the trivial group.
(3)	In $\mathbf{Ring}$ the initial object is $\mathbb{Z}$, while the terminal object is the trivial ring. Note that there is a unique unitary ring homomorphism $\mathbb{Z}\to R$ for every $\mathrm{ring}R$ with identity $1^{\prime }$, which is defined by $$f(n)=n\cdot 1^{\prime },\forall x\in \mathbb{Z}$$ We first show that if $f$ does exist, then it is unique. So, suppose that $f:\mathbb{Z}\to R$ is a unitary ring homomorphism. Then $f(0)=0^{\prime }=0\cdot 1^{\prime }$, where $0^{\prime }$ is the zero element of $R$. For every $k\in {\mathbb{N}}^{\ast }$, we have:$$\begin{array}{c}f(k)=f(\underset{k\text{ times }}{\underbrace{1+\cdots +1}})=\underset{k\text{ times }}{\underbrace{f(1)+\cdots +f(1)}}=\underset{k\text{ times }}{\underbrace{1^{\prime }+\cdots +1^{\prime }}}=k\cdot 1^{\prime }\\ f(-k)=-f(k)=-\left(k\cdot 1^{\prime }\right)=(-k)\cdot 1^{\prime }\end{array}$$Hence $f(n)=n\cdot 1^{\prime }$ for every $n\in \mathbb{Z}$. Now we show that the function $f$ is a unitary ring homomorphism. For every $m,n\in \mathbb{Z}$, we have:$$\begin{array}{r}f(m+n)=(m+n)\cdot 1^{\prime }=m\cdot 1^{\prime }+n\cdot 1^{\prime }=f(m)+f(n),\\ f(m\cdot n)=(m\cdot n)\cdot 1^{\prime }=\left(m\cdot 1^{\prime }\right)\cdot \left(n\cdot 1^{\prime }\right)=f(m)\cdot f(n)\end{array}$$and $f(1)=1\cdot 1^{\prime }=1^{\prime }$. Hence $f$ is a unitary ring homomorphism.
(4)	View the poset $(\mathbb{Z},⩽)$ as a poset category. This category has neither an initial object, nor a terminal object.

2.3Products

定义 2.3.1. Let $\mathcal{C}$ be a category and $A,B$ objects of $\mathcal{C}$. A product diagram for $A$ and $B$ consists of an object $P$ and arrows, called canonical projections$$A\stackrel{p_1}{⟵}P\stackrel{p_2}{⟶}B$$satisfying the following universal mapping property: given any diagram of the form$$A\stackrel{f_1}{⟵}X\stackrel{f_2}{\to }B$$there is a unique arrow $u:X\to P$ such that$$p_1\circ u=f_1\text{and}{p}_2\circ u=f_2$$that is, the following diagram is commutative:

We denote the product of $A$ and $B$ by $\left(P,p_1,p_2\right)$. Sometimes $P$ is also denoted by $A\prod B$ or $A\times B$.

注 2.3.2. Sometimes (especially in the so-called additive categories), the canonical projections $p_1:A\times B\to A$ and $p_2:A\times B\to B$ are also denoted by $\left[\begin{array}{cc}1& 0\end{array}\right]:A\times B\to A$ and $\left[\begin{array}{cc}0& 1\end{array}\right]:A\times B\to B$ respectively. The unique arrow $u:X\to A\times B$ such that $p_2\circ u=f_1$ and $p_2\circ u=f_2$ is also denoted by$$\left[\begin{array}{c}{f}_1\\ f_2\end{array}\right]:X\to A\times B$$Then equalities involving compositions of arrows such as $p_1\circ u=f_1$ and $p_2\circ u=f_2$ may be rewritten in terms of matrix multiplications as $\left[\begin{array}{cc}1& 0\end{array}\right]\cdot \left[\begin{array}{c}{f}_1\\ f_2\end{array}\right]=f_1$ and $\left[\begin{array}{cc}0& 1\end{array}\right]\cdot \left[\begin{array}{c}{f}_1\\ f_2\end{array}\right]=f_2$.

注 2.3.3. Note that the canonical projections $p_1$ and $p_2$ need not be epimorphisms. For instance, consider the category described by the following graph:such that $g\circ p_2=h\circ p_2$. Then $\left(P,p_1,p_2\right)$ is a product of $A$ and $B$, but $p_2$ is not an epimorphism, because we have $g\circ p_2=h\circ p_2$ and $g\ne h$.

As usual, universal mapping property insures the following uniqueness result.

命题 2.3.4. The product is unique up to an isomorphism.

证明. Suppose that $\left(P,p_1,p_2\right)$ and $\left(Q,q_1,q_2\right)$ are products of objects $A$ and $B$ of a category $\mathcal{C}$. Since $\left(Q,q_1,q_2\right)$ is a product, there is a unique arrow $i:P\to Q$ such that $q_1\circ i=p_1$ and $q_2\circ i=p_2$. Since $\left(P,p_1,p_2\right)$ is a product, there is a unique arrow $j:Q\to P$ such that $p_1\circ j=q_1$ and $p_2\circ j=q_2$. Hence we have the following commutative diagram:

It follows that $p_1\circ (j\circ i)=p_1$ and $p_2\circ (j\circ i)=p_2$. But we also have $p_1\circ 1_P=p_1$ and $p_2\circ 1_P=p_2$, and by the uniqueness condition of universal mapping property we have $j\circ i=1_P$. Similarly, one shows that $i\circ j=1_Q$. Hence $i:P\to Q$ is an isomorphism.

$\square$

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

定义 2.3.5. A product of a family ${\left(A_i\right)}_{i\in I}$ of objects of a category $\mathcal{C}$ consists of an object $P$, also denoted by $\prod _{i\in I}{A}_i$, and a family ${\left(p_i\right)}_{i\in I}$ of arrows, where $p_i:P\to A_i$ for every $i\in I$, satisfying the following universal mapping property: given any object $X$ of $\mathcal{C}$ and any family ${\left(f_i\right)}_{i\in I}$ of arrows, where $f_i:X\to A_i$ for every $i\in I$, there is a unique arrow $u:X\to P$ such that $p_i\circ u=f_i$ for every $i\in I$.

2.4Examples of Products

The Category $\mathbf{Set}$

The product of two sets $A$ and $B$ is$$\left(A\times B,p_1,p_2\right)$$where $p_1:A\times B\to A$ is the function defined by $p_1(a,b)=a$, and $p_2:A\times B\to B$ is the function defined by $p_2(a,b)=b$.

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

that is, $p_1\circ u=f_1$ and $p_2\circ u=f_2$. These equalities are equivalent to $p_1(u(x))=f_1(x)$ and $p_2(u(x))=f_2(x)$ for every $x\in X$. This means that$$u(x)=\left(f_1(x),f_2(x)\right),\forall x\in X$$Note that $u$ is uniquely determined by this definition.

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

The Category $\mathbf{Grp}$

The product of two groups $\left(G_1,\cdot \right)$ and $\left(G_2,\cdot \right)$ is$$\left(\left(G_1\times G_2,\cdot \right),p_1,p_2\right)$$where $p_1:G_1\times G_2\to G_1$ is the function defined by $p_1\left(g_1,g_2\right)=g_1$, and $p_2:G_1\times G_2\to G_2$ is the function defined by $p_2\left(g_1,g_2\right)=g_2$.

Note that $G_1\times G_2$ is a group with respect to the operation defined by$$\left(x_1,x_2\right)\cdot \left(x_1^{\prime },x_2^{\prime }\right)=\left(x_1\cdot x_1^{\prime },x_2\cdot x_2^{\prime }\right),\forall \left(x_1,x_2\right),\left(x_1^{\prime },x_2^{\prime }\right)\in G_1\times G_2$$

Let $(X,\cdot )$ be a group and let $f_1:X\to G_1$ and $f_2:X\to G_2$ be group homomorphisms. We look for a unique group homomorphism $u:X\to G_1\times G_2$ such that the following diagram is commutative:

that is, $p_1\circ u=f_1$ and $p_2\circ u=f_2$. These equalities are equivalent to $p_1(u(x))=f_1(x)$ and $p_2(u(x))=f_2(x)$ for every $x\in X$. This means that

$$u(x)=\left(f_1(x),f_2(x)\right),\forall x\in X$$Note that $u$ is uniquely determined by this definition.

We still need to prove that $u$ is a group homomorphism. For every $x_1,x_2\in X$ we have$$\begin{array}{rl}u\left(x_1\cdot x_2\right)& =\left(f_1\left(x_1\cdot x_2\right),f_2\left(x_1\cdot x_2\right)\right)=\left(f_1\left(x_1\right)\cdot f_1\left(x_2\right),f_2\left(x_1\right)\cdot f_2\left(x_2\right)\right)\\ & =\left(f_1\left(x_1\right),f_2\left(x_1\right)\right)\cdot \left(f_1\left(x_2\right),f_2\left(x_2\right)\right)=u\left(x_1\right)\cdot u\left(x_2\right)\end{array}$$hence $u$ is a group homomorphism.

The construction of a product may be easily generalized to an arbitrary family of groups.

Poset Categories

Let $(L,⩽)$ be a lattice. Hence every two elements of $L$ have an infimum (i.e., greatest lower 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 product of two elements $x,y\in L$ is$$\left(\inf (x,y),p_1,p_2\right)$$where $p_1:\inf (x,y)\to x$ and $p_2:\inf (x,y)\to y$ are the unique arrows having the given domains and codomains.

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

This means that $z⩽\inf (x,y)$. But this is true, because $z$ is a lower bound of $x$ and $y$, while $\inf (x,y)$ is the greatest lower bound of $x$ and $y$.

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

The construction of a product 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.

2.5Categories with Products

定义 2.5.1. A category $\mathcal{C}$ is said to have (binary) products if any family of (two) objects of $\mathcal{C}$ has a product.

例 2.5.2. We have seen that $\mathbf{Set}$ and $\mathbf{Grp}$ have (binary) products, while poset categories may not have products.

Let $\mathcal{C}$ be a category with binary products, and let $f:A\to B$ and $f^{\prime }:A^{\prime }\to B^{\prime }$ be arrows in $\mathcal{C}$. We want to define a product $f\times f^{\prime }:A\times A^{\prime }\to B\times B^{\prime }$ of $f$ and $f^{\prime }$.

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

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

One may prove the following result.

命题 2.5.3. Let $\mathcal{C}$ be a category with binary products. Then we have a covariant functor $R:\mathcal{C}\times \mathcal{C}\to \mathcal{C}$ defined by$$R\left(C,C^{\prime }\right)=C\times C^{\prime }$$for every object $\left(C,C^{\prime }\right)$ of $\mathcal{C}\times \mathcal{C}$, and$$R\left(f,f^{\prime }\right)=f\times f^{\prime }:A\times A^{\prime }\to B\times B^{\prime }$$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 products, 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 product.

命题 2.5.4. In any category $\mathcal{C}$ with binary products, we have$$A\times (B\times C)\cong (A\times B)\times C,$$where $A,B,C$ are objects of $\mathcal{C}$.

2.6Hom-sets

In this section, assume that all categories are locally small, that is, ${\mathrm{Hom}}_{\mathcal{C}}(A,B)$ is a set for every $A,B\in \mathcal{C}$.

Let $A$ be an object and $f:B\to B^{\prime }$ an arrow in a category $\mathcal{C}$. We define$$\begin{array}{rl}& f_{\ast }={\mathrm{Hom}}_{\mathcal{C}}(A,f):{\mathrm{Hom}}_{\mathcal{C}}(A,B)\to {\mathrm{Hom}}_{\mathcal{C}}\left(A,B^{\prime }\right)\\ & f_{\ast }(g)=f\circ g,\forall g\in {\mathrm{Hom}}_{\mathcal{C}}(A,B)\end{array}$$

One may show the following property.

命题 2.6.1. Let $\mathcal{C}$ be a category and let $A$ be an object of $\mathcal{C}$. Then we have a covariant functor ${\mathrm{Hom}}_{\mathcal{C}}(A,-):\mathcal{C}\to \mathbf{Set}$, called the covariant representable functor, defined by$$B\mapsto {\mathrm{Hom}}_{\mathcal{C}}(A,B)$$on every object $B$ of $\mathcal{C}$ and$$f\mapsto f_{\ast }={\mathrm{Hom}}_{\mathcal{C}}(A,f):{\mathrm{Hom}}_{\mathcal{C}}(A,B)\to {\mathrm{Hom}}_{\mathcal{C}}\left(A,B^{\prime }\right)$$for every arrow $f:B\to B^{\prime }$ in $\mathcal{C}$.

One may show the following property.

命题 2.6.2. Let $\mathcal{C}$ be a category with binary products. Then for every object $A$ of $\mathcal{C}$, the covariant functor ${\mathrm{Hom}}_{\mathcal{C}}(A,-):\mathcal{C}\to \mathbf{Set}$ preserves binary products, that is, for every $C,D\in \mathcal{C}$, there is a bijection (i.e., isomorphism in $\mathbf{Set}$):$${\mathrm{Hom}}_{\mathcal{C}}(A,C\times D)\cong {\mathrm{Hom}}_{\mathcal{C}}(A,C)\times {\mathrm{Hom}}_{\mathcal{C}}(A,D).$$

定义 2.6.3. A contravariant functor between two categories $\mathcal{C}$ and $\mathcal{D}$ is a mapping of objects of $\mathcal{C}$ to objects of $\mathcal{D}$ and of arrows of $\mathcal{C}$ to arrows of $\mathcal{D}$, denoted by$$F:\mathcal{C}\to \mathcal{D}$$satisfying the axioms:

(i)	For every $f:A\to B$ in $\mathcal{C}$, we have $F(f):F(B)\to F(A)$ in $\mathcal{D}$.
(ii)	For every object $A$ of $\mathcal{C}$, we have $F\left(1_A\right)=1_{F(A)}$.
(iii)	For every composable pair of arrows $f:A\to B$ and $g:B\to C$ in $\mathcal{C}$, we have $$F(g\circ f)=F(f)\circ F(g)$$

Let $A$ be an object and $f:B\to B^{\prime }$ an arrow in a category $\mathcal{C}$. We define$$\begin{array}{rl}& f^{\ast }={\mathrm{Hom}}_{\mathcal{C}}(f,A):{\mathrm{Hom}}_{\mathcal{C}}\left(B^{\prime },A\right)\to {\mathrm{Hom}}_{\mathcal{C}}(B,A),\\ & f^{\ast }(g)=g\circ f,\forall g\in {\mathrm{Hom}}_{\mathcal{C}}\left(B^{\prime },A\right).\end{array}$$

命题 2.6.4. Let $\mathcal{C}$ be a category and let $A$ be an object of $\mathcal{C}$. Then we have a contravariant functor ${\mathrm{Hom}}_{\mathcal{C}}(-,A):\mathcal{C}\to \mathbf{Set}$, called the contravariant representable functor, defined by$$B\mapsto {\mathrm{Hom}}_{\mathcal{C}}(B,A)$$on every object $B$ of $\mathcal{C}$ and$$f\mapsto f^{\ast }={\mathrm{Hom}}_{\mathcal{C}}(f,A):{\mathrm{Hom}}_{\mathcal{C}}\left(B^{\prime },A\right)\to {\mathrm{Hom}}_{\mathcal{C}}(B,A)$$on every arrow $f:B\to B^{\prime }$ in $\mathcal{C}$.

证明.  

(i)	For every arrow $f:B\to B^{\prime }$ in $\mathcal{C}$, we have the following function (arrow in $\mathbf{Set}$): $$f^{\ast }={\mathrm{Hom}}_{\mathcal{C}}(f,A):{\mathrm{Hom}}_{\mathcal{C}}\left(B^{\prime },A\right)\to {\mathrm{Hom}}_{\mathcal{C}}(B,A)$$
(ii)	For every object $B$ of $\mathcal{C}$, we have $${\mathrm{Hom}}_{\mathcal{C}}\left(1_B,A\right)=1_B^{\ast }=1_{\mathrm{Hom}}^{\mathcal{C}}(B,A)$$
(iii)	Let $f:B\to B^{\prime }$ and $g:B^{\prime }\to B^{\prime \prime }$ be arrows in $\mathcal{C}$. Then we have the following functions (arrows in $\mathbf{Set}$) $$(g\circ f{)}^{\ast },f^{\ast }\circ g^{\ast }:{\mathrm{Hom}}_{\mathcal{C}}\left(B^{\prime \prime },A\right)\to {\mathrm{Hom}}_{\mathcal{C}}(B,A)$$For every function $h\in {\mathrm{Hom}}_{\mathcal{C}}\left(B^{\prime \prime },A\right)$, we have $$(g\circ f{)}^{\ast }(h)=h\circ (g\circ f)=(h\circ g)\circ f=g^{\ast }(h)\circ f=f^{\ast }\left(g^{\ast }(h)\right)=\left(f^{\ast }\circ g^{\ast }\right)(h)$$Hence we have $(g\circ f{)}^{\ast }=f^{\ast }\circ g^{\ast }$. It follows that $${\mathrm{Hom}}_{\mathcal{C}}(g\circ f,A)=(g\circ f{)}^{\ast }=f^{\ast }\circ g^{\ast }={\mathrm{Hom}}_{\mathcal{C}}(f,A)\circ {\mathrm{Hom}}_{\mathcal{C}}(g,A)$$

Hence ${\mathrm{Hom}}_{\mathcal{C}}(-,A):\mathcal{C}\to \mathbf{Set}$ is a contravariant functor.

$\square$