5. Naturality

5.1Fully Faithful Functors

定义 5.1.1. Let $\mathcal{C}$ and $\mathcal{D}$ be locally small categories. Let $F:\mathcal{C}\to \mathcal{D}$ be a functor and for every objects $A,B$ of $\mathcal{C}$ consider the map$$F_{A,B}:{\mathrm{Hom}}_{\mathcal{C}}(A,B)\to {\mathrm{Hom}}_{\mathcal{D}}(F(A),F(B)),F_{A,B}(f)=F(f).$$Then $F$ is called:

(1)	faithful if for every objects $A,B$ of $\mathcal{C}$ the map $F_{A,B}$ is injective.
(2)	full if for every objects $A,B$ of $\mathcal{C}$ the map $F_{A,B}$ is surjective.
(3)	fully faithful if it is full and faithful, that is, for every objects $A,B$ of $\mathcal{C}$ the map $F_{A,B}$ is bijective.

定义 5.1.2. Let $\mathcal{C}$ be a category with the class of objects ${\mathcal{C}}_0$ and the class of arrows ${\mathcal{C}}_1$. By a subcategory of $\mathcal{C}$ we mean a collection $\mathcal{U}$ of some objects ${\mathcal{U}}_0\subseteq {\mathcal{C}}_0$ and some arrows ${\mathcal{U}}_1\subseteq {\mathcal{C}}_1$ closed under taking composition, under taking domains and codomains, and under taking identity. Hence there is a functor $i:\mathcal{U}\to \mathcal{C}$ called the inclusion functor.

By a full subcategory of $\mathcal{C}$ we mean a collection $\mathcal{U}$ of some objects of $\mathcal{C}$ together with all the arrows between them. Hence the inclusion functor $i:\mathcal{U}\to \mathcal{C}$ is fully faithful in this case.

例 5.1.3. (1) The inclusion functor $i:{\mathbf{Set}}_{fin}\to \mathbf{Set}$ is fully faithful, where ${\mathbf{Set}}_{fin}$ is the category of finite sets and functions. Note that the map$$i_{A,B}:{\mathrm{Hom}}_{{\mathbf{Set}}_{fin}}(A,B)\to {\mathrm{Hom}}_{\mathbf{Set}}(i(A),i(B)),i_{A,B}(f)=i(f)=f$$is bijective. The category ${\mathbf{Set}}_{fin}$ is a full subcategory of $\mathbf{Set}$.

(2) The forgetful functor $U:\mathbf{Grp}\to \mathbf{Set}$ is defined on objects (groups) $(G,\cdot )$ of $\mathbf{Grp}$ by the underlying set $G$ of $(G,\cdot )$, and on arrows (group homomorphisms) $f:A\to B$ in $\mathbf{Grp}$ by the underlying function $f:A\to B$. Then $U$ is faithful, but not full. Note that for every groups $(A,\cdot )$ and $(B,\cdot )$, the map$$U_{A,B}:{\mathrm{Hom}}_{\mathbf{Grp}}(A,B)\to {\mathrm{Hom}}_{\mathbf{Set}}(U(A),U(B)),U_{A,B}(f)=U(f)=f$$is injective, but not surjective. If $U_{A,B}$ were surjective, then for every function $\phi :U(A)\to U(B)$, there would exist a group homomorphism $f:A\to B$ such that $f=U_{A,B}(f)=\phi$, which is false. For instance, any function between two groups which does not preserve the identity element cannot be a group homomorphism.

(3) The inclusion functor $i:\mathbf{Ring}\to \mathbf{Rng}$ is clearly faithful, but not full. Note that there are non-unitary homomorphisms in $\mathbf{Rng}$ between unitary rings, and thus they are not arrows in $\mathbf{Ring}$. For instance, if $R$ is any unitary ring, then $\left|{\mathrm{Hom}}_{\mathbf{Ring}}(\mathbb{Z},R)\right|=1$, but $\left|{\mathrm{Hom}}_{\mathbf{Rng}}(\mathbb{Z},R)\right|>1$ (the trivial function $\mathbb{Z}\to R$ is a non-unitary homomorphisms in $\mathbf{Rng}$), hence the map$$i_{\mathbb{Z},R}:{\mathrm{Hom}}_{\mathbf{Ring}}(\mathbb{Z},R)\to {\mathrm{Hom}}_{\mathbf{Rng}}(i(\mathbb{Z}),i(R)),i_{\mathbb{Z},R}(f)=i(f)=f$$cannot be surjective. The category $\mathbf{Ring}$ is a subcategory of $\mathbf{Rng}$, but not a full subcategory.

(4) Let $(M,\cdot )$ and $(N,\cdot )$ be monoids, and view them as categories with one object. Note that a functor between monoid categories is just a monoid homomorphism. Moreover, a faithful functor is the same as an injective monoid homomorphism, while a full functor is the same as a surjective monoid homomorphism. These offer immediate examples of functors having both, one or none of the properties of being faithful or full. For instance, every surjective monoid homomorphism which is not injective gives an example of a full functor which is not faithful.

(5) Let $\mathcal{C}$ and $\mathcal{D}$ be locally small categories, and let $D$ be an object of $\mathcal{D}$. The constant functor $F:\mathcal{C}\to \mathcal{D}$ determined by $D$ is defined on objects by $F(C)=D$, and on arrows by $F(f)=1_D$. In general, the constant functor is neither faithful, nor full (if $\mathcal{C}$ and $\mathcal{D}$ have each at least two arrows).

引理 5.1.4. Let $F:\mathcal{C}\to \mathcal{D}$ be a fully faithful functor. Then $F$ is “essentially injective” on objects in the sense that $F(A)\cong F(B)$ implies $A\cong B$.

证明. Let $g:F(A)\to F(B)$ be an isomorphism. Consider $g^{-1}:F(B)\to F(A)$. Since $F$ is fully faithful, there are (unique) arrows $f:A\to B$ such that $F(f)=g$ and $f^{\prime }:B\to A$ such that $F\left(f^{\prime }\right)=g^{-1}$. We have $F\left(f\circ f^{\prime }\right)=F(f)\circ F\left(f^{\prime }\right)=g\circ g^{-1}=1_{F(B)}=F\left(1_B\right)$, which implies that $f\circ f^{\prime }=1_B$, because $F$ is fully faithful. Similarly, we have $f^{\prime }\circ f=1_A$. Hence $f:A\to B$ is an isomorphism. Thus, $F$ is essentially injective on objects.

$\square$

注 5.1.5. (1) A fully faithful functor need not be injective on objects. For instance, let $\mathcal{C}$ be a category with two objects $C_1$ and $C_2$, and two non-trivial morphisms $f:C_1\to C_2$ and $g:C_2\to C_1$ such that $g\circ f=1_{C_1}$ and $f\circ g=1_{C_2}$, and let $\mathcal{D}$ be a category with one object $D$ and one morphism $1_D$. Then the functor $F:\mathcal{C}\to \mathcal{D}$ given by $F\left(C_1\right)=F\left(C_2\right)=D$ and $F(f)=1_D$ is fully faithful, but it is not injective on objects.

(2) A fully faithful functor need not be “essentially surjective” on objects in the sense that for every object $D$ of $\mathcal{D}$, there is some object $C$ of $\mathcal{C}$ such that $F(C)\cong D$. For instance, the inclusion functor $i:\mathbf{Ab}\to \mathbf{Grp}$ is fully faithful, but not essentially surjective on objects.

定义 5.1.6. Let $\mathcal{C}$ be a locally small category. An object $C$ of $\mathcal{C}$ is called a:

(1)	generator of $\mathcal{C}$ if the covariant functor ${\mathrm{Hom}}_{\mathcal{C}}(C,-):\mathcal{C}\to$ Set is faithful.
(2)	cogenerator of $\mathcal{C}$ if the contravariant functor ${\mathrm{Hom}}_{\mathcal{C}}(-,C):\mathcal{C}\to$ Set is faithful.

命题 5.1.7. Let $\mathcal{C}$ be a locally small category, and let $C$ be an object of $\mathcal{C}$. Then:

(1)	$C$ is a generator of $\mathcal{C}$ if and only if for every arrows $f,g:X\to Y$ in $\mathcal{C}$ with $f\ne g$, there is an arrow $c:C\to X$ in $\mathcal{C}$ such that $f\circ c\ne g\circ c$.
(2)	$C$ is a cogenerator of $\mathcal{C}$ if and only if for every arrows $f,g:X\to Y$ in $\mathcal{C}$ with $f\ne g$, there is an arrow $c:Y\to C$ in $\mathcal{C}$ such that $c\circ f\ne c\circ g$.

证明. (1) The object $C$ is a generator of $\mathcal{C}$ if and only if the functor ${\mathrm{Hom}}_{\mathcal{C}}(C,-):\mathcal{C}\to \mathbf{Set}$ is faithful if and only if for every arrows $f,g:X\to Y$ in $\mathcal{C}$ with $f\ne g$, we have ${\mathrm{Hom}}_{\mathcal{C}}(C,f)\ne {\mathrm{Hom}}_{\mathcal{C}}(C,g)$ if and only if for every arrows $f,g:X\to Y$ in $\mathcal{C}$ with $f\ne g$, there is an arrow $c:C\to X$ in $\mathcal{C}$ such that ${\mathrm{Hom}}_{\mathcal{C}}(C,f)(c)\ne {\mathrm{Hom}}_{\mathcal{C}}(C,g)(c)$ if and only if for every arrows $f,g:X\to Y$ in $\mathcal{C}$ with $f\ne g$, there is an arrow $c:C\to X$ in $\mathcal{C}$ such that $f\circ c\ne g\circ c$.

$\square$

例 5.1.8. (1) The terminal object $1$ is a generator of $\mathbf{Set}$. Recall that $1$ is any single element set, say $\{a\}$. Let $X,Y$ be sets and $f,g:X\to Y$ functions such that $f\ne g$. Then there is $x_0\in X$ such that $f\left(x_0\right)\ne g\left(x_0\right)$. We look for a function $c:\{a\}\to X$ such that $f\circ c\ne g\circ c$. We have $f\circ c\ne g\circ c$ if and only if $f(c(a))\ne g(c(a))$. Just take $c(a)=x_0$. Actually, by a similar reasoning, it is easy to see that any non-empty set is a generator of $\mathbf{Set}$.

Every set $C$ with $|C|⩾2$ is a cogenerator of $\mathbf{Set}$. Let $a,b\in C$. Let $X,Y$ be sets and $f,g:X\to Y$ functions such that $f\ne g$. Then there is $x_0\in X$ such that $f\left(x_0\right)\ne g\left(x_0\right)$, and thus $|Y|⩾2$. We look for a function $c:Y\to C$ such that $c\circ f\ne c\circ g$. We have $c\circ f\ne c\circ g$ if and only if there is $x\in X$ such that $c(f(x))\ne c(g(x))$. We may define $c:Y\to C$ by $c\left(f\left(x_0\right)\right)=a$ and $c(y)=b$ for every $y\ne f\left(x_0\right)$. In particular, $c\left(g\left(x_0\right)\right)=b$ and $c\left(f\left(x_0\right)\right)=a\ne b=c\left(g\left(x_0\right)\right)$. Hence $C$ is a cogenerator of $\mathbf{Set}$ by the dual of Proposition 5.1.7.

(2) The field $K$ is a generator of the category $\mathbf{Vect}(K)$. We show that the covariant functor $F={\mathrm{Hom}}_K(K,-):\mathbf{Vect}(K)\to \mathbf{Set}$ is faithful. We claim that$$F(V)={\mathrm{Hom}}_K(K,V)\cong V,\forall V\in \mathbf{Vect}(K).$$To this end, define $\phi :{\mathrm{Hom}}_K(K,V)\to V$ by $\phi (f)=f(1)$, and $\psi :V\to {\mathrm{Hom}}_K(K,V)$ by $\psi (v)=t_v$, where $t_v:K\to V$ is given by $t_v(k)=kv$. Then $\phi$ and $\psi$ are $K$-linear maps. We have $\phi (\psi (v))=\phi \left(t_v\right)=t_v(1)=v$ for every $v\in V$, hence $\phi \circ \psi =1_V$. Also, we have $\psi (\phi (f))=\psi (f(1))=t_{f(1)}=f$ for every $f\in {\mathrm{Hom}}_K(K,V)$, hence $\psi \circ \phi =1_{{\mathrm{Hom}}_K(K,V)}$. Therefore, $\phi$ is a $K$-isomorphism. Now it is clear that for every $K$-vector spaces $A$ and $B$, the map$$F_{A,B}:{\mathrm{Hom}}_{\mathbf{Vect}(K)}(A,B)\to {\mathrm{Hom}}_{\mathbf{Set}}(F(A),F(B)),F_{A,B}(f)={\mathrm{Hom}}_K(K,f)$$is injective, hence $F={\mathrm{Hom}}_K(K,-):\mathbf{Vect}(K)\to \mathbf{Set}$ is faithful. Hence $K$ is a generator of $\mathbf{Vect}(K)$.

One may also show that $K$ is a cogenerator of the category $\mathbf{Vect}(K)$.

5.2Naturality

例 5.2.1. Let $\mathcal{C}$ be a category with binary products. For every objects $A,B,C$ of $\mathcal{C}$ we have an isomorphism $h:(A\times B)\times C\cong A\times (B\times C)$. This is an isomorphism regardless of what objects $A,B,C$ are. For every arrow $f:A\to A^{\prime }$ in $\mathcal{C}$, we have the following commutative diagram

where $h_A$ and $h_{A^{\prime }}$ are isomorphisms. In fact, we have an isomorphism between “constructions”:$$(-\times B)\times C\cong -\times (B\times C).$$Actually, this is an “isomorphism of functors”.

定义 5.2.2. Let $F,G:\mathcal{C}\to \mathcal{D}$ be functors. By a natural transformation between $F$ and $G$, denoted by $\nu :F\to G$, we mean a family ${\left({\nu }_C:F(C)\to G(C)\right)}_{C\in \mathcal{C}}$ of arrows in $\mathcal{D}$ such that for every arrow $f:C\to C^{\prime }$ in $\mathcal{C}$ we have the following commutative diagram

hence $G(f)\circ {\nu }_C={\nu }_{C^{\prime }}\circ F(f)$. Here ${\nu }_C$ is called the component of $\nu$ at $C$.

例 5.2.3. Let $\mathcal{C}$ be the category $\mathbf{Set}$. Consider the functor $F=\times :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$ defined on objects $(A,B)$ of $\mathcal{C}\times \mathcal{C}$ by $F(A,B)=A\times B$, and on arrows $(\alpha ,\beta )$ in $\mathcal{C}\times \mathcal{C}$ by $F(\alpha ,\beta )=\alpha \times \beta$. Also, consider the functor $G=\overline{\times }:\mathcal{C}\times \mathcal{C}\to \mathcal{C}$ defined on objects $(A,B)$ of $\mathcal{C}\times \mathcal{C}$ by $G(A,B)=B\times A$, and on arrows $(\alpha ,\beta )$ in $\mathcal{C}\times \mathcal{C}$ by $G(\alpha ,\beta )=\beta \times \alpha$.

Define a “twist” natural transformation whose component at an object $(A,B)$ of $\mathcal{C}\times \mathcal{C}$ is$$t_{(A,B)}:F(A,B)=A\times B\to G(A,B)=B\times A,t_{(A,B)}(a,b)=(b,a).$$Then for every arrow $(\alpha ,\beta )$ in $\mathcal{C}\times \mathcal{C}$ consider the following diagram:

The diagram is commutative, because we have:$$\begin{array}{rl}\left[(\beta \times \alpha )\circ t_{(A,B)}\right](a,b)& =(\beta \times \alpha )(b,a)=(\beta (b),\alpha (a))\\ & =t_{\left(A^{\prime },B^{\prime }\right)}(\alpha (a),\beta (b))\\ & =\left[t_{\left(A^{\prime },B^{\prime }\right)}\circ (\alpha \times \beta )\right](a,b)\end{array}$$for every $(a,b)\in A\times B$. Hence $t:F\to G$ is a natural transformation. Here all $t_{(A,B)}$ are bijections (isomorphisms in $\mathbf{Set}$).

命题 5.2.4. Let $F,G,H:\mathcal{C}\to \mathcal{D}$ be functors. Let $\nu :F\to G$ and $\Phi :G\to H$ be natural transformations. Then the family ${\left({\Phi }_C\circ {\nu }_C:F(C)\to H(C)\right)}_{C\in \mathcal{C}}$ of arrows in $\mathcal{D}$ defines a natural transformation, denoted by $\Phi \circ \nu :F\to H$ and called the composite of the natural transformations $\Phi$ and $\nu$.

证明. For every arrow $f:C\to C^{\prime }$ in $\mathcal{C}$ consider the following diagram:

Since $\nu :F\to G$ and $\Phi :G\to H$ are natural transformations, the two squares are commutative. Then the outer rectangle is also commutative. Hence the family $((\Phi \circ \nu {)}_C={\Phi }_C\circ {\nu }_C:F(C)\to$ $H(C){)}_{C\in \mathcal{C}}$ of arrows in $\mathcal{D}$ defines a natural transformation $\Phi \circ \nu :F\to H$.

$\square$

定义 5.2.5. Let $\mathcal{C}$ and $\mathcal{D}$ be categories. The functor category $\mathrm{Fun}(\mathcal{C},\mathcal{D})$ is defined by:

•	objects: the (covariant) functors $F:\mathcal{C}\to \mathcal{D}$
•	arrows: the natural transformations between functors from $\mathcal{C}$ to $\mathcal{D}$
•	composition: the composite of natural transformations
•	identity arrow: for every object $F$ of $\mathrm{Fun}(\mathcal{C},\mathcal{D})$, the identity arrow is the natural transformation $1_F:F\to F$ with components $(1_F{)}_C=1_{F(C)}:F(C)\to F(C)$ for every object $C$ of $\mathcal{C}$.

定义 5.2.6. A natural isomorphism is a natural transformation which is an isomorphism in $\mathrm{Fun}(\mathcal{C},\mathcal{D})$. We denote a natural isomorphism between functors $F,G:\mathcal{C}\to \mathcal{D}$ by $F\cong G$.

命题 5.2.7. A natural transformation $\alpha :F\to G$ is a natural isomorphism if and only if each component ${\alpha }_C:F(C)\to G(C)$ is an isomorphism.

证明. Let $\alpha :F\to G$ be a natural transformation between two functors $F,G:\mathcal{C}\to \mathcal{D}$.

Suppose that $\alpha :F\to G$ is a natural isomorphism. Then it has an inverse ${\alpha }^{-1}:G\to F$. Hence ${\alpha }^{-1}\circ \alpha =1_F$ and $\alpha \circ {\alpha }^{-1}=1_G$. Then for every object $C$ of $\mathcal{C}$, we have $({\alpha }^{-1}{)}_C\circ {\alpha }_C=({\alpha }^{-1}\circ \alpha {)}_C=1_{F(C)}$ and ${\alpha }_C\circ ({\alpha }^{-1}{)}_C=(\alpha \circ {\alpha }^{-1}{)}_C=1_{G(C)}$, which imply that ${\alpha }_C$ is an isomorphism.

Conversely, suppose that each component ${\alpha }_C:F(C)\to G(C)$ is an isomorphism. For every object $C$ of $\mathcal{C}$ define ${\alpha }_C^{-1}={\left({\alpha }_C\right)}^{-1}$. Let us show that this defines a natural transformation ${\alpha }^{-1}:G\to F$. To this end, let $f:C\to C^{\prime }$ be an arrow in $\mathcal{C}$. We show that the following diagram is commutative:

Since $\alpha :F\to G$ is a natural transformation, we have the commutative diagram:

Then ${\alpha }_{C^{\prime }}\circ F(f)=G(f)\circ {\alpha }_C$, which implies that $F(f)\circ {\left({\alpha }_C\right)}^{-1}={\left({\alpha }_{C^{\prime }}\right)}^{-1}\circ G(f)$, that is, $F(f)\circ {\alpha }_C^{-1}={\alpha }_{C^{\prime }}^{-1}\circ G(f)$. Hence the first square is commutative, and thus ${\alpha }^{-1}:G\to F$ is a natural transformation. For every object $C$ of $\mathcal{C}$ we have $({\alpha }^{-1}\circ \alpha {)}_C={\alpha }_C^{-1}\circ {\alpha }_C=1_{F(C)}$ and $(\alpha \circ {\alpha }^{-1}{)}_C={\alpha }_C\circ {\alpha }_C^{-1}=1_{G(C)}$. Hence ${\alpha }^{-1}\circ \alpha =1_F$ and $\alpha \circ {\alpha }^{-1}=1_G$. This shows that $\alpha$ is a natural isomorphism.

$\square$

例 5.2.8. The twist natural transformation from Example 5.2.3 is a natural isomorphism.

例 5.2.9. Consider the category $\mathbf{Vect}(\mathbb{R})$ of real vector spaces and $\mathbb{R}$-linear maps. For a real vector space $V$ denote $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, which is called the dual space of $V$.

Let $f:V\to W$ be an $\mathbb{R}$-linear map. Then it induces a dual $\mathbb{R}$-linear map $f^{\ast }:W^{\ast }\to V^{\ast }$ defined by $f^{\ast }(A)=A\circ f$. Hence we obtain a contravariant functor$$(-{)}^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(-,\mathbb{R}):\mathbf{Vect}(\mathbb{R}{)}^{\mathrm{op}}\to \mathbf{Vect}(\mathbb{R}).$$

We define a natural transformation $\eta :1_{\mathbf{Vect}(\mathbb{R})}\to (-{)}^{\ast \ast }$, where for every $V\in \mathbf{Vect}(\mathbb{R})$, we have ${\eta }_V:V\to V^{\ast \ast }$ defined by ${\eta }_V(v)=e{v}_v:V^{\ast }\to \mathbb{R}$ and $e{v}_v(A)=A(v)$ for every $A\in V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$.

Let us first show that each ${\eta }_V$ is an arrow in the category $\mathbf{Vect}(\mathbb{R})$, that is, an $\mathbb{R}$-linear map. Let $k_1,k_2\in \mathbb{R}$ and $v_1,v_2\in V$. For every $A\in V^{\ast }$, we have$$\begin{array}{rl}e{v}_{k_1{v}_1+k_2{v}_2}(A)& =A\left(k_1{v}_1+k_2{v}_2\right)=k_{1}A\left(v_1\right)+k_{2}A\left(v_2\right)\\ & =k_{1}e{v}_{v_1}(A)+k_{2}e{v}_{v_2}(A)=\left[k_{1}e{v}_{v_1}+k_{2}e{v}_{v_2}\right](A),\end{array}$$whence we have $e{v}_{k_1{v}_1+k_2{v}_2}=k_{1}e{v}_{v_1}+k_{2}e{v}_{v_2}$. It follows that:$${\eta }_V\left(k_1{v}_1+k_2{v}_2\right)=e{v}_{k_1{v}_1+k_2{v}_2}=k_{1}e{v}_{v_1}+k_{2}e{v}_{v_2}=k_1{\eta }_V\left(v_1\right)+k_2{\eta }_V\left(v_2\right).$$Hence each ${\eta }_V$ is an $\mathbb{R}$-linear map. One may show that ${\eta }_V$ is a $K$-isomorphism if and only if $V$ is finite-dimensional.

Now we check the commutativity of the following diagram:

Let $v\in V$. We have $\left({\eta }_W\circ f\right)(v)=e{v}_{f(v)}:W^{\ast }\to \mathbb{R}$ and $\left(f^{\ast \ast }\circ {\eta }_V\right)(v)=e{v}_v\circ f^{\ast }:W^{\ast }\to \mathbb{R}$. For every $A\in W^{\ast }$ we have $\left(e{v}_v\circ f^{\ast }\right)(A)=e{v}_v(A\circ f)=(A\circ f)(v)=e{v}_{f(v)}(A)$, hence $e{v}_{f(v)}=e{v}_v\circ f^{\ast }$. It follows that ${\eta }_W\circ f=f^{\ast \ast }\circ {\eta }_V$. This shows that $\eta$ is a natural transformation.

5.3Equivalence of Categories

例 5.3.1. Let ${\mathbf{Ord}}_{fin}$ be the category of finite ordinal numbers, whose objects are the sets $0,1,\dots ,n$, where $0=\varnothing$ and $n=\{0,\dots ,n-1\}$, and the arrows are the functions between these sets. For each finite set we select an ordinal $|A|$ that is its cardinal and an isomorphism $A\cong |A|$. Then for each function $f:A\to B$ between finite sets, we have a function $|f|$ by completing the square

In this way, we obtain a functor $F=|-|:{\mathbf{Set}}_{fin}\to {\mathbf{Ord}}_{fin}$. In fact, all the above maps are in ${\mathbf{Set}}_{fin}$. We have the inclusion functor $i:{\mathbf{Ord}}_{fin}\to {\mathbf{Set}}_{fin}$. For every finite set $A$ we have an isomorphism ${\nu }_A:A\to (i\circ F)(A)=i(|A|)$ and the above commutative square implies that $i(|f|)\circ {\nu }_A={\nu }_B\circ f$. Hence we have a natural isomorphism$$\nu :1_{{\mathbf{Set}}_{fin}}\to i\circ F=i\circ |-|$$between the functors $1_{{\mathbf{Set}}_{fin}},i\circ F:{\mathbf{Set}}_{fin}\to {\mathbf{Set}}_{fin}$. Also, we have$$F\circ i=|i(-)|=1_{{\mathbf{Ord}}_{fin}}:{\mathbf{Ord}}_{fin}\to {\mathbf{Ord}}_{fin}$$This is because, for any finite ordinal $n$, $|i(n)|=n$, and we can assume that we take ${\nu }_n=1_n:n\to |i(n)|$, so that also $|i(f)|=f:n\to m$. Note that the two categories are very similar, but they are not the same, and not even isomorphic.

The above example is an equivalence of categories in the following sense.

定义 5.3.2. An isomorphism of categories consists of a pair of functors $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ such that $F\circ G=1_{\mathcal{D}}$ and $G\circ F=1_{\mathcal{C}}$.

An equivalence of categories consists of a pair of functors $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$, and a pair of natural isomorphisms $1_{\mathcal{D}}\cong F\circ G$ and $G\circ F\cong 1_{\mathcal{C}}$. Here $F$ and $G$ are called pseudo-inverses and we write $\mathcal{C}\simeq \mathcal{D}$.

A duality between categories $\mathcal{C}$ and $\mathcal{D}$ is an equivalence between $\mathcal{C}$ and the opposite category ${\mathcal{D}}^{\mathrm{op}}$ of $\mathcal{D}$.

例 5.3.3. (1) Two poset categories are equivalent if and only if the posets are isomorphic.

(2) Let $A$ be a set and let $\mathcal{C}$ be the poset category of the poset $(\mathcal{P}(A),\subseteq )$, where $\mathcal{P}(A)$ is the power set of $A$, which consists of all subsets of $A$. Recall that the only arrows are given by the inclusions. We show that there is an isomorphism of categories between $\mathcal{C}$ and its opposite category ${\mathcal{C}}^{\mathrm{op}}$. This also shows that there is a duality between $\mathcal{C}$ and itself.

Let $F:\mathcal{C}\to {\mathcal{C}}^{\mathrm{op}}$ be the functor defined on objects $X$ of $\mathcal{C}$ by $F(X)=A\setminus X$ (set difference) and on arrows (inclusions) by $F(X\to Y)=(A\setminus X)\to (A\setminus Y)$. The latter is an arrow in ${\mathcal{C}}^{\mathrm{op}}$, because $(A\setminus Y)\to (A\setminus X)$ is an arrow in $\mathcal{C}$, being the inclusion, because $X\to Y$ is an inclusion.

Also, let $G:{\mathcal{C}}^{\mathrm{op}}\to \mathcal{C}$ be the functor defined on objects by $G(X)=A\setminus X$ and on arrows by $G(Y\to X)=(A\setminus Y)\to (A\setminus X)$. The latter is an arrow in $\mathcal{C}$, being inclusion, because $X\to Y$ is an inclusion as $Y\to X$ is an arrow in ${\mathcal{C}}^{\mathrm{op}}$.

It is easy to see that $(G\circ F)(X)=X$ for every object $X$ of $\mathcal{C}$ and $(G\circ F)(X\to Y)=X\to Y$ for every arrow (inclusion) in $\mathcal{C}$. Hence $G\circ F=1_{\mathcal{C}}$. Also, one shows that $F\circ G=1_{{\mathcal{C}}^{\mathrm{op}}}$. Therefore, the categories $\mathcal{C}$ and ${\mathcal{C}}^{\mathrm{op}}$ are isomorphic.

注 5.3.4. (1) The equivalence of categories is the more relevant out of the two concepts, the isomorphism of categories being too restrictive.

(2) Every isomorphism of categories is an equivalence of categories, but not conversely. In Example 5.3.1 we have seen that we have an equivalence of categories ${\mathbf{Ord}}_{fin}\simeq {\mathbf{Set}}_{fin}$ given by the above functors $F$ and $i$. But $i\circ F\ne 1_{{\mathbf{Set}}_{fin}}$, hence the functors $F$ and $i$ do not define an isomorphism of categories.

A useful practical characterization of equivalences of categories is the following one.

定理 5.3.5. Let $F:\mathcal{C}\to \mathcal{D}$ be a functor. Then the following are equivalent:

(1)	$F$ is (part of) an equivalence of categories.
(2)	$F$ is fully faithful, essentially injective on objects and essentially surjective on objects.
(3)	$F$ is fully faithful and essentially surjective on objects.

证明. (1) $⟹$ (2) Suppose that $F$ is an equivalence of categories. Then there is a functor $E:\mathcal{D}\to \mathcal{C}$ and natural isomorphisms $\alpha :1_{\mathcal{D}}\to F\circ E$ and $\beta :E\circ F\to 1_{\mathcal{C}}$. Then for every object $C$ of $\mathcal{C}$, ${\beta }_C:(E\circ F)(C)\to C$ is an isomorphism, and for every arrow $f:C\to C^{\prime }$ in $\mathcal{C}$ we have the following commutative diagram:

We show that $F$ is faithful. Let $f,g$ be arrows in $\mathcal{C}$ such that $F(f)=F(g)$. Then $(E\circ F)(f)=$ $(E\circ F)(g)$, whence ${\beta }_{C^{\prime }}\circ (E\circ F)(f)={\beta }_{C^{\prime }}\circ (E\circ F)(g)$. Using the above commutative square for $f$ and $g$, we have $f\circ {\beta }_C={\beta }_{C^{\prime }}\circ (E\circ F)(f)={\beta }_{C^{\prime }}\circ (E\circ F)(g)=g\circ {\beta }_C$, whence $f=g$, because ${\beta }_C$ is an isomorphism. Hence $F$ is faithful.

We show that $F$ is full. Let $h:F(C)\to F\left(C^{\prime }\right)$ be an arrow in $\mathcal{D}$. We have$$C\stackrel{{\beta }_C^{-1}}{\to }(E\circ F)(C)\stackrel{E(h)}{\to }(E\circ F)\left(C^{\prime }\right)\stackrel{{\beta }_{C^{\prime }}}{\to }{C}^{\prime }.$$Denote $f={\beta }_{C^{\prime }}\circ E(h)\circ {\beta }_C^{-1}$. Then $f\circ {\beta }_C={\beta }_{C^{\prime }}\circ E(h)$, hence the diagram

is commutative. By this diagram and the first diagram, we must have $E(h)=(E\circ F)(f)$. Since $E$ is faithful, we deduce that $h=F(f)$. Hence $F$ is full.

Note that $F$ is essentially injective on objects by Lemma 5.1.4.

Finally, for every object $D$ of $\mathcal{D}$ we have $D\cong (F\circ E)(D)=F(E(D))$ by using the isomorphism ${\alpha }_D$. Hence $F$ is essentially surjective on objects.

(2) $⟹$ (3) This is clear.

(3) $⟹$ (1) Suppose that $F$ is fully faithful and essentially surjective on objects. Then for every object $D$ of $\mathcal{D}$ we may choose an object $C$ of $\mathcal{C}$ (unique up to isomorphism by Lemma 5.1.4) such that $D\cong F(C)$. Hence to every object $D$ of $\mathcal{D}$ we may associate an object $E(D)=C$ (unique up to isomorphism) and an arrow ${\alpha }_D:D\to F(C)=(F\circ E)(D)$ which is an isomorphism.

Let $g:D\to D^{\prime }$ be an arrow in $\mathcal{D}$. Consider the arrow ${\alpha }_{D^{\prime }}\circ g\circ {\alpha }_D^{-1}:(F\circ E)(D)\to (F\circ E)(D^{\prime })$. Since $F$ is fully faithful, there is a unique arrow $f:E(D)\to E\left(D^{\prime }\right)$ such that ${\alpha }_{D^{\prime }}\circ g\circ {\alpha }_D^{-1}=F(f)$. Then we define $E(g)=f$. Let us show that this assignment yields a functor. Let $g:D\to D^{\prime }$ and $g^{\prime }:D^{\prime }\to D^{\prime \prime }$ be arrows in $\mathcal{D}$. Then there are unique arrows $f:E(D)\to E\left(D^{\prime }\right)$ and $f^{\prime }:E(D^{\prime })\to E(D^{\prime \prime })$ such that ${\alpha }_{D^{\prime }}\circ g\circ {\alpha }_D^{-1}=F(f)$ and ${\alpha }_{D^{\prime \prime }}\circ g^{\prime }\circ {\alpha }_{D^{\prime }}^{-1}=F(f^{\prime })$. It follows that:$$\begin{array}{rl}{\alpha }_{D^{\prime \prime }}\circ \left(g^{\prime }\circ g\right)\circ {\alpha }_D^{-1}& =\left({\alpha }_{D^{\prime \prime }}\circ g^{\prime }\circ {\alpha }_{D^{\prime }}^{-1}\right)\circ \left({\alpha }_{D^{\prime }}\circ g\circ {\alpha }_D^{-1}\right)\\ & =F\left(f^{\prime }\right)\circ F(f)\\ & =F\left(f^{\prime }\circ f\right).\end{array}$$Thus the arrow associated to $g^{\prime }\circ g$ is $f^{\prime }\circ f$, that is, $E\left(g^{\prime }\circ g\right)=f^{\prime }\circ f$. Then we have $F\left(E\left(g^{\prime }\circ g\right)\right)=$ $F\left(f^{\prime }\circ f\right)=F\left(E\left(g^{\prime }\right)\circ E(g)\right)$, which implies $E\left(g^{\prime }\circ g\right)=E\left(g^{\prime }\right)\circ E(g)$, because $F$ is faithful. Also, $(F\circ E)\left(1_D\right)={\alpha }_D\circ 1_D\circ {\alpha }_D^{-1}=1_{(F\circ E)(D)}=F\left(1_{E(D)}\right)$. This implies that $E\left(1_D\right)=1_{E(D)}$, because $F$ is faithful. Hence $E:\mathcal{D}\to \mathcal{C}$ is a functor.

We have $(F\circ E)(g)\circ {\alpha }_D=F(f)\circ {\alpha }_D={\alpha }_{D^{\prime }}\circ g\circ {\alpha }_D^{-1}\circ {\alpha }_D={\alpha }_{D^{\prime }}\circ g$, hence the following diagram is commutative:

This shows that we have a natural transformation $\alpha :1_{\mathcal{D}}\to F\circ E$, which is a natural isomorphism, because each component ${\alpha }_D$ is an isomorphism.

In order to construct a natural transformation $\beta :E\circ F\to 1_{\mathcal{C}}$, for every object $C$ of $\mathcal{C}$ consider the isomorphism ${\alpha }_{F(C)}:F(C)\to (F\circ E)(F(C))$. Then ${\alpha }_{F(C)}^{-1}:(F\circ E)(F(C))\to F(C)$. Since $F$ is fully faithful, there is a unique arrow ${\beta }_C:(E\circ F)(C)\to C$ such that ${\alpha }_{F(C)}^{-1}=F\left({\beta }_C\right)$, and a unique arrow ${\gamma }_C:C\to (E\circ F)(C)$ such that ${\alpha }_{F(C)}=F\left({\gamma }_C\right)$. We have$$F\left({\beta }_C\circ {\gamma }_C\right)=F\left({\beta }_C\right)\circ F\left({\gamma }_C\right)={\alpha }_{F(C)}^{-1}\circ {\alpha }_{F(C)}=1_{(F\circ E\circ F)(C)}=F\left(1_{(E\circ F)(C)}\right),$$which implies that ${\beta }_C\circ {\gamma }_C=1_C$, because $F$ is faithful. Similarly, one has ${\gamma }_C\circ {\beta }_C=1_{(E\circ F)(C)}$. Hence each ${\beta }_C$ is an isomorphism.

We show that $\beta$ is a natural transformation. Let $f:C\to C^{\prime }$ be an arrow in $\mathcal{C}$. Since ${\alpha }^{-1}:F\circ E\to 1_{\mathcal{D}}$ is a natural transformation, we have the following commutative diagram:

It follows that $F\left(f\circ {\beta }_C\right)=F(f)\circ {\alpha }_{F(C)}^{-1}={\alpha }_{F\left(C^{\prime }\right)}^{-1}\circ (F\circ E\circ F)(f)=F({\beta }_{C^{\prime }}\circ (E\circ F)(f))$. Since $F$ is faithful, we have $f\circ {\beta }_C={\beta }_{C^{\prime }}\circ (E\circ F)(f)$. Then the following diagram is commutative:

Hence $\beta :E\circ F\to 1_{\mathcal{C}}$ is a natural transformation, which is a natural isomorphism, because each component ${\beta }_C$ is an isomorphism.

Hence $F$ is an equivalence of categories.

$\square$

We have a similar result for isomorphisms of categories.

定理 5.3.6. Let $F:\mathcal{C}\to \mathcal{D}$ be a functor. Then the following are equivalent:

(1)	$F$ is (part of) an isomorphism of categories.
(2)	$F$ is fully faithful, injective on objects and surjective on objects.

证明. We give a direct proof, which is a simplification of the proof for equivalence of categories.

(1) $⟹$ (2) Suppose that $F$ is an isomorphism of categories. Then there is a functor $E:\mathcal{D}\to \mathcal{C}$ such that $F\circ E=1_{\mathcal{D}}$ and $E\circ F=1_{\mathcal{C}}$. If $f,g$ are arrows in $\mathcal{C}$ such that $F(f)=F(g)$, then $(E\circ F)(f)=(E\circ F)(g)$, whence $f=g$. Hence $F$ is faithful. If $h:F(C)\to F\left(C^{\prime }\right)$ is an arrow in $\mathcal{D}$, then $h=F(E(h))$. Hence $F$ is full. If $A,B$ are objects of $\mathcal{C}$ such that $F(A)=F(B)$, then $(E\circ F)(A)=(E\circ F)(B)$, whence $A=B$. Hence $F$ is injective on objects. If $D$ is an object of $\mathcal{D}$, then $D=(F\circ E)(D)=F(E(D))$. Hence $F$ is surjective on objects.

(2) $⟹$ (1) Suppose that $F$ is fully faithful, injective on objects and surjective on objects. Then for every object $D$ of $\mathcal{D}$ we have an object $C$ of $\mathcal{C}$ such that $D=F(C)$. If $C^{\prime }$ is another object of $\mathcal{C}$ such that $D=F\left(C^{\prime }\right)$, then we have $F(C)=F\left(C^{\prime }\right)$, which implies that $C=C^{\prime }$, because $F$ is injective on objects. Hence to every object $D$ of $\mathcal{D}$ we may associate a unique object $E(D)=C$. Now let $g:D\to D^{\prime }$ be an arrow in $\mathcal{D}$. Then $(F\circ E)(g):(F\circ E)(D)\to (F\circ E)\left(D^{\prime }\right)$. Since $F$ is fully faithful, there is a unique arrow $f:E(D)\to E\left(D^{\prime }\right)$ such that $g=F(f)$. Then we define $E(g)=f$. Let us show that $E$ defines a functor. Let $g:D\to D^{\prime }$ and $g^{\prime }:D^{\prime }\to D^{\prime \prime }$ be arrows in $\mathcal{D}$. Then there are unique arrows $f:E(D)\to E\left(D^{\prime }\right)$ and $f^{\prime }:E\left(D^{\prime }\right)\to E\left(D^{\prime \prime }\right)$ such that $g=F(f)$ and $g^{\prime }=F\left(f^{\prime }\right)$. It follows that $g^{\prime }\circ g=F\left(f^{\prime }\right)\circ F(f)=F\left(f^{\prime }\circ f\right)$. Thus the arrow associated to $g^{\prime }\circ g$ is $f^{\prime }\circ f$, that is, $E\left(g^{\prime }\circ g\right)=f^{\prime }\circ f$. Then we have $F\left(E\left(g^{\prime }\circ g\right)\right)=$ $F\left(f^{\prime }\circ f\right)=F\left(E\left(g^{\prime }\right)\circ E(g)\right)$, which implies $E\left(g^{\prime }\circ g\right)=E\left(g^{\prime }\right)\circ E(g)$, because $F$ is faithful. Also, $(F\circ E)\left(1_D\right)=1_D=1_{(F\circ E)(D)}=F\left(1_{E(D)}\right)$. This implies that $E\left(1_D\right)=1_{E(D)}$, because $F$ is faithful. Hence $E:\mathcal{D}\to \mathcal{C}$ is a functor.

Finally, we have $(F\circ E)(g)=F(E(g))=F(f)=g$ for every arrow $g$ in $\mathcal{D}$, and $(E\circ F)(f)=$ $E(F(f))=f$ for every arrow $f$ in $\mathcal{C}$. Hence $F$ is an isomorphism of categories.

$\square$

例 5.3.7. (1) Let $\mathcal{C}$ be a category with two objects $C_1$ and $C_2$, and two non-trivial morphisms $f:C_1\to C_2$ and $g:C_2\to C_1$ such that $g\circ f=1_{C_1}$ and $f\circ g=1_{C_2}$, and let $\mathcal{D}$ be a category with one object $D$ and one morphism $1_D$. Then the functor $F:\mathcal{C}\to \mathcal{D}$ given by $F\left(C_1\right)=F\left(C_2\right)=D$ and $F(f)=1_D$ is fully faithful and essentially surjective on objects. Hence it is an equivalence of categories, but it is not an isomorphism of categories.

(2) Let $U:\mathbf{Vect}(K)\to \mathbf{Ab}$ the forgetful functor. Let us show that $U$ is a faithful functor, which is essentially surjective on objects, but it is not full.

Recall that a $K$-linear map between two $K$-vector spaces is a group homomorphism between the underlying abelian groups which respects the scalar multiplication. So two distinct $K$-linear maps should already be distinct as group homomorphisms of underlying groups and respect the scalar multiplication. Hence $U$ is faithful.

Given the field $K$ and an abelian group $G$, one may define the trivial $K$-vector space ${}_{K}G$ on $G$ by taking $k\cdot g=g$ for every $g\in G$. Then $U\left({}_{K}G\right)=G$, which shows that $U$ is essentially surjective on objects.

Note that the endomorphism ring of the canonical $K$-vector space $K$ is isomorphic to $K$. But in general the endomorphism ring of the underlying abelian group $(K,+)$ is not isomorphic to $(K,+,\cdot )$ (e.g., take $K=\mathbb{C})$. Hence $U$ is not full.

例 5.3.8. Let $\mathbf{FDVect}(K)$ be the category of non-zero finite-dimensional $K$-vector spaces and $K$-linear maps. Note that $\mathbf{FDVect}(K)$ is a full subcategory of the category $\mathbf{Vect}(K)$. Consider also the category $\mathbf{Mat}(K)$ defined as follows:

•	objects: the non-zero natural numbers
•	arrows: the matrices in $M_{m,n}(K)$ for $m,n\in {\mathbb{N}}^{\ast }$
•	composition: the multiplication of matrices
•	identity arrow: the identity matrix

We show that there is an equivalence of categories between $\mathbf{FDVect}(K)$ and $\mathbf{Mat}(K)$.

Let $F:\mathbf{FDVect}(K)\to \mathbf{Mat}(K)$ be defined on objects $V$ of $\mathbf{FDVect}(K)$ by $F(V)={\dim }_K(V)$, and on arrows $f:V\to V^{\prime }$ by $F(f)=[f{]}_{B{B}^{\prime }}$, where $B$ and $B^{\prime }$ are some fixed bases of $V$ and $V^{\prime }$ respectively. In this way we obtain a functor $F$.

We show that $F$ is faithful. Let $f,g:V\to V^{\prime }$ be arrows in $\mathbf{FDVect}(K)$, that is, $K$-linear maps, such that $F(f)=F(g)$. Then $[f{]}_{B{B}^{\prime }}=[g{]}_{B{B}^{\prime }}$. But $K$-linear maps are determined by their matrices, hence $f=g$. Thus, $F$ is faithful.

We show that $F$ is full. Let $A\in M_{m,n}(K)$, that is, an arrow in $\mathbf{Mat}(K)$ for some $m,n\in {\mathbb{N}}^{\ast }$. Consider a $K$-linear map $f:K^n\to K^m$ such that $[f{]}_{B{B}^{\prime }}=A$. Then $F(f)=A$. Thus, $F$ is full.

We show that $F$ is essentially surjective on objects. Let $n\in {\mathbb{N}}^{\ast }$, that is, an object in $\mathbf{Mat}(K)$. Consider the object $V=K^n$ of $\mathbf{FDVect}(K)$. Then $n={\dim }_K(V)=F(V)$. Thus, $F$ is essentially surjective on objects.

Therefore, $F$ is an equivalence of categories by Theorem 5.3.6. Note that $F$ is not an isomorphism of categories.