4. Limits and Colimits

4.1Subobjects

定义 4.1.1. Let $X$ be an object of a category $\mathcal{C}$. A subobject of $X$ is a monomorphism $m:M\to X$.

Given two subobjects $m:M\to X$ and $m^{\prime }:M^{\prime }\to X$, we define an arrow $f:m\to m^{\prime }$ as an arrow $f:M\to M^{\prime }$ such that $m^{\prime }\circ f=m$, that is, the following diagram is commutative:

In this way, we obtain a category whose objects are subobjects of a given object $X$ and whose arrows are defined as above. We denote this category by ${\mathrm{Sub}}_{\mathcal{C}}(X)$.

Suppose that there is an arrow $g:M\to M^{\prime }$ such that $m^{\prime }\circ g=m$. We also have $m^{\prime }\circ f=m$. Then $m^{\prime }\circ f=m^{\prime }\circ g$, which implies $f=g$, because $m^{\prime }$ is a monomorphism. Hence there is a unique arrow between two subobjects.

We may view ${\mathrm{Sub}}_{\mathcal{C}}(X)$ as a poset category. We define the following relation between the objects of ${\mathrm{Sub}}_{\mathcal{C}}(X)$:$$m⩽m^{\prime }⟺\exists f:m\to m^{\prime }$$

We say that $m$ and $m^{\prime }$ are equivalent, and denote it by $m\equiv m^{\prime }$, if they are isomorphic as subobjects, that is, $m⩽m^{\prime }$ and $m^{\prime }⩽m$. This means that there are arrows $f:M\to M^{\prime }$ and $f^{\prime }:M^{\prime }\to M$ such that $m^{\prime }\circ f=m$ and $m\circ f^{\prime }=m^{\prime }$. It follows that $m=m^{\prime }\circ f=m\circ f^{\prime }\circ f$, which implies that $f^{\prime }\circ f=1_M$, because $m$ is a monomorphism. Similarly, one has $f\circ f^{\prime }=1_{M^{\prime }}$. Hence $f$ is an isomorphism and $M\cong M^{\prime }$. This shows that equivalent subobjects have isomorphic domains.

We sometimes abuse notation and language by calling $M$ the subobject when the monomorphism $m:M\to X$ is clear.

It is convenient to pass from the poset $\left({\mathrm{Sub}}_{\mathcal{C}}(X),⩽\right)$ to the poset obtained by factoring out the equivalence relation “$\equiv$”. In this way, a subobject is an equivalence class of monomorphisms under mutual inclusion.

If $m⩽m^{\prime }$, then the arrow $f:M\to M^{\prime }$ such that $m^{\prime }\circ f=m$ is also a monomorphism, hence $M$ is a subobject of $M^{\prime }$. In this way we have a functor$$\mathrm{Sub}(M^{\prime })\to \mathrm{Sub}(X)$$defined by composition by $f$. Note that the composite of monomorphisms is also a monomorphism.

4.2Pullbacks

定义 4.2.1. Let $\mathcal{C}$ be a category and let

be arrows in $\mathcal{C}$ (that is, a cospan). A pullback (or fibered product or cartesian square) of $f$ and $g$ consists of arrows in $\mathcal{C}$

(that is, a span) such that $f\circ p_1=g\circ p_2$ with the following universal mapping property: given any object $Z$ and any arrows $z_1:Z\to A$ and $z_2:Z\to B$ in $\mathcal{C}$ such that $f\circ z_1=g\circ z_2$, there is a unique arrow $u:Z\to P$ such that $p_1\circ u=z_1$ and $p_2\circ u=z_2$, hence the following diagram is commutative:

A pullback of arrows $f:A\to C$ and $g:B\to C$ in $\mathcal{C}$ is denoted by $\left(P,p_1,p_2\right)$ or $A{\times }_{C}B$.

DuallMy, we define pushouts.

定义 4.2.2. Let $\mathcal{C}$ be a category and let

be arrows in $\mathcal{C}$ (that is, a span). A pushout (or fibered coproduct or cocartesian square) of $f$ and $g$ consists of arrows in $\mathcal{C}$

(that is, a cospan) such that $q_1\circ f=q_2\circ g$ with the following universal mapping property: given any object $Z$ and any arrows $z_1:A\to Z$ and $z_2:B\to Z$ in $\mathcal{C}$ such that $z_1\circ f=z_2\circ g$, there is a unique arrow $u:Q\to Z$ such that $u\circ q_1=z_1$ and $u\circ q_2=z_2$, hence the following diagram is commutative:

A pushout of arrows $f:C\to A$ and $g:C\to B$ in $\mathcal{C}$ is denoted by $\left(Q,q_1,q_2\right)$ or $A{\oplus }_{C}B$.

定理 4.2.3. Let $\mathcal{C}$ be a category with binary products and equalizers. Consider the arrows $f:A\to C$ and $g:B\to C$ and the following diagram

where ${\pi }_1$, ${\pi }_2$ are the canonical projections of the product $A\times B$, $e$ is an equalizer of $f\circ {\pi }_1$ and $g\circ {\pi }_2$, and $p_1={\pi }_1\circ e$ and $p_2={\pi }_2\circ e$. Then $\left(E,p_1,p_2\right)$ is a pullback of $f$ and $g$.

证明. Since $e$ is an equalizer of $f\circ {\pi }_1$ and $g\circ {\pi }_2$ we have $f\circ {\pi }_1\circ e=g\circ {\pi }_2\circ e$. It follows that $f\circ p_1=f\circ {\pi }_1\circ e=g\circ {\pi }_2\circ e=g\circ p_2$.

Now let $Z$ be an object and $z_1:Z\to A$ and $z_2:Z\to B$ arrows in $\mathcal{C}$ such that $f\circ z_1=g\circ z_2$. We look for a unique arrow $v:Z\to E$ such that $p_1\circ v=z_1$ and $p_2\circ v=z_2$, that is, the following diagram is commutative:

Consider the following diagram

By universal mapping property of the product $\left(A\times B,{\pi }_1,{\pi }_2\right)$, there is a unique arrow $u:Z\to A\times B$ such that ${\pi }_1\circ u=z_1$ and ${\pi }_2\circ u=z_2$.

We have $f\circ {\pi }_1\circ u=f\circ z_1=g\circ z_2=g\circ {\pi }_2\circ u$. Now consider the following diagram

By universal mapping property of the equalizer $(E,e)$, there is a unique arrow $v:Z\to E$ such that $e\circ v=u$. Then we have $p_1\circ v={\pi }_1\circ e\circ v={\pi }_1\circ u=z_1$ and $p_2\circ v={\pi }_2\circ e\circ v={\pi }_2\circ u=z_2$.

For uniqueness, suppose that there is also an arrow $v^{\prime }:Z\to E$ such that $p_1\circ v^{\prime }=z_1$ and $p_2\circ v^{\prime }=z_2$. It follows that:$$\begin{array}{rl}& {\pi }_1\circ \left(e\circ v^{\prime }\right)=p_1\circ v^{\prime }=z_1=p_1\circ v={\pi }_1\circ (e\circ v),\\ & {\pi }_2\circ \left(e\circ v^{\prime }\right)=p_2\circ v^{\prime }=z_2=p_2\circ v={\pi }_2\circ (e\circ v).\end{array}$$By universal mapping property of the product, we have $e\circ v^{\prime }=e\circ v$, and thus $v^{\prime }=v$, since $e$ is a monomorphism.

$\square$

Dually, we have the following theorem.

定理 4.2.4. Let $\mathcal{C}$ be a category with binary coproducts and coequalizers. Consider the arrows $f:C\to A$ and $g:C\to B$ and the following diagram

where $j_1$, $j_2$ are the canonical injections of the coproduct $A\oplus B$, $q$ is a coequalizer of $j_1\circ f$ and $j_2\circ g$, and $q_1=q\circ j_1$ and $q_2=q\circ j_2$. Then $\left(Q,q_1,q_2\right)$ is a pushout of $f$ and $g$.

定义 4.2.5. We say that a category has pullbacks if every arrows $f:A\to C$ and $g:B\to C$ have a pullback. Dually, we say that a category has pushouts if every arrows $f:C\to A$ and $g:C\to B$ have a pushout.

推论 4.2.6. If a category has binary products and equalizers, then it has pullbacks. Dually, if a category has binary coproducts and coequalizers, then it has pushouts.

例 4.2.7. Let us see that the converses in the above corollary do not hold in general. Let $(G,\cdot )$ be a non-trivial group. View it as a category with one object $G$, with the elements of $G$ as the arrows and the composition given by the product of elements (recall monoid category). This category has pullbacks and pushouts $(x,y\in G$ have pullback and pushout given by $x^{-1}$ and $y^{-1}$ ), but it does not have (binary) products, (binary) coproducts, equalizers or coequalizers.

Next we present some examples of pullbacks and pushouts in certain categories.

The Category $\mathbf{Set}$

(1) Pullbacks. Let $f:A\to C$ and $g:B\to C$ be functions. We construct their pullback by using binary products and equalizers, as described above. Using the above notation, $(E,e)$ is an equalizer of $f\circ {\pi }_1$ and $g\circ {\pi }_2$, where ${\pi }_1:A\times B\to A$ and ${\pi }_2:A\times B\to B$ are the canonical projections defined by ${\pi }_1(a,b)=a$ and ${\pi }_2(a,b)=b$. By the construction of an equalizer in Set we have$$\begin{array}{rl}E& =\left\{(a,b)\in A\times B\mid \left(f\circ {\pi }_1\right)(a,b)=\left(g\circ {\pi }_2\right)(a,b)\right\}\\ & =\{(a,b)\in A\times B\mid f(a)=g(b)\}\end{array}$$and $e:E\to A\times B$ is the inclusion function. By Theorem 4.2.3, $\left(E,p_1,p_2\right)$ is a pullback of $f$ and $g$, where $p_1={\pi }_1\circ e:E\to A$ and $p_2={\pi }_2\circ e:E\to B$ are given by$$\begin{array}{rl}& p_1(a,b)={\pi }_1(e(a,b))={\pi }_1(a,b)=a,\\ & p_2(a,b)={\pi }_2(e(a,b))={\pi }_2(a,b)=b.\end{array}$$

Now let us analyze a particular pullback in $\mathbf{Set}$. Let $f:A\to C$ be a function, $B\subseteq C$ and $i:B\to C$ the inclusion function. Consider the inverse image of $B$ through $f$, that is,$$\stackrel{-1}{f}(B)=\{a\in A\mid f(a)\in B\}$$Also, define $\overline{f}:\stackrel{-1}{f}(B)\to B$ by $\overline{f}=f{|}_{\stackrel{-1}{f}(B)}$, and let $j:f(B)\to A$ be the inclusion function. Let us show that $\left(f(B),j,\overline{f}\right)$ is a pullback of $f$ and $i$.

If $a\in \stackrel{-1}{f}(B)$, then $f(a)\in B$, and thus $\overline{f}(a)\in B$. We have $i\circ \overline{f}=f\circ j$. Now let $z_1:Z\to A$ and $z_2:Z\to B$ be functions such that $f\circ z_1=i\circ z_2$. Define $u:Z\to \stackrel{-1}{f}(B)$ by $u(x)=z_1(x)$. Since $f\left(z_1(x)\right)=i\left(z_2(x)\right)=z_2(x)\in B$, we have $z_1(x)\in \stackrel{-1}{f}(B)$, and thus $u$ is well defined. Then we have $j\circ u=z_1$. Also, for every $x\in Z$ we have$$(\overline{f}\circ u)(x)=\overline{f}\left(z_1(x)\right)=f\left(z_1(x)\right)=i\left(z_2(x)\right)=z_2(x),$$hence $\overline{f}\circ u=z_2$.

For uniqueness, suppose that there is also a function $u^{\prime }:Z\to \stackrel{-1}{f}(B)$ such that $j\circ u^{\prime }=z_1$ and $\overline{f}\circ u^{\prime }=z_2$. Then we have $j\circ u^{\prime }=j\circ u$, hence $u^{\prime }=u$, because $j$ is a monomorphism.

(2) Pushouts. Let $f:C\to A$ and $g:C\to B$ be functions. We construct their pushout by using binary coproducts and coequalizers, as described above. Consider the coproduct of $A$ and $B$, namely $\left(A\bigsqcup B,j_1,j_2\right)$, where $A\bigsqcup B=\{(a,1)\mid a\in A\}\cup \{(b,2)\mid b\in B\}$ is the disjoint union of $A$ and $B$, and $j_1:A\to A\bigsqcup B$ and $j_2:B\to A\bigsqcup B$ are the canonical injections defined by $j_1(a)=(a,1)$ and $j_2(b)=(b,2)$. Consider the functions $j_1\circ f,j_2\circ g:C\to A\bigsqcup B$. As in the construction of a coequalizer in $\mathbf{Set}$, consider the relation $r=(A\bigsqcup B,A\bigsqcup B,R)$, whose graph $R$ contains all pairs $(a,1),(b,2)$ for which there is $c\in C$ such that $f(c)=a$ and $g(c)=b$. Let $\overline{r}=(A\bigsqcup B,A\bigsqcup B,\overline{R})$ be the smallest equivalence relation on $A\bigsqcup B$ containing $r$. Consider the partition $Q=(A\bigsqcup B)/\overline{R}$ and the function $q:A\bigsqcup B\to Q$ defined by $q(x)=\overline{R}⟨x⟩$. Then we know that $(Q,q)$ is a coequalizer of $j_1\circ f$ and $j_2\circ g$. Let $q_1=q\circ j_1:A\to Q$ and $q_2=q\circ j_2:B\to Q$. Then $\left(Q,q_1,q_2\right)$ is a pushout of $f$ and $g$ by Theorem 4.2.4.

Poset Categories

Let $(L,⩽)$ be a lattice, which may be viewed as a poset category. The pullback of a pair of arrows $a⩽c$ and $b⩽c$ is $\left(\inf (a,b),p_1,p_2\right)$, where $p_1:\inf (a,b)\to a$ and $p_2:\inf (a,b)\to b$ are the unique arrows having the given domains and codomains. Note that it coincides with the product of $a$ and $b$!

The pushout of a pair of arrows $c⩽a$ and $c⩽b$ is $\left(\sup (a,b),q_1,q_2\right)$, where $q_1:a\to \sup (a,b)$ and $q_2:b\to \sup (a,b)$ are the unique arrows having the given domains and codomains. Note that it coincides with the coproduct of $a$ and $b$!

If a poset $(L,⩽)$ is not a lattice, two elements of $L$ might not have a pullback or a pushout.

例 4.2.8. The category $\mathbf{Set}$ has pullbacks and pushouts, while poset categories may not have pullbacks and pushouts.

4.3Properties of Pullbacks

One may prove the following property.

命题 4.3.1. Consider the following commutative diagram in a category:

in which the right square is a pullback. Then the left square is a pullback if and only if the outer rectangle is a pullback.

定理 4.3.2. A category $\mathcal{C}$ has finite products and equalizers if and only if $\mathcal{C}$ has pullbacks and terminal object.

证明. Assume that $\mathcal{C}$ has finite products and equalizers. Then we have seen that $\mathcal{C}$ has pullbacks. Also, note that the empty product is the terminal object.

Conversely, assume that $\mathcal{C}$ has pullbacks and terminal object, denoted by $1$. Let $A$ and $B$ be objects of $\mathcal{C}$. Then there are unique arrows $f:A\to 1$ and $g:B\to 1$. Consider their pullback

We show that $\left(P,p_1,p_2\right)$ is a product of $A$ and $B$. Let $f_1:X\to A$ and $f_2:X\to B$ be arrows in $\mathcal{C}$. We have $g_1\circ f_1,g_2\circ f_2:X\to 1$. Since $1$ is a terminal object, we must have $g_1\circ f_1=g_2\circ f_2$. By universal mapping property of the pullback there is a unique arrow $u:X\to P$ such that $p_1\circ u=f_1$ and $p_2\circ u=f_2$. This shows that $\left(P,p_1,p_2\right)$ is a product of $A$ and $B$.

Now let $f,g:A\to B$ be arrows in $\mathcal{C}$. Consider the arrows $\left[\begin{array}{c}f\\ g\end{array}\right]:A\to B\times B$ and $\left[\begin{array}{c}1\\ 1\end{array}\right]:B\to$ $B\times B$. Consider their pullback

We show that $(E,e)$ is an equalizer of $f$ and $g$. We have $\left[\begin{array}{c}f\\ g\end{array}\right]\cdot [e]=\left[\begin{array}{c}1\\ 1\end{array}\right]\cdot [h]$, whence $f\circ e=h=g\circ e$. Now let $Z$ be an object and $z:Z\to A$ an arrow in $\mathcal{C}$ such that $f\circ z=g\circ z$. It follows that$$\left[\begin{array}{c}f\\ g\end{array}\right]\cdot [z]=\left[\begin{array}{c}f\circ z\\ g\circ z\end{array}\right]=\left[\begin{array}{c}f\circ z\\ f\circ z\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right]\cdot [f\circ z].$$By universal mapping property of the pullback there is a unique arrow $u:Z\to E$ such that $e\circ u=z$ and $h\circ u=f\circ z$.

For uniqueness, suppose that there is also an arrow $u^{\prime }:Z\to E$ such that $e\circ u^{\prime }=z$. Then $e\circ u=e\circ u^{\prime }$, whence $\left[\begin{array}{c}f\\ g\end{array}\right]\cdot [e\circ u]=\left[\begin{array}{c}f\\ g\end{array}\right]\cdot \left[e\circ u^{\prime }\right]$. Hence$$\left[\begin{array}{c}h\circ u\\ h\circ u\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right]\cdot [h\circ u]=\left[\begin{array}{c}1\\ 1\end{array}\right]\cdot \left[h\circ u^{\prime }\right]=\left[\begin{array}{c}h\circ u^{\prime }\\ h\circ u^{\prime }\end{array}\right].$$It follows that $h\circ u=h\circ u^{\prime }$, hence $f\circ z=f\circ e\circ u=h\circ u^{\prime }$. Thus we have $e\circ u^{\prime }=z$ and $h\circ u^{\prime }=f\circ z$. By the uniqueness property of the pullback, we must have $u^{\prime }=u$. This shows that $(E,e)$ is an equalizer of $f$ and $g$.

$\square$

Dually, we have the following theorem.

定理 4.3.3. A category $\mathcal{C}$ has finite coproducts and coequalizers if and only if $\mathcal{C}$ has pushouts and initial object.

4.4Limits

定义 4.4.1. Let $\mathcal{J}$ and $\mathcal{C}$ be categories. A diagram of type $\mathcal{J}$ in $\mathcal{C}$ is a functor $D:\mathcal{J}\to \mathcal{C}$. We write the objects of $\mathcal{J}$ by $i,j$ etc. and we call $\mathcal{J}$ the index category. We denote the values of the functor $D$ in the form $D_i,D_j$ etc.

A cone to a diagram $D$ consists of an object $C$ of $\mathcal{C}$ and a family of arrows $c_j:C\to D_j$ in $\mathcal{C}$, one for each object $j\in \mathcal{J}$ such that for every arrow $\alpha :i\to j$ in $\mathcal{J}$, the following triangle is commutative:

A morphism of cones (or arrow of cones) $\nu :\left(C,c_j\right)\to \left(C^{\prime },c_j^{\prime }\right)$ is an arrow in $\mathcal{C}$ from $C$ to $C^{\prime }$ making the following triangles commutative $(j\in \mathcal{J})$:

In this way we obtain a category $\mathrm{Cone}(D)$ with:

•	objects: the cones to $D$
•	arrows: the morphisms of cones to $D$
•	composition: composition of arrows in $\mathcal{C}$, compatible with commutative triangles
•	identity arrow: the identity morphism of a cone

注 4.4.2. We are here thinking of the diagram $D$ as a “picture of $\mathcal{J}$ in $\mathcal{C}$”. A cone to such a diagram $D$ is then imagined as a many-sided pyramid over the “base” $D$, and a morphism of cones is an arrow between the apexes of such pyramids.

定义 4.4.3. A limit for a diagram $D:\mathcal{J}\to \mathcal{C}$ is a terminal object in $\mathrm{Cone}(D)$. A finite limit is a limit for a diagram on a finite index category $\mathcal{J}$.

A limit will be denoted by $p_i:\underset{⟵}{\lim }{D}_j\to D_i$.

命题 4.4.4. The limit is uniquely determined up to isomorphism.

注 4.4.5. Detailing the definition, the limit of a diagram $D$ has the following universal mapping property: given any cone $\left(C,c_j\right)$ to $D$, there is a unique arrow $u:C\to \underset{⟵}{\lim }{D}_j$ such that for all $j$ we have $p_j\circ u=c_j$. Thus, the limiting cone $(\underset{⟵}{\lim }{D}_j,p_j)$ can be thought of as the “closest” cone to the diagram $D$, and indeed any other cone $\left(C,c_j\right)$ comes from it just by composing with an arrow at the vertex, namely $u:C\to \underset{⟵}{\lim }{D}_j$.

例 4.4.6. (1) Let $\mathcal{J}$ be the category with two objects, denoted 1 and 2 , and no nonidentity arrow:$$12$$

A diagram on $\mathcal{J}$ is a functor $D:\mathcal{J}\to \mathcal{C}$, which takes the objects $1$, $2$ into objects $D_1$, $D_2$.

A cone to $D$ is an object $C$ together with arrows $c_1:C\to D_1$ and $c_2:C\to D_2$. Hence it is a span as follows:$$D_1\stackrel{c_1}{⟵}C\stackrel{c_2}{\to }{D}_2$$

A limit for $D$ is a terminal object in $\mathrm{Cone}(D)$, that is, for every cone to $D$ (span)$$D_1\stackrel{c_1^{\prime }}{⟵}{C}^{\prime }\stackrel{c_2^{\prime }}{\to }{D}_2$$there is a unique morphism between the two cones, that is, there is a unique arrow $u:C^{\prime }\to C$ in $\mathcal{C}$ such that $c_1\circ u=c_1^{\prime }$ and $c_2\circ u=c_2^{\prime }$.

(2) Let $\mathcal{J}$ be the category having two objects, denoted $1$ and $2$, and two non-identity arrows as follows:

A diagram on $\mathcal{J}$ is a functor $D:\mathcal{J}\to \mathcal{C}$, which takes an object $i$ into an object $D_i$ for $i=1,2$, and the arrows $\alpha ,\beta :1\to 2$ into arrows $D_{\alpha },D_{\beta }:D_1\to D_2$.

A cone to $D$ is an object $C$ together with arrows $c_1:C\to D_1$ and $c_2:C\to D_2$ such that the following diagram is commutative:

hence $D_{\alpha }\circ c_1=c_2$ and $D_{\beta }\circ c_1=c_2$, and thus $D_{\alpha }\circ c_1=D_{\beta }\circ c_1$.

A limit for $D$ is a terminal object in $\mathrm{Cone}(D)$, that is,

For every object $C^{\prime }$ and arrow $c_1^{\prime }:C^{\prime }\to D_1$ in $\mathcal{C}$, there is a unique arrow $u:C^{\prime }\to C$ such that $c_1\circ u=c_1^{\prime }$.

Hence the limit is an equalizer of $D_{\alpha }$ and $D_{\beta }$.

(3) Let $\mathcal{J}=\varnothing$. Then $\underset{⟵}{\lim }{D}_j\cong 1$ is a terminal object.

(4) A limit on the following finite category $\mathcal{J}$ will be a pullback:

注 4.4.7. Products, equalizers, terminal objects and pullbacks are all examples of finite limits.

注 4.4.8. Dually, one may define the notion of colimit, which will be a limit in the opposite category. A colimit will be denoted by $q_i:D_i\to \underset{⟶}{\lim }{D}_j$. The colimit is uniquely determined up to an isomorphism. Coproducts, coequalizers, initial objects and pushouts are all examples of finite colimits.

One may get the following property and its dual.

定理 4.4.9. A category has all finite limits if and only if it has finite products and equalizers (if and only if it has pullbacks and a terminal object).

定理 4.4.10. A category has all finite colimits if and only if it has finite coproducts and coequalizers (if and only if it has pushouts and an initial object).

4.5Preservation of Limits

定义 4.5.1. A covariant functor $F:\mathcal{C}\to \mathcal{D}$ is said to preserve limits of type $\mathcal{J}$ if whenever $p_j:L\to D_j$ is a limit for a diagram $D:\mathcal{J}\to \mathcal{C}$, $F\left(p_j\right):F(L)\to F\left(D_j\right)$ is a limit for a diagram $F\circ D:\mathcal{J}\to \mathcal{D}$.

Briefly, we write$$F(\underset{⟵}{\lim }{D}_j)\cong \underset{⟵}{\lim }F(D_j).$$

One has the following results.

定理 4.5.2. Let $\mathcal{C}$ be a locally small category and let $A$ be an object of $\mathcal{C}$. Then the covariant representable functor ${\mathrm{Hom}}_{\mathcal{C}}(A,-):\mathcal{C}\to \mathbf{Set}$ preserves all limits, that is,$${\mathrm{Hom}}_{\mathcal{C}}(A,\underset{⟵}{\lim }{D}_j)\cong \underset{⟵}{\lim }{\mathrm{Hom}}_{\mathcal{C}}\left(A,D_j\right).$$

定理 4.5.3. Let $\mathcal{C}$ be a locally small category and let $A$ be an object of $\mathcal{C}$. Then the contravariant representable functor ${\mathrm{Hom}}_{\mathcal{C}}(-,A):\mathcal{C}\to \mathbf{Set}$ maps all colimits to limits, that is,$${\mathrm{Hom}}_{\mathcal{C}}(\underset{⟶}{\lim }{D}_j,A)\cong \underset{⟵}{\lim }{\mathrm{Hom}}_{\mathcal{C}}\left(D_j,A\right).$$