PlanetMath: properties of a comma category

Let $\mathcal{A},\mathcal{B},\mathcal{C}$ be categories, and $F:\mathcal{A}\to \mathcal{C}$ and $G:\mathcal{B}\to \mathcal{C}$ be functors between them, and $(F\downarrow G)$ the comma category.

Proof. [Sketch of Proof] We first define functors $P$ and $Q$ .

For $P:(F\downarrow G)\to \mathcal{A}$ , set $P(A,B,f):=A$ and $P(x,y):=x$ .
For $Q:(F\downarrow G)\to \mathcal{B}$ , set $Q(A,B,f):=B$ and $Q(x,y):=y$ .

We leave it for the reader to check that $P$ and $Q$ preserve morphism composition and therefore are indeed covariant functors.

Now, given $F\! \circ\! P,G\!\circ\! Q:(F\downarrow G)\to \mathcal{C}$ , we set $$\eta_{(A,B,f)}:=f$$ for every object $(A,B,f)$ in $(F\downarrow G)$ . To see that $\eta$ is natural, start with a morphism $(x,y):(A,B,f)\to (C,D,g)$ . Then the rest of the proof is done by looking at two commutative diagrams and realizing that they are exactly the same:

$\displaystyle \xymatrix @+=1.5cm{F(A) \ar[r]^f \ar[d]_{F(x)} & G(B) \ar[d]^{G(y... ...D,g) \ar[r]_{\eta_{(C,D,g)}} & G\!\circ\! Q(C,D,g) \ar@{}''1'';''2''\vert-{=} }$

where the first diagram comes from the definition of $(x,y)$ . $\qedsymbol$

Proposition 2 If we have another diagram

$\displaystyle \xymatrix @+=2cm{\mathcal{A} \ar[d]_F & \mathcal{D} \ar[r]^S \ar[... ...mathcal{B} \ar[d]^G \\ \mathcal{C} && \mathcal{C} \ar@{=>}''1'' ;''2''_{\tau} }$

Then there is a unique functor $H:\mathcal{D}\to (F\downarrow G)$ with the commutative diagram

$\displaystyle \xymatrix @+=1.5cm{ & \mathcal{D} \ar[d]^H \ar[dl]_R \ar[dr]^S & \\ \mathcal{A} & (F\downarrow G) \ar[r]_Q \ar[l]^P & \mathcal{B} }$

Furthermore, we have the horizontal composition of natural transformations $\tau=\eta\circ 1_H$ :

$\displaystyle \UseAllTwocells \xymatrix @+=3cm{\mathcal{D} \ruppertwocell<5>^{... ...quad \eta}} \rlowertwocell<-4.5>_{\stackrel{}{G\circ Q}}{\omit} & \mathcal{C}}$

Proof. $H:\mathcal{D}\to (F\downarrow G)$ can be obtained as follows:

for any object $X$ in $\mathcal{D}$ , set $H(X):=(R(X),S(X),\tau_X)$ , and
for any morphism $u:X\to Y$ in $\mathcal{D}$ , set $H(u):=(R(u),S(u))$ . This is well-defined because

$\displaystyle \xymatrix @+=2cm{F(R(X)) \ar[d]_{\tau_X} \ar[r]^{F(R(u))} & F(R(Y)) \ar[d]^{\tau_Y} \ G(S(X)) \ar[r]^{G(S(u))} & G(S(Y))}$
is a commutative diagram due to the naturality of $\tau$ .

To see that $H$ is a (covariant) functor, let $u:X\to Y$ and $v:Y\to Z$ be morphisms in $\mathcal{D}$ . Then $H(v\circ u)=(R(v\circ u),S(v\circ u))=(R(v)\circ R(u),S(v)\circ S(u))=(R(v),S(v))\circ (R(u),S(u))=H(v)\circ H(u)$ .

Next, we check the commutativity conditions, which are clear: $P\!\circ\! H(X)=R(X)$ , $P\!\circ\! H(u) = R(u)$ , and $Q\!\circ\! H(X)=S(X)$ , $Q\!\circ\! H(u)=S(u)$ . Finally, $(\eta\circ 1_H)_X = \eta_{H(X)}\circ (F\circ P)((1_H)_X) = \eta_{H(X)}\circ (F\circ P)(1_{H(X)}) = \eta_{H(X)}\circ F(1_{R(X)}) = \eta_{H(X)}\circ 1_{F\circ R(X)} = \tau_X\circ 1_{F\circ R(X)} = \tau_X$ .

This shows the existence of $H$ .

Now we prove the uniqueness of $H$ . Suppose $K:\mathcal{D}\to (F\downarrow G)$ is another such a functor (satisfying the diagram above). This boils down to showing that $K(X)=H(X)$ and $K(u)=H(u)$ for any object $X$ and morphism $u$ in $\mathcal{D}$ .

If $(A,B,f)=K(X)$ , then $A=P(A,B,f)=P(K(X))=R(X)$ and $B=Q(A,B,f)=Q(K(X))=S(X)$ . This shows that $f:F(A)\to G(B)$ is a morphism from $F(R(X)$ to $G(S(X))$ . Finally, $\tau_X = (\eta \circ 1_K)_X = \eta_{K(X)} \circ (F\circ P)((1_K)_X) = f\circ (F\circ P)((1_K)_X) = f\circ (F\circ P)(1_{K(X)}) = f\circ F(1_{P(K(X))}) = f\circ F(1_{R(X)}) = f\circ 1_{F(R(X))} = f$ . Therefore, $K(X)=(A,B,f)=(R(X),S(X),\tau_X)=H(X)$ .
If $(x,y)=K(u)$ , then $x=P(x,y)=P(K(u))=R(u)$ and $y=Q(x,y)=Q(K(u))=S(u)$ . Therefore, $K(u)=(x,y)=(R(u),S(u))=H(u)$ .

This shows that $H$ is unique, and the proof is complete. $\qedsymbol$