I am going through some notes Physics, Topology, Logic and Computation: A Rosetta Stone, on category theory.

We first define the opposite category:

Given a category $C$, we define the opposite category $C^{op}$ as the category with the same objects as $C$, but a morphism $f:X\rightarrow Y$ in $C^{op}$ is a morphism $f:Y\rightarrow X$ in C, and the composition $gf$ in $C^{op}$ is the composite $fg$ in $C$.

We then define the hom functor:

For any category $C$, the hom functor $$\mathrm{hom}:C^{op}\times C\rightarrow \mathrm{Set}$$ sends any object $(X,Y)\in C^{op}\times C$ to the set $\mathrm{hom}(X,Y)$, and sends any morphism $(f,g)\in C^{op}\times C$ to the function $$\mathrm{hom}(f,g):\mathrm{hom}(X,Y)\rightarrow \mathrm{hom}(X',Y')$$ $$h\rightarrow ghf$$ when $f:X'\rightarrow X$ and $g:Y\rightarrow Y'$ are morphisms in $C$.

My question is, how is $\mathrm{hom}$ a functor? If $f$ maps from $X'\rightarrow X$ and $g$ from $Y\rightarrow Y'$, then shouldn't $\mathrm{hom}(f,g)$ map from $\mathrm{hom}(X',Y)$ to $\mathrm{hom}(X,Y')$? At first I thought this was a typo, but I can't seem to make it work if I switch round $X$ and $X'$.


A morphism from $(X,Y)$ to $(X',Y')$ in $C^{op}\times C$ consists of a morphism $X\to X'$ in $C^{op}$ and a morphism $Y\to Y'$ in $C$, which is the same as a morphism $X'\to X$ in $C$ and a morphism $Y\to Y'$ in $C$. So when they write that $f:X'\to X$, they are thinking of $f$ as a morphism in $C$, rather than $C^{op}$; in $C^{op}$, $f$ is going the expected direction $X\to X'$. When they write $ghf$, they are also computing this composition in $C$ (and you can check that this composition makes sense: in $C$, you have $X'\stackrel{f}\to X\stackrel{h}\to Y\stackrel{h}\to Y'$, so the composition is defined and is a map $X'\to Y'$).


Choose some $h \in \hom(X,Y)$. Then we have functions $f:X' \to X$, $h:X \to Y$, and $g:Y \to Y'$, so the composition $ghf$ maps $X' \to Y'$. Therefore $ghf \in \hom(X',Y')$.

So we have a rule that starts with any morphism $h \in \hom(X,Y)$ and builds from it a morphism $ghf \in \hom(X',Y')$. That means we have a mapping $\hom(X,Y) \to \hom(X',Y')$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.