Question on Hom functors

Suppose we have functors $\text{Hom}(\cdot, A):\mathcal{C}^{\text{op}}\to\mathbf{Set}, \text{Hom}(B,\cdot):\mathcal{C}\to\mathbf{Set}$, I am confused of how to compose these two functors to form the hom functor $\text{Hom}(\cdot,\cdot):\mathcal C\times\mathcal C^{\text{op}}\to\mathbf{Set}.$ Any help is appreciated!

Solution 1:

Let me first answer the question you asked, and then try to explain the relevance of the single variable hom functors.

Answer:

It's not really a composition of the functors, but rather a separate functor. I'll assume you are familiar with the definition of the product category $\newcommand\C{\mathcal{C}}\newcommand\op{\mathrm{op}}\C^\op\times \C$. Then we define the $\newcommand\Hom{\operatorname{Hom}}\Hom_\C$ functor from $\C^\op\times\C\to \newcommand\Set{\mathbf{Set}}\Set$ by $\Hom_\C(X,Y) = \C(X,Y)$, where I'm using $\C(X,Y)$ to denote the set of morphisms from $X$ to $Y$ in $\C$ to distinguish it from the $\Hom_\C$ functor. Then for a morphism in the product category $(f^\op,g) : (X,Y)\to (X',Y')$ where $f:X'\to X$ and $g:Y\to Y'$ are morphisms in $\C$, we define $\Hom_\C(f^\op,g)(\alpha) = g\circ \alpha \circ f$, which makes sense, since $\alpha : X\to Y$ is a morphism, so the composite is $$\newcommand\toby\xrightarrow X'\toby{f}X\toby{\alpha}Y\toby{g}Y'. $$

You can check that this is indeed a functor (preserves identities and compositions).

What do the single variable $\Hom$ functors have to do with it?

Well, in general, suppose we have a purported functor $\newcommand\D{\mathcal{D}}\newcommand\E{\mathcal{E}} G:\C\times \D\to \E$ note that the morphisms in $\C\times \D$ are pairs of the form $(a,b) : (x,y)\to (x',y')$ where $a:x\to x'$ is a morphism in $\C$ and $b:y\to y'$ is a morphism in $\D$. And then note that $$(a,b) = (a,1_{y'})\circ (1_x,b) = (1_{x'},b)\circ (a,1_y).$$

In other words $\C\times \D$ is generated by two sets of commuting morphisms, those with the identity in the first coordinate and those with the identity in the second coordinate. It therefore suffices to check functoriality for these two separate sets of morphisms.

I.e. it suffices to show

1. For all $x\in\C$, $G(x,-):\D\to \E$ is a functor,
2. For all $y\in\D$, $G(-,y):\C\to \E$ is a functor,
3. and finally that $G(a,y')\circ G(x,b)=G(x',b)\circ G(a,y)$ where $a:x\to x'$ and $b:y\to y'$ are as above.

Condition 3 is easy to check for the $\Hom$ functor, so the fact that the single variable $\Hom$ functors are functors tells you that the two variable one is as well.