Definition of functor through diagram

Question

I am studying commutative algebra and right now I'm learning a bit of category theory. I have encountered many times diagrams (commutative ones as well: e.g. generally to express universal mapping properties). However, I don't quite understand their essence. To exemplify I provide a simple exercise:

(Definition of functor for some context) Let $\mathcal{C},\mathcal{C}'$ be categories. A functor $F:\mathcal{C}\to\mathcal{C}'$ is a rule that assigns an object $A\in\mathcal{C}$ to $F(A)\in \mathcal{C}'$ and a morphism $\alpha:A\to B$ of $\mathcal{C}$ to $F(\alpha):F(A)\to F(B)$ of $\mathcal{C}'$, preserving both composition and identity, i.e.

1. $F(\beta\alpha)=F(\beta)F(\alpha)$ for maps $\alpha:A\to B$ and $\beta:B\to C$;

2. $F(1_A)=1_{F(A)}$ for each object $A\in C$.

Question: Show that property $(1)$ is equivalent to the commutativity of the following diagram:

$$\begin{array} A\text{Hom}_\mathcal{C}(B,C) & \longrightarrow & \text{Hom}_{\mathcal{C}'}(F(B),F(C)) \\ \downarrow & & \downarrow \\ \text{Hom}_\mathcal{C}(A,C) & \longrightarrow & \text{Hom}_{\mathcal{C}'}(F(A),F(C)) \end{array}. $$

Take $\beta\in\text{Hom}_\mathcal{C}(B,C)$.

My question is: what do these arrows mean? Intuitively, my guess is that the top and bottom arrows correspond to the functor $F$ itself. But what do the left and right arrows mean? After understanding this diagram I think I can solve this. Any help? Thanks in advance!

Answer 1

Take a morphism $\alpha: A\rightarrow B$ in $\mathcal{C}$. Then there is an induced morphism

$\alpha^*: \text{Hom}_{\mathcal{C}}(B,C) \rightarrow \text{Hom}_{\mathcal{C}}(A,C)$

for any object $C$, given by $(f: B\rightarrow C)\mapsto (f\circ \alpha: A\rightarrow B\rightarrow C)$ (precomposition).

So in your situation, take $\alpha: A\rightarrow B$ a morphism in $\mathcal{C}$. Then the top and bottom arrows of your diagram are described by application of your functor $F$, and the sides are given by precomposition with $\alpha$ and $F\alpha$ respectively. Then, taking $\beta: B\rightarrow C$, the upper and right hand side give

$(F\alpha)^* \circ F\beta = F\beta \circ F\alpha $

and the left hand and upper side give

$(F \circ \alpha^*)\beta = F(\beta \circ \alpha) $.

Therefore, the commutativity of your diagram is equivalent to your condition 1.

Indeed, if $\mathcal{C}$ is a locally small category, for some fixed object $A$ there is a functor $\text{Hom}_{\mathcal{C}}(-,A)$ from the category $\mathcal{C}$ to the category $\text{Set}$ of sets and functions between them; it sends an object $B$ to the set of morphisms from $B$ to $A$, and a morphism $\alpha: B\rightarrow C$ to the function $\alpha^*$ which is precomposition with $\alpha$.