Natural transformations as categorical homotopies I was reading this question with the same title at MathOverflow, which defines natural transformations in the following way:

given two functors $\mathcal F,\mathcal G \colon \mathcal C \to \mathcal D$ a natural transformation is a functor $\varphi \colon \mathcal C \times 2 \to \mathcal D$, where $2$ is the arrow category $0 \to 1$, such that $\varphi(-,0)=\mathcal F$ and $\varphi(-,1)=\mathcal G$.

This relates $\varphi$ to $\mathcal F$ and $\mathcal G$ on objects, but what about arrows? Why don't we also need to specify that $\varphi(-, id_0) = \mathcal F$ and $\varphi(-, id_1) = \mathcal G$?
 A: There are two directions to this proof.
One direction is that given a functor $\varphi: \mathcal C \times 2 \to \mathcal D$, there is a corresponding natural transformation $\varphi(-, 0) \to \varphi(-, 1)$. $\varphi(-, 0)$ is a whole functor $\mathcal C \to \mathcal D$. The action on objects is obvious (simply evaluate $\varphi$ at the pair $(c, 0)$. If you haven't seen this before, the action on morphisms might not be obvious. Morphisms in $\mathcal C \times 2$ are defined to be pairs of morphisms in $\mathcal C$ and $2$, so a priori, $\varphi(f, 0)$ doesn't make any sense. However, it's typical with functors of multiple variables that an object is also shorthand for the identity at that object. That is, $\varphi(f, 0)$ is $\varphi(f, id_0): \varphi(c, 0) \to \varphi(c', 0)$.
Then, the natural transformation $\varphi(-, 0) \to \varphi(-, 1)$ is simply $\alpha_c := \varphi(c, \to)$, where $\to$ is the unique arrow $0 \to 1$ in $2$.

The other direction is that given a natural transformation $\alpha: \mathcal F \to \mathcal G$, there is a corresponding functor $\varphi: \mathcal C \times 2 \to \mathcal D$ such that $\varphi(-, 0) = \mathcal F$ and $\varphi(-, 1) = \mathcal G$. The behavior of $\varphi$ on objects is determined by the conditions that it's equal to the given functors at $0$ and $1$. For example, $\varphi(c, 0) = \mathcal F(c)$.
That leaves the action of $\varphi$ on morphisms. $\varphi(f, \to): \varphi(c, 0) \to \varphi(c', 1)$, i.e. $\mathcal F(c) \to \mathcal G(c')$. The natural choice is then the diagonal of the commutative diagram
$$
\require{AMScd}
\begin{CD}
\mathcal F(c) @>{\mathcal F(f)}>> \mathcal F(c')\\
@V{\alpha_c}VV @VV{\alpha_{c'}}V \\
\mathcal G(c) @>>{\mathcal G(f)}> \mathcal G(c')
\end{CD}
$$

Finally, one should really show that going one direction then the other leaves you where you left off. Once functorality of $\varphi$ and naturality of $\alpha$ are proven, that gives a bijection between functors of that certain form and natural transformations.
