Naturality in two variables is equivalent to naturality in each variable

category-theoryfunctorsnatural-transformations

Exercise from Leinster:

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I managed to prove one direction as follows. Suppose the mentioned family is a natural transformation $F\implies G$. Then for all arrows $f:A\to A'$ and $g:B\to B'$, $$\tag{1} \alpha_{A',B'}\circ F(f,g)=G(f,g)\circ\alpha_{A,B}.$$
Then

  • taking $A=A',f=1_A$, we get $$\tag{2}\alpha_{A,B'}\circ F(1_A,g)=G(1_A,g)\circ \alpha_{A,B},$$ i.e., $(\alpha_{A,B}:F^A(B)\to G^A(B))_{B\in\mathscr B}$ is a natural transformation $F^A\implies G^A$;
  • taking $B=B',g=1_B$, we get $$\tag{3}\alpha_{A',B}\circ F(f,1_B)=G(f,1_B)\circ \alpha_{A,B},$$ i.e., $(\alpha_{A,B}:F_B(A)\to G_B(A))_{A\in\mathscr A}$ is a natural transformation $F_B\implies G_B$.

How do I prove the other direction? I don't see how to deduce $(1)$ from $(2)$ and $(3)$.

Let $(f,g)=(1_A',g)\circ (f,1_B)$. Then $$G(f,g)\circ\alpha_{A,B}=G(1_{A'},g)\circ\color{blue}{G(f,1_B)\circ \alpha_{A,B}}=G(1_{A'},g)\circ \color{blue}{\alpha_{A',B}\circ F(f,1_B)}\\= \color{navy}{G(1_{A'},g)\circ \alpha_{A',B}}\circ F(f,1_B)=\color{navy}{\alpha_{A',B'}\circ F(1_{A'},g)}\circ F(f,1_B)=\alpha_{A',B'}\circ F(f,g)$$

The first equality follows because $G$ is a functor.

The second equality follows from $(3)$.

The third equality follows by interchanging colors.

The fourth equality follows from $(2)$ with $A$ replaced by $A'$.

The fifth equality follows because $F$ is a functor.

Best Answer

Hint: In the product category $\mathscr{A}\times \mathscr{B}$, an arbitrary arrow $(f,g)\colon (A,B)\to (A',B')$ can be factored as $(1_{A'},g)\circ (f,1_B)$.

Spelling it out (mouse over to reveal "spoiler"):

\begin{align*} G(f,g) \circ \alpha_{A,B} &= G(1_{A'},g)\circ G(f,1_B)\circ \alpha_{A,B} & \text{by functoriality of $G$}\\ &= G(1_{A'},g)\circ \alpha_{A',B}\circ F(f,1_B) & \text{by naturality of $\alpha_{-,B}$}\\ &= \alpha_{A',B'}\circ F(1_{A'},g) \circ F(f,1_B) & \text{by naturality of $\alpha_{A',-}$}\\ &= \alpha_{A',B'}\circ F(f,g) & \text{by functoriality of $F$} \end{align*}

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