I want to prove the following.
Work in a finite-product category.
Given morphisms $f :X \rightarrow A$ and $g:X \rightarrow B$, the following are equivalent:
- The map $(f,g) : X \rightarrow A \times B$ is an isomorphism.
- $(X,f,g)$ form a terminal cone to $\{A,B\}$
I can do the forward direction, but the backward one seems a bit tricky. Help, anyone?
Proof.
$(\Rightarrow).$ (Some details are omitted.) Suppose $(f,g)$ is an isomorphism. Then for any cone $(X',f',g')$ to $\{A,B\}$, we can build $(f',g') : X' \rightarrow A \times B$, which gives us a map $(f,g)^{-1} \circ (f',g') : X' \rightarrow X$. Suppose we have another map $\varphi : X' \rightarrow X$ such that $f \circ \varphi = f'$ and $g \circ \varphi = g'$. Then $(f \circ \varphi,g \circ \varphi) = (f',g').$ So $(f,g) \circ \varphi = (f',g')$. Hence $\varphi = (f,g)^{-1} \circ (f',g')$.
$(\Leftarrow).$ Since $(X,f,g)$ is terminal, there's a unique morphism $\varphi : A \times B \rightarrow X$ such that $f \circ \varphi = \pi_0$ and $g \circ \varphi = \pi_1$. I claim that $\varphi$ and $(f,g)$ are inverses. We have:
$$(f,g) \circ \varphi = (f \circ \varphi,g \circ \varphi) = (\pi_0,\pi_1) = \mathrm{id}$$
It remains to show that $$\varphi \circ (f,g) = \mathrm{id}.$$
It's not clear to me how to finish this off. Obviously uniqueness of $\varphi$ needs to come into play, but how?
$\mathrm{id}$ is the unique morphism $X\to X$ satisfying $f\circ\mathrm{id} = f$ and $g\circ\mathrm{id} = g$ so ...