For certain commutative squares involving product projections: when are they pushouts

category-theory

In a category with finite products the following diagram commutes for every $f:a\rightarrow b$.
$\require{AMScd}$

\begin{CD}
a \times b @>{\pi_1}>> b \\
@V{id\times f}VV @VV{f}V \\
a\times c @>{\pi_1}>> c
\end{CD}

(If $a$ is terminal then the square is trivially a pushout.
In the category of sets and functions, if $a$ is empty then the square is a pushout if and only if $f$ is an iso.)

Question: assuming that the square is a pushout, can we deduce anything about $a$ or $f$?

Best Answer

In $\mathsf{Set}$, this square is a pushout if and only if:

  • $a$ is empty and $f$ is a bijection,
  • $a$ is a singleton, or
  • $|a|>1$ and $f$ is surjective.

This condition is sufficiently complicated that I very much doubt you'll be able to find a satisfying generalization to an arbitrary category with finite products.

Edit: Now let's consider your question in the category $\mathsf{Set}^{\text{op}}$. This is equivalent to the question of when the following square is a pullback in $\mathsf{Set}$: $\require{AMScd}$ \begin{CD} c @>{i_2}>> a\sqcup c \\ @V{f}VV @VV{\text{id}\sqcup f}V \\ b @>{i_2}>> a\sqcup b \end{CD} The answer is: Always!

While you can certainly find sufficient general conditions for your square to be a pushout in a category with finite products (certainly whenever $a$ is terminal or $f$ is an iso, the square is a pushout), but the example of $\mathsf{Set}^{\text{op}}$ convinces me that there will be no nontrivial necessary condition in general. That is, assuming the square is a pushout gives no further information about $f$ or $a$ in general.

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