In the category Set of sets and functions, consider the functor F(X) = X \* X where \* is the product (its action on arrows is just F(f) = f \* f). Does this functor preserve pushouts? Or at least pushouts of pairs of epimorpisms?



The "binary" product preserves pushouts? 
General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi. 


It's very easy to see the answer is no for general pushouts: since a coproduct $A + B$ is the pushout of a pair $A \leftarrow 0 \to B$, and since the squaring functor preserves $0$, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, $a^2 + b^2$ is generally not equal to $(a+b)^2$; therefore squaring cannot preserve coproducts. 


But in general, we can say that for the category of sets, given a pair of functions 


$$A \stackrel{f\_1}{\leftarrow} P \stackrel{f\_2}{\to} B,$$


the canonical arrow 


$$\phi: A^2 +\_{P^2} B^2 \to (A +\_P B)^2$$ 


is **monic**. (Lemma to be proved.) Assuming this, suppose for example that $f\_2: P \to B$ is epic. Then the pushout of $f\_2$ along $f\_1$ is also epic (well-known fact); let $A \to C$ denote this epi (so $C$ is shorthand for $A +\_P B$). Then $A^2 \to C^2$ is also epic, and since this obviously factors as 


$$A^2 \to A^2 +\_{P^2} B^2 \stackrel{\phi}{\to} C^2$$ 


we see $\phi$ is also epi. But it is monic by the lemma, hence $\phi$ is an isomorphism, as we wanted to show. 


To prove the lemma, it helps to have a clear picture of how pushouts are formed in $Set$. The pushout $C$ is the set of equivalence classes on $A + B$ where $x \in A + B$ is deemed equivalent to $x' \in A + B$ iff there is a zig-zag path 


$$x = x\_0 \stackrel{f\_{i\_1}}{\leftarrow} p\_0 \stackrel{f\_{i\_2}}{\to} x\_1 \leftarrow \ldots x\_{n-1} \stackrel{f\_{i\_{n-1}}}{\leftarrow} p\_{n-1} \stackrel{f\_{i\_n}}{\to} x\_n = x'$$ 


where for each $k$, either $p\_k$ belongs to $P$ and the arrows out of $p\_k$ alternate between $f\_1$ and $f\_2$, or we are in a "holding pattern" where $p\_k$ belongs to $A$ or $B$ and the two arrows out of $p\_k$ are both identities. Now it is not hard to convince yourself that given $(x, y)$ in $A^2$ or $B^2$, and $(x', y')$ in $A^2$ or $B^2$, if there is a zig-zag path from $x$ to $x'$, and a zig-zag path from $y$ to $y'$, then there is a zig-zag path from $(x, y)$ to $(x', y')$ with respect to the pair of maps 


$$A \times A \stackrel{f\_1 \times f\_1}{\leftarrow} P \times P \stackrel{f\_2 \times f\_2}{\to} B \times B;$$ 


all we do is pair together zig-zag paths in the separate $x$- and $y$-components. (Note that if the zig-zag to get from $y$ to $y'$ is longer than the zig-zag from $x$ to $x'$, we can always insert a holding pattern in the $x$-component so that the zig-zag in the $y$-component can "catch up", i.e., so that the lengths of the zig-zags match up.) But this means precisely that the map $\phi$ is monic. 


**Edit:** In answer to vincenzoml's questions in the comments, I wrote up a proof of the more general desired result which applies to a general $\infty$-pretopos, [here](http://ncatlab.org/toddtrimble/published/Lemma+on+wide+pushouts). 


