# Has every set a universal map with respect to the “squaring functor”?

Exercise $26C$ of Herrlich & Strecker Category theory asks to show the following:

Show that every set has a universal map with respect to the "squaring functor".

Recall that a universal map with respect to an object $B\in \mathcal B$ and a functor $F:\mathcal A\rightarrow \mathcal B$ is a pair $(\eta_B,A_B)$ where $\eta_B:B\rightarrow F(A_B)$ and $A_B$ is an object of $\mathcal A$ such that for any object $A'$ of $\mathcal A$ and any morphism $f:B\rightarrow F(A')$, there exists a unique $\overline{f}:A_B\rightarrow A$ such that $f=F(\overline{f})\circ \eta_B$.

The squaring functor, $()^2:$Set$\rightarrow$ Set, is defined by $(A)^2=A\times A$ and $(f)^2=f\times f$.

I tried to do this with $A_B=B$ and $\eta_B:B\rightarrow B\times B$ given by $\eta_B(b)=(b,b)$, where $B$ is any set, but with no success.

I would like to get hints to prove this exercise.

Thanks

Answer: You should take $A_B=B\sqcup B$ and $\eta_B(b)=((b,1),(b,2))$.
Let X be a set, $f\colon B\to X\times X$ - mapping, such that $f(b)=(f_1(b),f_2(b))$. Then $\overline f\colon B\sqcup B\to X$ defined by $\overline f(b,i)=f_i(b)$ for $i=1,2$. It's easy to check that $(\overline f\times\overline f)\circ\eta_B=f$ and $\overline f$ is a unique map with this property.
Actually, it works in arbitrary category with finite products and coproducts. The main idea is that two morphisms $f_1,f_2\colon a\to b$ determine morphism $a\to b\times b$ as well as morphism $a\sqcup a\to b$.