A misleading commutative diagram

Question

Let $U$ be a set, let $\phi$ be an involutive bijection of $U$ with itself.

Let $A$, $B$ be subsets of $U$.

Consider the commutative diagram $A \overset{\phi}{\leftrightarrow} B$ describing a bijection between $A$ and $B$ (that is $\phi$ restricted to $A\times B$).

This diagram is misleading into thinking that $\phi: A\rightarrow B$ is a self-inverse morphism (that is $\phi\circ\phi=1$), what cannot be true (and even is undefined) if $A\ne B$.

Please advise on how to make a less misleading diagram on this topic.

Answer 1

I think people will not be confused by

$$\require{AMScd} A\stackrel\phi\leftrightarrow B $$

and that to make people think that $\phi^2 = 1$ you would need something like this:

$$\begin{CD} A @>{\phi}>> B \\ @VV{\text{id}_A}V @VV{\text{id}_B}V \\ A@<{\phi}<< B \end{CD} $$

or the analogous triangular diagram obtained by merging the two $B$s (or the two $A$s), or similarly something like this: $$A{{\stackrel\phi\rightarrow}\atop{\stackrel\leftarrow\phi}}B$$

Answer 2

Perhaps it would help to clearly indicate that the two $\phi$'s are really different functions (with different domains), like

$$A{{\stackrel{\phi|_A}\rightarrow}\atop{\stackrel\leftarrow{\phi|_B}}}B$$