If $f^{-1}(j)\leq f(j)$ for all $j$ then $f=f^{-1}$

Question

Let $f$ denote the bijective map from $I=\{1,2,3,4,5\}$ to $I$. If $f^{-1}(j)\leq f(j)$ for all $j$, then $f=f^{-1}$.

I have solved the problem using induction. I was wondering if there are some other ways to prove the result.

Answer 1

Notice that every $j$ is in the form $f^{-1}(k)$ for some $k$. With this in mind, we notice that the hypothesis yields that $f^2(j)\ge j$ for all $j$, because $j=f^{-1}(k)\le f(k)=f^2(j)$. But the only bijective map $\{1,2,\cdots,n\}\to \{1,2,\cdots,n\}$ which dominates $id$ is the identity itself, therefore $f\circ f=id$ q.e.d.