Bijective mapping between finite sets - check my proof Let $m, n \in \mathbb{N}$ and $[m]:= 1, ..., m$, $[n]:= 1, ..., n$ be finite sets. 
Show: "there exists a bijective mapping $f: m\to n$" $\Leftrightarrow m=n$ 
My approach:
"$\Leftarrow$": If $m=n$ is assumed then we simply set $f=id(\cdot)$.
"$\Rightarrow$": As $f$ is bijective we have $\vert f([m])\vert = m \leq n$ and $f^{-1}([n])=n\leq m \Rightarrow m=n$. 
(I am not sure if I have to add that $f([m]) \subseteq [n]$ and $f^{-1}([n]) \subseteq [m]$ as this is just a fact of the definition of functions)
Is it correct?

It's correct.
If you want, you can add in "We have $f[[m]]\subseteq [n]$. As $f[[m]]$ has cardinality $[m]$ (by injectivity) and injects into $[n]$, we have $m\le n$. The other direction is similar and hence we get $m=n$."

Here's an alternative proof that I wrote earlier.
