Characterization of the identity mapping
Task number: 2815
Let \(f: X \to X\) and suppose that for all \(g: X\to X\) it is true that \(f\circ g=g\circ f\). Show that \(f\) must be the identity mapping.
Hint
If \(f\) is not the identity mapping, look for a contradiction concerning \(g\).
Resolution
We first note that the identity mapping \(id\) fulfills \(id\circ g=g=g\circ id\).
We proceed by contradiction: assume that there exist distinct \(x\) and \(y\) such that \(f(x)=y\).
If \(f(y)=y\), we choose \(g\) such that \(g(y)=x\). Then \((f\circ g)(y)=f(g(y))=f(x)=y\), but \((g\circ f)(y)=g(f(y))=g(y)=x\), which is a contradiction.
If \(f(y)\ne y\), we choose \(g\) such that \(g(x)=g(y)=y\). Then \((f\circ g)(x)=f(g(x))=f(y)\ne y\), but \((g\circ f)(x)=g(f(x))=g(y)=y\), which is a contradiction.
