Counterexample. pp no need to be identity when p≫f=fp \gg f = f [math-000X]

The counterexample in the category of sets is

p:2→2p=notf:2→1f(b)=⋆\begin{align*} &p : 2 \to 2 \\ &p = not \\ &f : 2 \to 1 \\ &f(b) = \star \end{align*}

We have f(p(x))=f(x)f(p(x)) = f(x) for all x:2x : 2, but pp is not identity. Where 1,21, 2 represent the set with 11 and 22 elements, respectively.