<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>p</mi></mrow><annotation encoding="application/x-tex">p</annotation></semantics></math>p no need to be identity when <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>p</mi><mo>≫</mo><mi>f</mi><mo>=</mo><mi>f</mi></mrow><annotation encoding="application/x-tex">p \gg f = f</annotation></semantics></math>p≫f=f

The counterexample in the category of sets is

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

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