Let $f: \mathbb{R} \rightarrow \mathbb{R}$. Is it true that $f \circ f = f \iff f \upharpoonright rng(f) = id_{rng(f)}$?

Question

I think the answer is no, because we can take $f(x) = -x$. Then $f \circ f = f$, but $f \upharpoonright rng(f) = - id_{rng(f)} \neq id_{rng(f)}$.

Answer 1

Your example satisfies a similar equation: $f\circ f={\rm id}$.

In fact, $f\circ f=f\iff\forall x:f(f(x))=f(x) \iff\forall y\in{\rm range}(f):f(y)=y$.