The outline is actually already given in Topics in DG by Michor. To prove the following:
1.15. Theorem. Let $M$ be a connected manifold and suppose that $f : M → M$ is smooth with $f\circ f = f$. Then the image $f(M)$ of $f$ is a submanifold of $M$.
The author proposes to find an open neighbourhood $U$ of $f(M)$ such that the derivative $T_xf$ has constant rank throughout $U$. Now I understand how this is done. But I don't know how to use the following corollary, as the author instructs, to complete the proof:
Corollary. Let $f : M → N$ be $C^∞$ with $T_xf$ of constant rank $k$ for all $x ∈ M$. Then for each $b ∈ f(M)$ the set $f^{−1}(b) ⊂ M$ is a submanifold of $M$ of dimension $\text{dim}M − k$.
I feel that he hints us to represent $f(M)$ via sets of the form of $f^{-1}(b),\,b\in f(U)=f(M)$, but I don't think it is possible. Am I missing something?
My thoughts: at each $x\in f(M)$, find a chart $(U,\phi)$ of $M$ centered at $x$, then $g: U\to M, g(y) = \phi^{-1}(\phi(f(y))-\phi(y))$ satisfies that $g^{-1}(x)=f(M)\cap U$. So $f(M)\cap U$ is a submanifold of $M$ by the corollary. Now how to get a global result?
PS I don't think this is the right track, since I feel the very purpose of introducing the rank theorem here is to avoid using charts.
PPS my thoughts are totally absurd since $g$ doesn't have the constant rank property any more, in fact it totally vanishes inside $f(M)$.
I'm NOT asking for proof of the corollary, but for how to use the corollary to prove the theorem. Thanks in advance!