Proving that something is a submanifold

Question

How can one prove that a connected set $S$ of a smooth manifold $M$ is a submanifold if there is a smooth map $f \colon M \to S$, such that $f_{|S}=id_S$. Shouldn't there be already a smooth structure on $S$ if you claim that $f$ is a smooth map, or they consider it as a smooth map $f \colon M \to M$? And why connectedness of $S$ is important? I don't have an idea to start with.

Answer 1

I just realize hot to do it, with the help of comments, of course.Since $f \circ f=f$ on $S$, we have that $T_xf \circ T_xf=T_xf$ which means $im T_xf=ker(Id-T_xf)$. From Rank-nullity theorem and fact that rank cannot locally fall, we conclude that the rank of $f$ is locally constant, but since $S$ is connected, it is actually constant on $S$. Having constant rank is an open condition (as it can be easily deduced), so there is a neighborhood of $S$ on which $f$ has constant rank, and now by Constant rank theorem, S is a submanifold.