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I need help with the following problem:

Let $M$ be a connected smooth manifold. Let $f\colon M\to M$ be a smooth mapping satisfying $f(f(x))=f(x)$ for each $x \in M$. Show that $f(M)$ is an embedded submanifold of $M$.

I'd like to prove that $\operatorname{rank}(df_x)$ is locally constant. If so, $\operatorname{rank}(d_xf)$ would be a constant and then I could solve this problem. But I failed after several tries. Beside, I noticed that $df_x$ is a projection for each $x$ of $f(M)$, and I doubt this conclusion is useful though I'm still confused.

Thanks for your help and advice!

Seirios
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Kanae Shinjo
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    Is this homework? What have you tried? This is a challenging problem, but you need to make some effort and work on it, not just expect us to write you a textbook. – Ted Shifrin Oct 03 '13 at 03:14
  • Thanks. It's not my homework but I've been stuck in it for days. I'd like to prove that the rank(df_x) is locally constant, but failed. Beside, I notice that df_x is a projection for each x of f(M), but Idon't know how to use this conclusion. – Kanae Shinjo Oct 03 '13 at 04:07
  • Interesting! Not to be a doubter, but what makes you think this is true? – Robert Lewis Oct 03 '13 at 06:34
  • This is an exercise in a book. – Kanae Shinjo Oct 03 '13 at 07:11
  • OK, it must be true! ;) But seriously folks, I've had a chance to mull it over and and I can see it at this point ; e.g. let $M = M_1 \times M_2$ and take $f(x,y) = (x_0, y)$ for some $x_0 \in M_1$ etc. – Robert Lewis Oct 03 '13 at 07:24

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