This is an exercise in Mathematical Analysis by Zorich, in the subsection 12.1.

Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a smooth mapping satisfying condition $f\circ f=f$.

$\quad$a) Show that the set $f(\mathbb{R}^n)$ is a smooth surface in $\mathbb{R}^n$.

$\quad$b) By what property of the mapping $f$ is the dimension of the surface determined?

According to Zorich, the 'smooth surface' here has the same meaning as 'manifold' in Euclidean Space. So to prove this it's necessary to give it an atlas. But how to obtain the local charts?

The only idea in my mind is to perform the differentiation to get the matrix equality: $$f'(f(x))\cdot f'(x)=f'(x)$$ without any progress.

Thank you indeed if you can give me some help!


1 Answer 1


To get you started, here are some observations.

  1. If $x\in f(\mathbb{R}^n)$ then $f(x)=x$. In fact, that is if and only if.

  2. For such $x$, your matrix identity reduces to $f'(x)^2=f'(x)$.

  3. The rank of $f'(x)$ is in general semicontinuous as a function of $x$. You need to establish something more – continuity would give you that it is constant.

  • 2
    $\begingroup$ Thanks a lot, I got it. The continuity is because the trace equals the rank here. $\endgroup$
    – Wei Zhan
    Jul 21, 2013 at 10:54

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