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!