Let X be a connected n-dimensional differentiable manifold, and $f : X \longrightarrow X$ a differentiable map satisfying $f \circ f=f$.
(a) Show that there exists an open $U \subset X$ with $f(X)\subset U$ and a $k \in N$ such that $rkDf(p)=k$ for all $p \in U$.
(b) Show that f(X) is a k-dimensional submanifold of X
To a) I do know that since $f \circ f=f$ there must be at least one point, where f(x)=x, but why does this have to hold on an open set? Could it not be, that f sends all points of X to one point $x_0$? And for the rank, I do not know how to start trying to prove this.
To b) I wanted to use the Rank Theorem, which says that for an open neighbourhood U of a point p with $rk(D(h'\circ f \circ h^{-1})(h(p)))=k$ (h, h' being charts of our manifold) for all $q \in U$, there are charts (U'',h'',V'') for $p\in U''$, (U''',h''',V''') with f(p) in U''', such that f(U'') is in U''' and $(h'' \circ f \circ h'''^{-1})(x_1,x_2,...,x_n)=(x_1,x_2,..,x_k,0,...,0)\subset R^n$. But, therefore I need to prove that the rkDf(p)=k we found in a) implies that $rk(D(h'\circ f \circ h^{-1})(h(p)))=k$. I do not know how to get there.
I find it in general very hard to prove things about the rank. What is the proper way to start such a proof?
I think I can help you get started with (a). For starters, for any $p \in f(X)$ we have that $f(p) = p$ by the given property of $f \circ f = f$. Therefore, if we apply the chain rule to $f \circ f = f$, we get that $Df(p) = (Df)(f(p)) \circ Df(p) = Df(p) \circ Df(p)$. From here on you can prove that $rk(Df(p))$ is constant on $f(X)$, and then extend this to an open region in $X$ by using that $X$ is connected.
By the way, I highly suspect we are following the exact same course right now. I'm working on the same questions you posted right now as well, word for word. Shoot me a DM on twitter @Supahsemmie and maybe we can help eachother out!
Sorry if this isn't the right place for this, but I don't see any other ways to contact you right now since my account isn't allowed to place comments yet.