Given a one-t0-one function f that maps M onto an arbitrary set A, prove there is a unique way to make A a manifold s.t. f becomes a diffeomorphism.

Question

I'm really unsure of how to proceed, I've drawn a picture and can understand the general setting but don't know how to actually prove it.

Answer 1

This is not possible if $f$ is not a bijection. This is because, whatever manifold structure $A$ has, a diffeomorphism is, first and foremost, a bijection.

If $f$ IS a bijection though, first define the topology on $A$ by defining a subset $U\subset A$ to be open $\iff$ $f^{-1}(A)$ is open in $M$. Now $f,f^{-1}$ are inverse homeomorphisms with this new topological structure.

Next, if $(U_\alpha,\phi_\alpha)$ is a cover of $M$ by open sets $U$ and homeomorphisms $\phi_\alpha:U_\alpha\to \tilde U\subset \mathbb R^n$ gives the manifold structure on $M$, then define $(V_\alpha,\psi_\alpha)$ as $V_\alpha = f(U_\alpha)$ and define $\psi_\alpha$ to be $\phi_\alpha\circ f^{-1}$ . The $\psi_\alpha$ are homeomorphisms.

Now I will rest and ask you to work. Prove four things:

1. In the second paragraph, prove that $f$ and $f^{-1}$ are homeomorphisms with the topology I just defined on $A$.
2. In the third paragraph, prove that the $(V_\alpha,\psi_\alpha)$ are actually valid charts giving $A$ a manifold structure
3. Prove that $f$ gives a diffeomorphism with this manifold structure on $A$
4. Prove that this is the unique manifold structure that gives a diffeomorphism (First prove that the topology cannot be any different and next prove that the manifold structure cannot either.

Let me know if you have trouble somewhere or need additional help.