How to prove a graph is a manifold?

Question

Let $f$ be a $C^k$ map defined in an open set $U\in \mathbb R^n$, I'm trying to prove this is a manifold. I couldn't find a parametrization. I've tried $\varphi(x_1,\ldots,x_n)=(x_1,\ldots,x_n,f(x_1,\ldots,x_n))$, but I couldn't prove this is an homeomorphism. I need help here.

Answer 1

You've got the right parameterization. It's certainly a 1-1 map (why?), and certainly onto (why?), and certainly continuous (why?), and indeed, $C^k$ (why? What's the derivative matrix at $(x_1, \ldots, x_n)$ look like?).

Hint: The only possible remaining part is finding an inverse and proving that the inverse is nice as well. So: here's the inverse. Let $K$ denote the graph of $f$.

$$ p: K \to \Bbb R^n : (x_1, \ldots, x_n, u_{n+1}, \ldots, u_k) \mapsto (x_1, \ldots, x_n). $$

Can you explain why $p$ is continuous, 1-1 (on $K$), and $C^k$?