I am having trouble with showing graphs are manifolds. I would like to discuss the problem in the following specific example: Show that the following graph is a manifold. $$G_r(f) = \{(x,f(x)) \mid x \in \mathbb{R}^{N-1} \}$$ where $f \in C^1 : \mathbb{R}^{N-1} \rightarrow \mathbb{R}$.
Here is my attempt:
We define a map $g: G_r(f) \rightarrow \mathbb{R}^{N-1} \times \{0\}$ by $(x,y) \rightarrow (x,y-f(x))$. Then I claim that this map is a diffeomorphism and only need to check the last component of this map. Let $F(x,y) = y - f(x)$, then $F(x,y) = 0$ and $D_yF(x,y) = 1$. Now I would like to use implicit function theorem and inverse function theorem to show that $F$ is diffeomorphic. However, I realized that these two theorems only shows that $\exists h: h(x_0) = y_0$ but tells nothing about $F$. Can you give me a hint on this? or if this method is completely wrong?
I think you mean to study the set $$ X=\{(x,f(x)): x\in \mathbb{R}^n\}$$ where $f:\mathbb{R}^n\to \mathbb{R}$. $X$ as a space in $\mathbb{R}^{n+1}$ is as smooth as $f$ is. If $f$ is $\mathscr{C}^r$ then the map $x\mapsto (x,f(x))$ is a $\mathscr{C}^r$ diffeomorphism of $\mathbb{R}^n$ to $X$. To see this, note that the components of the map above are the identity and the map $f$. Both of these components are $\mathscr{C}^r$. This is readily seen to be a bijection by noting that it has inverse $\pi:X\to \mathbb{R}^n$ given by $\pi(x,f(x))=x.$ Moreover, $\pi$ is $\mathscr{C}^\infty$. This exhibits a global $\mathscr{C}^r$ diffeomorphism from $\mathbb{R}^n$ to $X$.
This is unsurprising if you draw the situation with $f(x,y)=x^2+y^2$.