Diffeomorhism of manifold

Question

This is one of the exam questions of the previous semester. I have studied these. But I didn't do this. Please show me how to solve this question. Thank you for help

Answer 1

Consider the functions $$\phi\colon\mathbb{R}^{n-1}\to f^{-1}(0)\colon (x^1,\ldots,x^{n-1})\mapsto(x^1,\ldots,x^{n-1},F(x^1,\ldots,x^{n-1}))$$ and $$\psi\colon f^{-1}(0)\to\mathbb{R}^{n-1}\colon(x^1,\ldots,x^n)\mapsto(x^1,\ldots,x^{n-1}).$$

$\psi$ and $\phi$ are both clearly differentiable and $\psi\circ\phi$ is clearly the identity. Now consider $(\phi\circ\psi)(x^1,\ldots,x^n) = (x^1,\ldots,x^{n-1},F(x^1,\ldots,x^{n-1}))$. Since $(x^1,\ldots,x^n)\in f^{-1}(0)$, we have $0=f(x^1,\ldots,x^n) = F(x^1,\ldots,x^{n-1})-x^n$, hence $x^n=F(x^1,\ldots,x^{n-1})$ and $(\phi\circ\psi)(x^1,\ldots,x^n)=(x^1,\ldots,x^n)$.

Thus, indeed $f^{-1}(0)$ is diffeomorphic to $\mathbb{R}^{n-1}$ as a manifold. This also immediately implies that $f^{-1}(0)$ is in fact a manifold.

Answer 2

Define a diffeomorphism $h:R^n\rightarrow R^n$ by $h(x^1,\dots,x^n)=(x^1,\dots,x^{n-1},f(x^1,\dots,x^n))$. Then $f^{-1}(0)$ is mapped to $R^{n-1}=\lbrace (x^1,\dots,x^n)\in R^n|x^n=0 \rbrace\subset R^n$ which is a submanifold. Hence $f^{-1}(0)$ is a manifold with a smooth structure induced by $h$ (and $h$ is defined by $f$) and diffeomorphic to $R^{n-1}$.