Apply Implicit Function Theorem to show that no $C^1$ function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ can be one to one near any point of its domain. Repeat the proof by using Inverse Mapping Theorem instead of IFT.
Assume $\frac{\partial f}{\partial x}(a,b)\neq0$ and define $g(x,y)=(f(x,y),y)$. Prove that $g$ is a one-to-one function near the point $(a,b)$.
I know I am supposed to give some of my work first, but I have spent hours on reading the book about IFT and I have no idea at all about this question :( can anyone at least give me some hints or explain the question ?
Assume by contradiction your function is 1-1 in a 'hood (neighborhood ) $N(x,y)$ of some $(x,y)$ in $\mathbb R^2$. Restrict to a compact neighborhood $K(x,y) \subset N(x,y)$. The restriction is then a continuous bijection between compact space $K(x,y)$ and a Hausdorff space $f(K(x,y))$, so it is a homeomorphism between an open set in $\mathbb R^2$ (which you can restrict to a ball in $K(x,y)$) and an open interval in the Real line. But there can be no homeomorphism between a ball in $\mathbb R^2$ and an open interval, for many different reasons, e.g., an open interval has a 1-point cutset, but an open $\mathbb R^2$-ball does not. You can use the premises in your problem together with this corollary to arrive at a contradiction.