I was reading definition of surface in differential geometry book which defined as follows
A subset $S\subset \mathbb R^3$ is regular surface if $\forall p\in S $ there is open set in S such that $p\in V $ and $\exists \phi :U\to V\in S$ where U is open set in $\mathbb R^2 $ such that map is surjective , smooth and homeomorphism with image and also $\forall q\in U, dX_q:\mathbb R^2\to \mathbb R^3$ is injective
TO better understand cocept I wanted to know counterexample of regular surface.
Please Help me
Any Help will be appreciated
If $f:\Bbb R\to\Bbb R^2$ is defined by $f(x)=(x^2,x^3)$ then the subset $$S:=f(\Bbb R)\times\Bbb R$$ is a counterexample.
One reason for this is that any tangent vector at $0_{\Bbb R^3}\in S$ must have the first two coordinates equal to zero. This is because if $c:\Bbb R\to \Bbb R^3$ is a smooth curve with value in $S$ such that $c(0)=0_{\Bbb R^3}$ then $c_1^\prime(0)=0$ because $c_1$ is positive, and because $c_2=(c_1)^{3/2}$ you have $c_2^\prime(0)=0$. Hence $$v=c^\prime(0)=(0,0,c_3^\prime(0))$$
But for a surface the set of tangent vectors at a point is a vector space of dimension $2$.
Note that if we define $\phi:\Bbb R^2\to \Bbb R^3$ by $$\phi(x,y)=(f(x),y)$$ then $\phi$ is smooth and is a homeomorphism onto its image $S$, with inverse $\phi^{-1}:(x,y,z)\mapsto (y^{1/3},z$). So $\phi$ has all properties you are looking for, except the last one: $d_0\phi$ is not one-to-one. This shows you why this property is important, which is not clear at first in my opinion.