Using the theory on differentiability for the properties of differentiability on surfaces

Almost all properties on the differentiability of maps on surfaces are proved by the analogous properties of differentiable maps between open sets of Euclidean spaces. I point out two of them

1. A parametrization of a surface is differentiable. Here we are saying that the parametrization $X_U\subset{\mathbb R}^2\rightarrow V\subset S$ is differentiable, where $V$ is an open set of a surface $S$. In order to clarify the notation, we stand for $Y$ the above map, and $X:U\rightarrow {\mathbb R}^3$ the parametrization. In fact, $Y$ is noting the restriction of $X$ into the codomain. Because $Y$ arrives to a surface, $Y$ is differentiable if $i\circ Y: U\rightarrow{\mathbb R}^3$ is smooth. But this map is just $X$, which it is smooth because is the second property of a parametrization.
2. The inverse of a parametrization is differentiable. Here we mean $X^{-1}:V\rightarrow U\subset{\mathbb R}^2$ is differentiable. Now $X^{-1}$ is a map whose domain is a surface, in fact, the open set $V$ of $S$, which is indeed a surface. By the definition, we have to prove that $X^{-1}\circ Z$ is smooth for some parametrization of $S$. Here we take $Z=X$. Then $X^{-1}\circ X$ is the identity map on the open set $U$, which is trivially smooth.

Surfaces constructed from curves: cylinders

We have defined some types of surfaces from curves, for example, generalized cylinders. Let $\alpha:I\rightarrow{\mathbb R}^3$ a curve contained in a plane $P$, which we suppose it is the plane $z=0$ and let $a\in {\mathbb R}^3$ be a vector that is not contained in $P$. The cylinder on base $\alpha$ in the direction of $a$ is the set $$S=\{\alpha(s)+ta:s\in I,t\in{\mathbb R}\}.$$ Of course, the parametrization is $$X:I\times{\mathbb R}\rightarrow{\mathbb R}^3, X(s,t)=\alpha(s)+ta.$$ If we prove that $S$ is a surface, it is immediate that $X$ is differentiable, $X_s=\alpha'(s)$, $X_t=a$ and both vectors are independent linearly. The difficulty appears when we want to prove that $X$ is a parametrization. The sets $I\times{\mathbb R}$ and $S$ are open in ${\mathbb R}^2$ and $S$, respectively. Also, it is immediate that $X$ is continuous. It remains to prove that $X$ is biyective and $X^{-1}$ is continuous. Of course, if $\alpha$ is not one-to-one, then $X$ is not, as in the next pictures (here the vector $a$ is $a=(1,1,1)$.

1. For this reason, we suppose two cases: $\alpha:I\rightarrow{\mathbb R}^3$ is an embedding or 
2. $\alpha:{\mathbb R}\rightarrow {\mathbb R}^3$ is a simple closed curve.

In the first case, $X$ is one-to-one. If $(x,y,z)=(\alpha_1(s)+ta_1,\alpha_2(s)+t a_2,ta_3)$, then $t=z/a_3$ and so $$s=\alpha^{-1}(x-\frac{z}{a_3} a_1,y-\frac{z}{a_3} a_2).$$ It is immediate that $X^{-1}$ is continuous.

In the second case, $\alpha$ is an embedding in an interval of length less than $T$, where $T>0$ is the period of $\alpha$. 

We have the next pictures for the simple closed curve $\alpha(s)=(3 \cos (s),\sin(s)+\cos(s)+\cos(2s)$ and $a=(0,0,1)$. 

Surfaces and topology

The surfaces that we are introduced present a variety of possibilities on its topology.

1. There are connected surfaces (sphere) and non-connected surfaces (hyperboloid of two sheets, with two connected components).
2. There are compact surfaces (sphere) and non-compact surfaces (plane).
3. There are closed surfaces (sphere) and non-closed surfaces (a hemisphere).
4. The boundary of the sphere ${\mathbb S}^2$  is the very sphere ${\mathbb S}^2$.
5. Every point of a surface has a neighborhood homeomorphic to ${\mathbb R}^2$. In fact, the coordinate open $V$ of $p\in S$ is homeomorphic to an open set $U\subset{\mathbb R}^2$ via the parametrization $X:U\rightarrow V$. Since $U$ is an open set, there exists a ball $B_r(q)$ around $q=X^{-1}(p)$ with $B_r(q)\subset U$. Then then restriction $$X_{| B_r(q)}:B_r(q)\rightarrow X(B_r(q))$$ is a homeomorphism, being $X(B_r(q))$ an open set of $V$, so, of $S$. This means that $X(B_r(q))$ is an open set around $p$ homeomorphic to ${\mathbb R}^2$. As a consequence, we conclude:
• The interior of a surface is empty.
• The surfaces 'have not boundary point', I mean, for example, the closed hemisphere $T=\{p\in{\mathbb S}^2: z(p)\geq 0\}$ is not a  surface because the above property fails at the points with $z(p)=0$.
• A point is not a surface.
• A surface has a non-countable set of points.
6. If $\phi:{\mathbb R}^3\rightarrow{\mathbb R}^3$ is a diffeomorphism and $S$ is a surface, then $\phi(S)$ is a surface which is a homeomorphic to $S$ thanks to the restriction $\phi_{|S}:S\rightarrow \phi(S)$.
7. An open set of a surface is a surface (proved).
8. Some closed sets of a surface are surfaces; other not. For example, if $S$ is the union of two disjoint spheres, then each sphere is closed and it is a surface. On the other hand, the closed hemisphere is closed in ${\mathbb S}^2$ and it is not a closed set.

Curves-maps; surfaces-sets

I remarked in the classroom the differences between the definition of a curve and a surface: a curve is a differentiable map and a surface is a subset of Euclidean space where there do exist parametrizations. I return again with it.

If a curve $\alpha:I\rightarrow{\mathbb R}^3$ is regular $t_0$, then $\alpha'(t_0)\not=0$. If we write in terms of the differential map of $\alpha$, it means that $(d\alpha)_t:{\mathbb R}\rightarrow {\mathbb R}^3$ is a non-zero linear map. This is equivalent to say that $\mbox{rank}(d\alpha)_t=1$, because $$(d\alpha)_{t_0}(1)=\frac{d}{ds}{\Big |}_{s=0}\alpha(t_0+s)=(x'(t_0),y'(t_0),z'(t_0))\not=(0,0,0).$$ Thus the rank  of $(d\alpha)_t$ is the maximum possible (it would be $0$ or $1$). Furthermore, by using the inverse function theorem, ``the curve is a graph locally around $t_0$''. In fact, it was proved that there exists $\epsilon>0$ such that $$\alpha:J=(t_0-\epsilon,t_0+\epsilon)\rightarrow \alpha(t_0-\epsilon,t_0+\epsilon)$$ coincides with the graph of a function, that is, there exists a differentiable function $f:K\subset {\mathbb R}\rightarrow {\mathbb R}$ such that $\{(x,f(x)):x\in K\}=\alpha(J)$. As a consequence, $\alpha:J\rightarrow\alpha(J)$ is homeomorphic to an interval of ${\mathbb R}$.

Then the map $\alpha$ would play (almost) the same role of parametrizations in a surface. If we want to give the definition of the  analogous $1$-dimensional case of a surface, then a subset $C\subset{\mathbb R}^3$ is a $1$-surface (=curve) if for each point $p\in C$ there exists $I\subset {\mathbb R}$ and a map $X:I\rightarrow V\subset C$ a homeomorphism, where $V$ is an open of $C$ around $p$, $X:I\rightarrow {\mathbb R}^3$ is differentiable and $X'(t)\not=(0,0,0)$.

The question is the definition of curve given in chapter $1$ is now a $1$-surface, more precisely, if the trace $\alpha(I)$ is a such $1$-surface. Then one would think `yes' by taking around each point $p\in C$ the corresponding restriction of $\alpha$ to the suitable interval $J$. However, the only problem is the following: Is $\alpha(J)$ an open set of $C$? because the other properties have been showed. We find the answer in the curve $\alpha(t)=(\cos(t),\sin(2t))$, $t\in {\mathbb R}$.

This curve self-intersects at the origin. Thus it can not be a $1$-surface because this point has not a neighbourhood which is homeomorphic to ${\mathbb R}$. By the inverse function theorem, around $t_0=0$, $\alpha(J)\cong J$ is a graph, but $\alpha(J)$ is not an open set of $\alpha({\mathbb R})$, which it happens exactly in our example, as one can see in the next picture: the red color line is $\alpha(J)$ is not an open set in $C$.