Why this definition for surface works?

Question

A set $M\subset \mathbb{R}^n$ is called surface of dimension $m$ and class $C^k$ when every point $p\in M$ is contained in some open $U\subset R^{n}$ such that $V = U\cap M$ is the image of a parametrization $\phi: V_0\to V$ of dimension $m$ and class $C^k$.

But what is a parametrization?

A parametrization of class $C^k$ and dimension $m$ of a set $V\subset \mathbb{R}^n$ is an imersion $\phi:V_0\to V$ of class $C^k$ which is, at the same time, an homeomorphism of the open $V_0\subset\mathbb{R}^m$ under $V$

But what is an imersion?

An imersion in the open $U\subset\mathbb{R}^m$ in the space $\mathbb{R}^n$ is a differentiable application $f:U\to\mathbb{R}^n$ such that for all $x\in U$, $f'(x):\mathbb{R}^m\to\mathbb{R}^n$ is an injective linear transformation

But why this defines a surface? What do we want to define as a surface and what we don't? My book simply throws these definitions at my face but I don't even understand why this is required to define a surface.

Is there a book that explains better why we should define surfaces this way, and why each piece of the definition is required?

UPDATE; Why a plane is a surface by this definition?

Answer 1

A surface of dimension $m$ is a surface which at every point locally looks like a piece of $\mathbb{R}^m$.

To make this definition rigorous, you need:

• A way of defining what "local" means. In our case, we'll say that a point "locally" has a behavior if there's an open set $U$ around it that has the behavior.
• A formal definition of what it means to be "locally like" $\mathbb{R}^m$. What we mean is that (for each point) there's a map $\phi$ from a neighborhood of $\mathbb{R}^m$ to a neighborhood of your point. The mapping must preserve some structure of the original space $\mathbb{R}^m$, which is what you would need in order to show that the piece of $\mathbb{R}^m$ and the piece of $M$ around your point have similar structure.
• Names for the "neighborhoods" or "pieces" involved. Let's use $p$ to refer to the point on your manifold $M$. We want to say that the point looks locally like a piece of $\mathbb{R}^m$. Let's use $V$ to refer to the piece of $\mathbb{R}^m$ which it resembles. Also, if your manifold $M$ is part of a larger space $\mathbb{R}^{N}$, let's use $U$ to refer to the piece of the ambient space $\mathbb{R}^N$ around your point. Then $V_0 \equiv U\cap M$ will be the piece of the manifold around your point. We want $\phi :V_0\rightarrow V$.
• A definition of what "structure" you want the parametrization ("resemblance") $\phi$ to preserve. Our definition says that it is in $C^k$, it's an immersion, and it's a homeomorphism.
  • It's a homeomorphism, which means that it matches the topological structure of $V_0\subseteq M$ to the topological structure of $V\subseteq \mathbb{R}^m$ in a reversible way. Roughly, this means that points that are close before the mapping are close afterwards, and you can "squish" $V$ into $V_0$ and back without cutting or gluing.
  • It's in $C^k$, which means that the map is not only continuous but differentiable; you can smoothly transform $V$ into $V_0$.
  • It's an immersion, which formally means that it's a function whose derivative is injective. Why do we want this definition? Consider a specific 2 dimensional surface such as a sphere. Every point on this surface has a two dimensional tangent plane made of all its tangent vectors. If $V_0$ is a neighborhood of a point on this sphere, and $V$ is a neighborhood of $\mathbb{R}^m$, a map $\phi$ will be an injective linear map just if the tangent space of the destination point $\phi(p)$ has the same dimension as the tangent space of the original point $p$---no less! If the dimension were less, it would mean that $\phi$ distorts the manifold too much at $p$, e.g. twisting it to a point or a line. See: Embedding, immersion

Drawing pictures of the situation helps, in my experience.

---

Intuitively, a plane is a surface because at every point it looks locally like $\mathbb{R}^2$. (Doesn't it?) Show how at every point $p$ on the plane, you can take an open neighborhood of $\mathbb{R}^2$ and map it onto a piece of the plane around $p$ in a structure-preserving way.