Manifold

Definition: locally Euclidean

locally Euclidean

A space $M$ is locally Euclidean of dimension $n$ if every point $x\in M$ has a neighborhood homeomorphic (to any open subset $U\subseteq {\mathbb{R}}^n$) or (to an open ball ${\mathbb{B}}^n\subseteq {\mathbb{R}}^n$) or (to ${\mathbb{R}}^n$)

They are all equivalent

Definition: manifold

Quote

An n-dimensional topological manifold is a second countable hausdorff space that is locally Euclidean of dimension $n$.

Examples

Definition: Chart and atlas

${\phi }_x$ is a homeomorphism to ${\mathbb{R}}^n$ $(U_x,{\phi }_x)$ is a coordinate chart An atlas is the collection of all coordinate charts $\mathcal{A}=\{(U_x,{\phi }_x)|x\in M\}$

NOTE

This can be imagined as an UV unwrapping process but locally.

Dimension of manifold

Quote

If $m\ne n$, a nonempty space cannot both be an m-manifold and an n-manifold

We assume there is a nonempty space such that there is a point, whose neighborhood is both $m$ and $n$ dimensional. This would mean there is a homeomorphism from $m$-dimensional space to $n$-dimensional space, which leads to contradiction

manifold with boundary (not a manifold)

An n-dimensional manifold with boundary is a second countable hausdorff space in which every point has a neighborhood homeomorphic to an open subset of ${\mathbb{R}}^n$ or ${\mathbb{H}}^n$

Definition: upper half-space

The upper half-space ${\mathbb{H}}^n\subseteq {\mathbb{R}}^n$ is $\mathbb{H}:=\{(x_1,\dots ,x_n\in {\mathbb{R}}^n)|x_n\ge 0\}$

interior point

the point has a neighborhood homeomorphic to ${\mathbb{R}}^n$

boundary point

the point has a neighborhood homeomorphic to ${\mathbb{H}}^n$ AND the image of it must lie on the boundary of ${\mathbb{H}}^n$

Warning

interior points of a manifold is NOT the topological interior of the set boundary points of a manifold is NOT the topological boundary of the set

I hate semantic pollution !!!

Theorem

Quote

No point of a manifold with boundary is both an interior point and a boundary point.

Assume there is such a point $x$. (same strategy as Dimension of manifold) This means this point has a homeomorphism $\phi$ to an open ball in ${\mathbb{R}}^n$ and a homeomorphism $\psi$ to an open subset in ${\mathbb{H}}^n$. $⟹$ there is a homeomorphism from $\phi (x)\to \psi (x)$ Then we remove the point $x$ from $\phi (x)$ and from $\psi (x)$ However, then we would have two sets such that one of them has a hole while the other one does not. There cannot be a homeomorphism between these two sets, which leads to contradiction. (We need homology to prove there is no such homeomorphism)