Topological Manifolds: Definition and Properties
In this post, I would like to explore some basic properties of Topological manifolds.
As a pre-concept of Differential manifolds, the definition and properties of topological manifolds is quite important. In fact, we will emphasis on what kind of impact the \emph{locally Euclidean} will make on the original topological space.
\section{The Definition of Topological Manifolds}
Let’s first give the definition of \emph{Topological Manifolds}.
\begin{defn}
Suppose that $M$ is a topological space. If
\begin{enumerate}
\item $M$ is Haussdorff. I.e., for any points $p,q$, $p\neq q$, then there are neighbourhoods $U, V$ of $p$ and $q$ respectively, such that $U\cap V=\emptyset$.
\item For each $x\in M$, there is a neighbourhood $U$ of $x$ and an integer $n\geq0$, such that $U$ is homeomorphic to $\R^n$.
\end{enumerate}
Then, we call that $M$ is a \emph{Topological Manifolds}.
\end{defn}
\begin{rem}
condition 2 also called “\emph{Locally Euclidean}”.
\end{rem}
\begin{rem}
In condition 2, the neighborhood can be replaced by \emph{open neiborhood of $M$}.
\end{rem}
\begin{proof}
Since every open neighborhood is also an neibourhood, thus, if we assumed that the existence of open neighborhood $U$ for every $x$, then we also have the existence of the neighborhood of $x$.
On the other hand, suppose that for each $x\in M$, there exists a neighborhood $U$ of $x$, then, by the definition of neighborhood, we must have an open set $V$, such that $x\in V\subset U$. Now, we need the easy verified facts: \emph{If $\phi$ is a homeomorphic between the topological space $M$ and $N$, and $U$ is a subset of $M$, then under the induced topology, $\phi$ is a homeomorphic between $U$ and $\phi(U)\subset N$.} Apply this fact, Since $\phi$ is a homemorphic between $U$ and $\R^n$, and $V\subset U$ is open, we know that $\phi(V)$ is also open in $\R^n$. Noted that $\phi(x)\in\phi(V)$, we can find an open ball $W$, $\phi(x)\in W\subset \phi(V)$, use again that $\phi^{-1}$ is a homeomorphic between $\R^n$ and $U$, we know that $\phi^{-1}(W)$ is open in $U$. It is apparently that $\phi^{-1}(W)\subset V\subset U$, and $\phi$ is a homeomorphic between $\phi^{-1}(W)$ and $W$. Now by the well known fact that any open ball in $\R^n$ is homeomorphic to $\R^n$, from which we complete the proof.
\end{proof}
\begin{rem}
In the difinition, we can require that $U$ in condition 2 is homeomorphic to an open subset of $\R^n$. (Why? see, Boothby p.6)
\end{rem}
\begin{rem}
Condition 1 is independent. Since we have the following example:
\end{rem}
\begin{examp}
Let $X=A_+\cup A_-\cup B$, $X\subset \R^2$, with
\begin{align*}
A_+&=\set{(x,y)|x\geq0,y=1},\\
A_-&=\set{(x,y)|x\geq0,y=-1},\\
B&=\set{(x,y)|x< 0, y=0}.
\end{align*}
Define the topology as follows, for the neighborhood of points other than $(0,\pm1)$, we use the induced topology from $\R^2$; and added $N^\pm_\eps=\set{(x,\pm1)|0\leq x< \eps}\cup\set{(x,0)|-\eps\leq x< 0}$ as neighborhood of $(0,\pm1)$.
It is a directly verification that the above definition do defined a topology for $X$ and it is locally Euclidean. Note that $X$ can’t be Hausdorff, since $(0,\pm1)$ can’t be separated by two non-intersection open set.
\end{examp}
\section{The Properties of Topological Manifolds}
Now, we’ll turn to the properties of topological manifolds.
\begin{prop}
Suppose that $M$ is a Topological Manifolds, then
\begin{enumerate}
\item $M$ is locally connected, i.e., for every point $x\in M$, and any neighborhood $U$ of $x$, there is a open connected set $V$, such that $x\in V\subset U$;
\item Since $M$ is locally connected, then every connected component of $M$ is open;
\item If $M$ is connected, then it is connected by arc, i.e., for any two points $p,q\in M$, there exist an continuous map (called arc or path) $\phi:[0,1]\to M$, such that $\phi(0)=p, \phi(1)=q$;
\item $M$ is locally compact, i.e., for every point $x\in M$, and any neighborhood $U$ of $x$, there is a open set $V$ with $\overline V$ is compact, such that $x\in V\subset \overline V\subset U$;
\item Since every Hausdorff locally compact space is regular, so $M$ is regular. (Recall that a space is regular, if the point and closed set can be separated by their prospectively neighborhoods.)
\emph{The above property is not true for infinite dimensional space}.
\item Manifold may not normal. (Recall that a space is normal if any two closed subset can be separated by their neighborhoods.)
\end{enumerate}
\end{prop}
Another important thing is about the second countability, we always need to assume that $M$ has a countable basis.
\begin{prop}
For any topological manifold, the following properties are equivalent:
\begin{enumerate}
\item Each component of $M$ is $\sigma$-compact(also called compact at infinity), i.e., there exist a family of compact sets $\set{K_m}_{m=1}^\infty$, such that $K_{i-1}\subset K_i^\circ\subset K_{i+1}$ and $M=\cup_{m=1}^\infty K_m$;
\item Each component of $M$ is second-countable;
\item $M$ is metrizable;
\item $M$ is paracompact, i.e., each open covering has a locally finite refinement.
\end{enumerate}
thus, any compact manifold is metrizable.
\end{prop}
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