Topological Properties of Manifolds

Munkres Topology

Given a finite $m\in {\mathbb{Z}}^{+}$, a $m$-manifold is a $T2$ second-countable topological space (i.e. Hausdorff with countable basis) such that each point $x$ has a neighborhood homeomorphic with an open set in ${\mathbb{R}}^m$.

We first explain some core terminology. The support of a map $\varphi :X\to \mathbb{R}$ where $X$ is some topological space is defined to be the closure of ${\varphi }^{-1}(\mathbb{R}-\{0\})$. Furthermore, given a finite open cover $\{U_1,U_2,\dots ,U_n\}$ of a space $X$, a family of functions ${\varphi }_i:X\to [0,1]$ ($1\le i\le n$) is said to be a partition of unity dominated by $\{U_i\}$ if the support of ${\varphi }_i$ is a subset of $U_i$ and if ${\sum }_{i=1}^n{\varphi }_i(x)=1$ for all $x\in X$.

Theorem 1) Let $U_1,U_2,\dots ,U_n$ be a finite open covering of a normal space $X$. Then, there exists a partition of unity ${\varphi }_i:X\to [0,1]$ ($1\le i\le n$) dominated by $\{U_i\}$.

Proof) For any finite open cover $\{U_i{\}}_{1\le i\le n}$, we can find a “smaller” open cover $\{V_i{\}}_{1\le i\le n}$ such that $\overline{V_i}\subset U_i$. This is a simple explicit construction: $X-{\bigcup }_{i=2}^n{U}_i$ is a closed subset of $U_1$, and by normality there is some open $V_1$ such that $(X-{\bigcup }_{i=2}^n{U}_i)\subset V_1$ and $\overline{V_1}\subset U_1$. Clearly, the new collection $\{V_1,U_2,\dots ,U_n\}$ is still an open cover of $X$. We do the same for $U_2$, taking $X-(V_1\cup {\bigcup }_{i=3}^n{U}_i)$ and some $V_2$ using normality such that $X-(V_1\cup {\bigcup }_{i=3}^n{U}_i)\subset V_2$ and $\overline{V_2}\subset U_2$.

Using this construction, take a “smaller” cover $\{V_i{\}}_{1\le i\le n}$ of $\{U_i{\}}_{1\le i\le n}$ such that $\overline{V_i}\subset U_i$ and then a “smaller” cover $\{W_i{\}}_{1\le i\le n}$ of $\{V_i{\}}_{1\le i\le n}$ such that $\overline{W_i}\subset V_i$. Using the Urysohn Lemma, we can create continuous functions ${\psi }_i:X\to [0,1]$ such that ${\psi }_i(\overline{W_i})=\{1\}$ and ${\psi }_i(X-V_i)=\{0\}$. Clearly, $\text{support}({\psi }_i)\subset \overline{V_i}\subset U_i$, and ${\psi }_i>0$ on $W_i$. As $\{W_i\}$ is an open cover of $X$, the function $\psi ={\sum }_{i=1}^n{\psi }_i$is a strictly positive continuous function defined on $X$. Taking ${\varphi }_i={\psi }_i/\psi$ gives us our desired partition of unity.

Theorem 2) Every $m$-manifold $X$ is regular and thus metrizable by Urysohn metrization.

Proof) The local Euclidean structure and Hausdorff condition is sufficient to show $X$ is regular; the local ${\mathbb{R}}^m$ condition is particularly powerful as it allows us to use ${\mathbb{R}}^m$ structure in $X$.

Take an arbitrary point $x\in X$ and neighborhood $W$ of $x$; it suffices to show there exists a neighborhood $V\subset W$ of $x$ such that $\overline{V}\subset W$. Let $U_1$ be a neighborhood of $x$ homeomorphic to an open subset of ${\mathbb{R}}^m$, and let $U_2=U_1\cap W$; then, $U_2$ is a neighborhood of $x$ contained in $W$ that is homeomorphic to an open subset of ${\mathbb{R}}^m$. Let $\varphi :U_2\to {\mathbb{R}}^m$ be this imbedding and let $y=\varphi (x)\in {\mathbb{R}}^m$, $V_1=\varphi (U_2)$.

Consider an open ball $V_2$ about $y$ contained in $V_1$ such that $\overline{V_2}\subset V_1$; this is possible as we are working in ${\mathbb{R}}^m$. Then, consider the set $U_3={\varphi }^{-1}(V_2)$, which is a neighborhood about $x$. Note that $\overline{V_2}$ is compact, and thus ${\varphi }^{-1}(\overline{V_2})$ is compact; we can invoke the fact that $X$ is Hausdorff to conclude ${\varphi }^{-1}(\overline{V_2})$ is not only closed in $U_2$ but in the entire space $X$. Thus, $\overline{U_3}\subset {\varphi }^{-1}(\overline{V_2})\subset U_2$, and $U_3$ is precisely the neighborhood $V$ of $x$ that we desired. We are done.

Theorem 3) If $X$ is a compact $m$-manifold, $X$ can be imbedded into some finite-dimensional Euclidean space ${\mathbb{R}}^N$.

Proof) Note that $X$ is Hausdorff and compact, and thus is normal. Using compactness, we can find a finite open cover $\{U_1,U_2,\dots ,U_n\}$ of $X$ such that each $U_i$ is homeomorphic to an open set in ${\mathbb{R}}^m$. Let $g_i:U_i\to {\mathbb{R}}^m$ be the corresponding embedding. Using Theorem 1 and normality, construct a partition of unity $\{{\varphi }_i{\}}_{1\le i\le n}$ dominated by $\{U_i\}$. The core idea is that we can concatenate the $U_i$ “coordinates” to create an embedding of $X$ into Euclidean space, with the ${\varphi }_i$s acting as multipliers to smooth out the function to be continuous.

Define $h_i:X\to {\mathbb{R}}^m$ as $h_i(x)=g_i(x){\varphi }_i(x)$ when $x\in U_i$ and $h_i(x)=(0,0,\dots ,0)$ when $x\in X-\text{support}({\varphi }_i)$. These two parts of the definition agree on the open set $U_i\cap X-\text{support}({\varphi }_i)$ and $h_i$ is continuous when restricted to either part; thus, $h_i$ is continuous. Taking $N=n(m+1)$, we define the map $F:X\to {\mathbb{R}}^N$ as follows, for any $x\in X$:

$$F(x)=({\varphi }_1(x),{\varphi }_2(x),\dots ,{\varphi }_n(x),h_1(x),\dots ,h_n(x))$$

As $F$ is a continuous map from a compact space $X$ to a Hausdorff space ${\mathbb{R}}^n$, to show $F$ is an imbedding it suffices to prove the injectivity of $F$. If $F(x)=F(y)$ with $x,y\in X$, there must be some ${\varphi }_i$ such that ${\varphi }_i(x)>0$, implying $x\in U_i$. $y$ must also satisfy ${\varphi }_i(y)={\varphi }_i(x)>0$, implying $y\in U_i$. Here, $h_i(x)=h_i(y)$ implies $g_i(x)=g_i(y)$, which forces $x=y$ as $g_i$ is an imbedding. Thus, we are done.

Remarks: Note the $h_i$s act as smooth functions on $X$ that pick up the $U_i$ “coordinate” of a point. Furthermore, note that in this proof we did not use the second-countability of $m$-manifolds. It is possible to show any $m$-manifold can be embedded into Euclidean space without compactness, but this requires considerably more machinery.