[Munkres Topology] 36. Imbeddings of Manifold

 This article is one of the posts in the Textbook Commentary Project.

Definition. An $m$-manifold is a Hausdorff space $X$ with a countable basis such that each point $x$ of $X$ has a neighborhood that is homeomorphic with an open subset of $\mathbb{R}^m$.

If $\phi:X\rightarrow \mathbb{R}$, then the support of $\phi$ is defined to be the closure of the set $\phi^{-1}(\mathbb{R}-\{0\})$. Thus if $x$ lies outside the support of $\phi$, there is some neighborhood of $x$ on which $\phi$ vanishes.

Definition. Let $\{U_1,\cdots ,U_n\}$ be a finite indexed open covering of the space $X$. An indexed family of continuous function $$\phi_i: X\rightarrow [0,1]\quad \mbox{for } i=1,\cdots, n,$$ is said to be a partition of unity dominated by $\{U_i\}$ if:

(1) (support $\phi_i$) $\subset U_i$ for each $i$. 

(2) $\sum^n_{i=1} \phi_i(x)=1$ for each $x$.

Theorem 36.1 (Existence of finite partitions of unity). Let $\{U_1,\cdots ,U_n\}$ be a finite open covering of the normal space $X$. Then there exists a partition of unity dominated by $\{U_i\}$.

Proof. Step 1. First, we prove that one can 'shrink' the covering $\{U_i\}$ to an open covering $\{ V_1,\cdots, V_n\}$ of $X$ such that $\bar{V}_i\subset U_i$ for each $i$.

We proceed by introduction. First, note that the set $$A=X-(U_2\cup \cdots \cup U_n)$$ is a closed subset of $X$. Because $\{U_1,\cdots, U_n\}$ covers $X$, the set $A$ is contained in the open set $U_1$. Using normality, choose an open set $V_1$ containing $A$ such that $\bar{V}_1\subset U_1$. Then the collection $\{V_1, U_2,\cdots, U_n\}$ covers $X$. See Figure 36.1.

In general, given open sets $V_1,\cdots, V_{k-1}$ such that the collection $$\{V_1,\cdots, V_{k-1}, U_k, U_{k+1},\cdots, U_n\}$$ covers $X$, let $$A=X-(V_1\cup \cdots \cup V_{k-1})-(U_{k+1}\cup \cdots \cup U_n).$$ Then $A$ is a closed subset of $X$ which is contained in the open set $U_k$. Choose $V_k$ to be an open set containing $A$ such that $\bar{V}_k\subset U_k$. Then $\{V_1,\cdots, V_{k-1},V_k,U_{k+1},\cdots, U_n\}$ covers $X$. At the $n$th step of the induction, our result is proved. 

Step 2. Now we prove the theorem. Given the open covering $\{U_1,\cdots, U_n\}$ of $X$, choose an open covering $\{V_1,\cdots, V_n\}$ of $X$ such that $\bar{V}_i\subset U_i$ for each $i$. Then choose an open covering $\{W_1,\cdots, W_n\}$ of $X$ such that $\bar{W}_i\subset V_i$ for each $i$ (use Step 1 again.). Using the Urysohn lemma, choose for each $i$ a continuous function $$\psi_i :X\rightarrow [0,1]$$ such that $\psi_i(\bar{W}_i)=\{1\}$ and $\psi_i (X-V_i)=\{0\}$. Since $\psi_i^{-1}(R-\{0\})$ is contained in $V_i$, we have $$(\mbox{support} \psi_i)\subset \bar{V}_i \subset U_i.$$ See Figure 36.2.

Because the collection $\{W_i\}$ covers $X$, the sum $\Psi(x)=\sum^n_{i=1} \psi_i(x)$ is positive for each $x$. Therefore, we may define, for each $j$, $$\phi_i(x)=\frac{\psi_i(x)}{\Psi(x)}.$$ (just normalize $\psi$.) It is easy to check that the functions $\phi_1,\cdot,\phi_n$ form the desired partition of unity.

Theorem 36.2. If $X$ is a compact $m$-manifold, then $X$ can be imbedded in $\mathbb{R}^N$ for some positive integer $N$.

Proof. Cover $X$ by finitely many open sets $\{U_1,\cdots, U_n\}$, each of which may be imbedded in $\mathbb{R}^m$. Choose imbeddings $g_i :U_i\rightarrow \mathbb{R}^m$ for each $i$. Being compact and Hausdorff, $X$ is normal. Let $\phi_1,\cdots, \phi_n$ be a partition of unity dominated by $\{U_i\}$; let $A_i=\mbox{support } \phi_i$. For each $i=1,\cdots,n$, define a function $h_i:X\rightarrow \mathbb{R}^m$ by the rule $$h_i(x)= \begin{cases} \phi_i(x)\cdot g_i(x) & \mbox{for } x\in U_i,\\ \mathbf{0}=(0,\cdot,0) & \mbox{for } x \in X-A_i.\end{cases}$$ ($\phi$ is coordinate of cover, $g$ is partition of cover, then $h$ is embedding of cover for each partition.) [Here $\phi_i(x)$ is a real number $c$ and $g_i(x)$ is a point $\mathbf{y}=(y_1,\cdots,y_m)$ of $\mathbb{R}^m$; the product $c\cdot \mathbf{y}$ denotes of course the point $(cy_1,\cdots,cy_m)$ of $\mathbb{R}^m$.] The function $h_i$ is well defined because the two definitions of $h_i$ agree on the intersection of their domains, and $h_i$ is continuous because its restrictions to the open sets $U_i$ and $X-A_i$ are continuous.

Now define $$F:X\rightarrow (\mathbb{R}\times \cdots \times \mathbb{R}\times \mathbb{R}^m\times \cdots \times \mathbb{R}^m)$$ by the rule $$F(x)=(\phi_1(x),\cdots,\phi_n(x),h_1(x),\cdots,h_n(x)).$$ ($F$ is ) Clearly, $F$ is continuous. To prove that $F$ is an imbedding we need only to show that $F$ is injective (because $X$ is compact). Suppose that $F(x)=F(y)$. Then $\phi_i(x)=\phi_i(y)$ and $h_i(x)=h_i(y)$ for all $i$. Now $\phi_i(x)>0$ for some $i$ [since $\sum \phi_i(x)=1$]. Therefore, $\phi_i(y)>0$ also, so that $x,y\in U_i$. Then $$\phi_i(x)\cdot g_i(x)=h_i(x)=h_i(y)=\phi_i(y)\cdot g_i(y).$$ Because $\phi_i(x)=\phi_i(y)>0$, we conclude that $g_i(x)=g_i(y)$. But $g_i:U_i\rightarrow \mathbb{R}^m$ is injective, so that $x=y$, as desired.