# Math4202 Topology II (Lecture 6) ## Manifolds ### Imbedding of Manifolds #### Definition for partition of unity Let $\{U_i\}_{i=1}^n$ be a finite open cover of topological space $X$. An indexed family of **continuous** function $\phi_i:X\to[0,1]$ for $i=1,...,n$ is said to be a **partition of unity** dominated by $\{U_i\}_{i=1}^n$ if 1. $\operatorname{supp}(\phi_i)=\overline{\{x\in X: \phi_i(x)\neq 0\}}\subseteq U_i$ (the closure of points where $\phi_i(x)\neq 0$ is in $U_i$) for all $i=1,...,n$ 2. $\sum_{i=1}^n \phi_i(x)=1$ for all $x\in X$ (partition of function to $1$) #### Existence of finite partition of unity Let $\{U_i\}_{i=1}^n$ be a **finite open cover** of a **normal** space $X$ (Every pair of closed sets in $X$ can be separated by two open sets in $X$). Then there exists a partition of unity dominated by $\{U_i\}_{i=1}^n$. _A more generalized version, If the space is paracompact, then there exists a partition of unity dominated by $\{U_i\}_{i\in I}$ with locally finite. (Theorem 41.7)_ We will prove for the finite partition of unity.
Proof for finite partition of unity Some intuitions: By definition for partition of unity, consider the sets $W_i,V_i$ defined as $$ W_i=f^{-1}_i((\frac{1}{2n},1])\subseteq f^{-1}_i([\frac{1}{2n},1])\subseteq V_i=f^{-1}_i((0,1])\subseteq \operatorname{supp}(f_i)\subseteq U_i $$ $$ V_1\subseteq \overline{V_1}\subseteq U_1 $$ Note that $V_i$ is open and $\overline {V_i}\subseteq U_i$. And $\bigcup_{i=1}^n V_i=X$. and $W_i$ is open and $\overline{W_i}\subseteq V_i$. And $\bigcup_{i=1}^n W_i=X$. --- **Step 1**: $\exists$ V_i$ ope subsets $i=1,\dots,n$ such that $\overline{V_i}\subseteq U_i$, and $\bigcup_{i=1}^n V_i=X$. For $i=1$, consider $A_1=X-(U_2\cup U_3\cup \dots \cup U_n)$. Therefore $A_1$ is closed, and $A_1\cup U_1=X$. So $A_1\subseteq U_1$. Note that $A_1$ and $X-U_1$ are disjoint closed subsets of $X$. Since $X$ is normal, we can separate disjoint closed subsets $A_1$ and $X-U_1$. So we have $A_1\subset V_1\subseteq \overline{V_1}\subseteq U_1$ (by [normal space proposition](https://notenextra.trance-0.com/Math4201/Math4201_L37/#proposition-of-normal-spaces)). For $i=2$, note that $V_1\cup\left( \bigcup_{i=2}^n U_i\right)=X$, Take $A_2=X-\left(V_1\cup\left( \bigcup_{i=3}^n U_i\right)\right)$ (skipping $U_2$). Then we have $V_2\subseteq \overline{V_2}\subseteq U_2$. For $i=j$, we have $$ A_j=X-\left(\left(\bigcup_{i=1}^{j-1}V_i\right)\cup \left(\bigcup_{i=j+1}^n U_i\right)\right) $$ and $\bigcup_{i=1}^n V_i=X$. Repeat the above construction for $\{V_i\}_{i=1}^n$. Then we have $\{W_i\}_{i=1}^n$ open and $W_i\subseteq \overline{W_i}\subseteq V_i\subseteq \overline{V_i}\subseteq U_i$. And $\bigcup_{i=1}^n W_i=X$. **Step 2**: Using [Urysohn's lemma](https://notenextra.trance-0.com/Math4201/Math4201_L37/#urysohn-lemma). To construct the partition of unity $\phi_i$. > [!NOTE] > > Suppose > > - $X$ be a normal space > - $Z_1,Z_2\subseteq X$ are closed > - $Z_1$ and $Z_2$ are disjoint > > Then: > > There exists $f:X\to[0,1]$ such that > > - $f(Z_1)=\{0\}$ and $f(Z_2)=\{1\}$ > - $f$ is continuous. Since $W_1\subseteq \overline{W_1}\subseteq V_1\subseteq \overline{V_1}\subseteq U_1$, Note that $\overline{W_1}$ and $X-V_1$ are two disjoint closed subsets of normal space $X$ Then we can have $f_1:X\to[0,1]$ such that $f_1(\overline{W_1})=\{0\}$ and $f_1(X-V_1)=\{1\}$. Then we have the remaining list of function $f_2,\dots,f_n$. Recall the definition for support of functions $\operatorname{supp}(f_i)=\overline{\{x\in X: f_i(x)>0\}}$. Since $f_i(x)=0$ for $x\in X-V_i$, we have $\operatorname{supp}(f_i)\subseteq \overline{V_i}$ Next we need to check $\sum_{i=1}^n f_i(x)=1$ for all $x\in X$. Note that $\forall x\in X$, since $\bigcup_{i=1}^n W_i=X$, then there exists $i$ such that $x\in W_i$, thus $f_i(x)=1$. And $\sum _{i=1}^n f_i(x)\geq 1$. Then we do normalization for our value. Set $F(x)=\sum_{i=1}^n f_i(x)$. Since $F(x)$ is sum of continuous functions, $F$ is continuous. Then we define $\phi_i=f_i/F(x)$, since $F(x)\geq 1$, we are safe to divide by $F(x)$ and $\phi_i(x)$ is continuous. And $\operatorname{supp}(\phi_i)=\operatorname{supp}(f_i)\subseteq \overline{V_i}\subseteq U_i$. And $\sum_{i=1}^n \phi_i(x)=\frac{\sum_{i=1}^n f_i(x)}{F(x)}=\frac{F(x)}{F(x)}=1$ for all $x\in X$.
### Some Extension #### Definition of paracompact space Locally finite: $\forall x\in X$, $\exists$ open $x\in U$ such that $U$ only intersects finitely many open sets in $\mathcal{B}$. A space $X$ is paracompact if every open cover $A$ of $X$ has a **locally finite** refinement $\mathcal{B}$ of $A$ that covers $X$. ## Algebraic Topology Homeomorphism: A topological space $X$ is homeomorphic to a topological space $Y$ if there exists a homeomorphism $f:X\to Y$ - $f$ is continuous - $f^{-1}$ is continuous - $f$ is bijective Equivalence relation: If $\sim$ satisfies the following: - $\sim$ is reflexive $\forall x\in X, x\sim x$ - $\sim$ is symmetric $\forall x,y\in X, x\sim y\implies y\sim x$ - $\sim$ is transitive $\forall x,y,z\in X, x\sim y, y\sim z\implies x\sim z$ Homeomorphism is an equivalence relation. - Reflexive: identity map - Symmetric: inverse map is also homeomorphism - Transitive: composition of homeomorphism is also homeomorphism Main Question: classify topological space up to homeomorphism. ### Invariant in Mathematics Quantities associated with topological spaces that don't change under homeomorphism. We want to use some algebraic tools to classify topological spaces.