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Topology · Blog Entry
Alright, so it is currently 11:41 PM (I wrote the following sections before this one), so I will make this one quick. I began with breakfast with my friend Bhargava, who is going to take topology with me. I had some nice French toast. After that, I studied a little bit until lunch time, where I wandered around the Princeton art museum with some of my friends, who wish to start a Princeton game development studio.
Then we got lunch at Choi Dining Hall. I ate some kung pao chicken, which I actually never had before. It was actually so good, I wanted seconds but I was too lazy to get myself seconds. All the while I talked to the game dev studio kids, one of them is named Jessica and we met on Discord a while back when the Princeton 2030 Discord server was starting to become a thing.
After that, I noticed that some of my classes had finally published their Canvas pages, so I started to look at the syllabi for them. Pretty exciting! First day of classes are tomorrow.
Then, I went to get ice cream with my friend Grace in the evening. Grace is not a math major, she is a classics major. I got some chocolate chip cookie ice cream with whipped cream and chocolate syrup on top. It was so good.
After that, I worked on my PHY 105 pset while calling my family. Yes, a pset is already out, and I knocked out 2/3rds of it. It was mostly review stuff, nothing that crazy. Then I hung out with some friends in my hall, such as Fay and Sophia. I flexed my piano "skills" (I don't have skills) and showed them my physics pset.
Now, I am lying in bed, reflecting on this day. It was a pretty packed day, and pretty exciting too. But tomorrow, that is going to be an entirely different beast. Tomorrow is the day my life changes for the next four years, because no longer am I in little league. Tomorrow, I am truly welcomed into Princeton.
Let us continue our discussion of paracompactness. Now, we will begin to see the motivations behind why paracompactness is such a useful tool. Formulated as a sort of alternative to compactness, it carries some useful properties that make it possible to study manifold theory more deeply. We begin with the following definition:
Suppose that $f: S \to \mathbb{R}$ is a mapping from some space $S$ to the real numbers. We say that the support of $f$ is the set $$\text{support}(f) = \text{closure}\{x \mid f(x)\not= 0\}$$
In other words, it is the set consisting of all the points in the space where $f$ is zero, or $f$ is basically "about" to be zero, in a sort of sense. As it turns out, this allows us to define a concept known as the partition of unity, which is a bunch of maps where at any point there is a neighborhood where only a finite number of maps are nonzero, and the sum of the images of the maps at each point is 1, hence the term partition of "unity."
Suppose $\{U_\alpha \mid \alpha \in A\}$ is a sort of open covering of the space $X$. Then, we define a partition of unity subordinate to this covering as a collection of maps $$\{f_\beta: X \to [0,1] \mid \beta \in B\}$$ such that the following hold:
I will omit the proof for this next theorem.
If $X$ is paracompact and $\textbf{U}$ is an open covering of $X$, then there exists a partition of unity subordinate to $\textbf{U}$. Furthermore, if $\{U_\alpha\}$ is a locally finite open covering of $X$, then there is an open covering $\{V_\alpha\}$ such that, for each $\alpha$, $\overline{V}_\alpha \subset U_\alpha$.
We now consider the last definition for today, which is on $\sigma$-compactness.
A space is $\sigma$-compact if it can be considered to be the union of a countable number of compact subspaces.
For example, the set of rational numbers is $\sigma$-compact. The next theorem, I will omit the proof, but the theorem after, I will supply the proof.
A locally compact, Hausdorff space is paracompact if and only if it is the disjoint union of open $\sigma$-compact subsets.
Recall that a locally compact space is a topological space such that each point has a compact neighborhood (in this case, a neighborhood is not necessarily open, there just needs to be some open set that is a subset of the neighborhood with the point in the open set. Anyways, moving on,
If $X$ is locally compact, Hausdorff, and second countable then its one-point compactification $X^+$ is metrizable and $X$ is $\sigma$-compact and paracompact.
Since $X$ is second countable, then we may have a countable basis $\textbf{B}$ for $X$. Now, suppose $C \subset X$ is compact. Since $X$ is locally compact, any point $x \in C$ has a compact neighborhood $N$ aroun dit, and therefore there exists some $U_x \in \textbf{B}$ such that $x \in U_x \subset N$. Therefore, since every point is a subset of some $U_x$ in the basis, then $C$ is covered by a finite union $U_{x_1} \cup \cdots \cup U_{x_n}$, since $C$ is compact. Let $V = X - \bigcup \overline{U}_{x_i}$, which is the set of all points in $X$ not covered by the closures of one of the $U_x$ open sets. Consider the set $V \cup \{\infty\}$, which is a neighborhood of $\infty$ in the one-point compactification of $X$, $X^+$, which is contained in the neighborhood $X^+-C$. Now consider all the possible sets $C$ and all the possible choices of finite collections of basis elements. There are many sets $V$. We can index them by finite subsets of $\textbf{B}$; that is, for each finite subset of $\textbf{B}$, there is some compact set $C$ associated with it as well as a set $V$ constructed from both. Therefore, since the set of finite subsets of $\textbf{B}$ (which is countable) is countable (one can see a book on set theory to prove this fact, but it is a very trivial corollary of Cantor's Diagonalization Theorem), then all the sets $V$ are countable as well. $V$ allows us to create a local basis around the point $\infty$, so that combined with $\textbf{B}$ gives us a countable basis of $X^+$. Therefore, $X^+$ is second-countable. $X$ is apparently $\sigma$-compact as well, though I can't seem to grasp why. Using the previous theorem (the backwards direction), $X$ is paracompact. Furthermore, note that a compact Hausdorff space is normal (a previous result). A normal space is also completely regular, and finally, by the Urysohn Metrization Theorem, a space that is second countable and completely regular is metrizable. Hence, $X^+$ is metrizable.
Homotopy is a pivotal tool in algebraic topology, and we begin to see it in its most basic form. A homotopy is essentially a bunch of mappings that vary continuously, from one mapping to another. More precisely, what I mean is this:
Suppose $X$ and $Y$ are topological spaces. A homotopy of maps from $X$ to $Y$ is a continuous map $F: X \times [0,1] \to Y$, where we recall that $\times$ is the Cartesian product.
Furthermore, two maps $f_0,f_1:X \to Y$ are homotopic if there exists some homotopy $F:X \times [0,1] \to Y$ such that $F(x,0)=f_0(x)$ and $F(x,1)=f_1(x)$ for all $x \in X$.
Note that homotopy defines an equivalence relation, where "$f$ is homotopic to $g$" is written as $f \simeq g$.
Suppose $f,g:X \to Y$, $h: X'\to X$ and $k: Y \to Y'$ are all continuous. We have that $$f \simeq g \implies f \circ h \simeq g \circ h \text{ and }k \circ f \simeq k \circ g$$
Suppose that $f \simeq g$. Then there exists some continuous $F: X \times I \to Y$, where $I = [0,1]$, such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$. Consider the map $G: X' \times I \to Y$, where $G(x',t)=F(h(x'),t)$. It is clear due to the continuity of $h$ and $F$ that $G$ is continuous as well, and that $G(x',0)=(f \circ h)(x')$ and $G(x',1)=(g \circ h)(x')$. Therefore, $f \circ h \simeq g \circ h$. A similar argument shows that $k \circ f = k \circ g$.
To continue our discussion of homotopy, we propose the following definition of homotopy equivalence, not between two maps, but between two topological spaces.
A homotopy equivalence is a continuous map $f: X \to Y$ between two topological spaces with a homotopy inverse $g: Y \to X$ such that $g \circ f \simeq 1_X$ and $f \circ g \simeq 1_Y$, where $1_X$ is the identity function on $X$. If this is true, we say that $X \simeq Y$, or that $X$ and $Y$ have the same homotopy type.
One will be able to check that this is an equivalence relation between topological spaces as well, for we have that $X \simeq X$, if $X \simeq Y$ then $Y \simeq X$, and if $X \simeq Y$ and $Y \simeq Z$ then $X \simeq Z$. We now define contractibility, which is essentially a space that can be "deformed" into just one point.
A topological space $X$ is contractible if it is homotopy equivalent to a topological space consisting of just one point.
We now consider and prove the following proposition.
A topological space $X$ is contractible if and only if the identity map $1_X: X \to X$ is homotopic to a map $r: X \to X$ whose image is just a single point.
Suppose $X$ is contractible. We wish to show that the identity map is homotopic to $r$. So, let $Y = \{ x_0\}=\text{im}(r)$. We have the retraction $r: X \to Y \subset S$, and the inclusion map $i: Y \to X$. Notice that $i \circ r \simeq 1_X$, which implies that $r \simeq 1_X$ as well, hence showing that $1_X$ is homotopic to a map $r$ whose image is a single point. Now, in the opposite direction, assume that $1_X$ is homotopic to $r$. Defining the inclusion map $i: Y \to X$, we obtain $r \circ i = 1_Y$ and $i \circ r \simeq 1_X$. Therefore, $X$ and $Y$ are homotopy equivalent, and therefore $X$ is contractible.
We will now show that $\mathbb{R}^n$ is contractible. Define $F: \mathbb{R}^n \times I \to \mathbb{R}^n$ where $F(x,t)=tx$. It is clear that $f_1(x) = F(x,1)= x=1_X$ and $f_0(x)=F(x,0)=0$. Therefore, $f_1 \simeq f_0$, so by Proposition 2, $\mathbb{R}^n$ is contractible.
Additionally, it follows that $S^{n-1}$, which is the $n-1$-dimensional sphere (for example, the 2-dimensional sphere is the sphere you are familiar with), and $\mathbb{R}^n - \{0\}$ are homotopy equivalent. Let $i: S^{n-1} \to \mathbb{R}^n - \{0\}$ be the inclusion map, and let $r: \mathbb{R}^n - \{0\} \to S^{n-1}$ be the following expression: $$ r(x) = \frac{x}{\|x\|} $$ We can clearly see that $r\circ i = 1$, and $i \circ r \simeq 1$ through the homotopy $F: (\mathbb{R}^n - \{0\}) \times I \to \mathbb{R}^n - \{0\}$, where $$ F(x,t)=tx+(1-t)\frac{x}{\|x\|}$$ It is clear that $f_0(x)=F(x,0)=(i \circ r)(x)$ and $f_1(x)=F(x,1)=1(x)$. Therefore, $i \circ r \simeq 1$ and $r \circ i \simeq 1$, so $S^{n-1}\simeq \mathbb{R}^n - \{0\}$.