# Abstract Nonsense

## Topological Manifolds (Pt. III)

Point of Post: This is a continuation of this post.

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Some Topological Properties of Topological Manifolds

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What we’d now like to do is discuss some of the nice topological properties that topological manifolds have. The first of the them is the fact that they have a countable cover to topological Euclidean balls (i.e. spaces homeomorphic to a ball $B_\delta(0)\subseteq\mathbb{R}^n$ for some $n$). The basic idea is simple, at each point $p$ we know that there is a neighborhood which is homeomomorphic to an open subset of $\mathbb{R}^n$. Now, this open subset of Euclidean space contains a Euclidean ball around the image of $p$–the preimage of this ball will be a topological Euclidean ball containing $p$. Now, we have an open cover of our manifold by topological Euclidean balls–the argument is then over once we recall that every second countable space is Lindelof and so we may find a countable subcover. Formally:

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Theorem: Let $M$ be a topological manifold. Then, $M$ admits a countable atlas for topological Euclidean balls.

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This is a good time to point out a non-example of a topological manifold. Now, while you may be trying to think of some crazy, pathological beast of a space, there is in fact a very nice non-manifold living (embedded) in your computer screen. Indeed, consider the topological space represented by the plus sign: $+$. More rigorously, we can consider the infinite analogue gotten by considering the union of the two lines $\mathbb{R}\times\{0\}$ and $\{0\}\times\mathbb{R}$ sitting inside $\mathbb{R}^2$ as a subspace. Why is $+$ not a topological manifold? Well, suppose for a second that it was, then by the above theorem there exists a neighborhood $U$ of the point $p$ where the two lines meet (the origin in the more rigorous version) which is homeomorphic to $B_\delta(0)\subseteq\mathbb{R}^n$ for some (clearly nonzero) $n$. That said, if $U\approx B_\delta(0)$ then $U-\{p\}$ is homeomorphic to $B_\delta(0)$ minus a point. That said, $U-\{p\}$ has four connected components whereas $B_\delta(0)$ minus a point has at most two connected components (two in dimension $n=1$ and zero in higher dimensions)–this is a contradiction.

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With this theorem we can prove many, very interesting, facts about the topology of a topological manifold. For example, we immediately get the following theorems:

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Theorem: Every topological manifold is locally simply connected.

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Theorem: Every topological manifold is locally compact. In fact, it has a countable basis of precompact Euclidean topological balls.

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Now, perhaps a more interesting (because it is not as obvious) fact is the following:

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Theorem: Let $M$ be a topological manifold. Then, $M$ is connected if and only if it’s path connected.

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This follows immediately from the following lemma:

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Lemma: Let $X$ be a locally path connected connected space, then $X$ is path connected.

Proof: Let $x_0\in X$ and let $\Omega$ be the set of all points in $X$ for which there is a path from $x_0$ to that point. Since $\Omega\ne\varnothing$ ($x_0\in \Omega$) it suffices to prove that it is both open and closed (and the result will then follow from $X$‘s connectedness). So, let $x\in\Omega$. By assumption there exists some neighborhood $U$ of it such that $U$ is path connected, we claim that $U\subseteq\Omega$. To see this, let $x'\in U$. Concatenating the paths from $x$ to $x'$, and from $x$ to $x_0$ we can then construct a path from $x'$ to $x_0$ and so $x'\in \Omega$. It follows that $\Omega$ is open.

Now, to show that $\Omega$ is closed we show it is invariant under closure, and since $\Omega\subseteq\overline{\Omega}$ it suffices to show the reverse inclusion. So, let $x\in\overline{\Omega}$. Then, by $X$‘s local connectedness there exists a neighborhood $U$ of $x$ which is path connected. But, since $x\in\overline{\Omega}$ there must be some point $x'\in\Omega\cap U$. So, using the exact same technique described in theorem (*) again we may connect the path from $x$ to $x'$ and  the path from $x'$ to $x_0$ to get a path from $x$ to $x_0$. It follows that $x\in\Omega$.

Thus, $\Omega$ is open and closed and since it’s non-empty it must be that $\Omega=X$. But, that means that every point of $X$ may be connected to $x_0$ by a path. This clearly implies the theorem. $\blacksquare$

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Last but not least we prove the, very interesting, fact that the fundamental group of a topological manifold is necessarily countable. This, once again, is due to the fact that our manifold has a countable cover by topological Euclidean balls. Note, since connectedness is equivalent to path connectedness, the independence of base point for fundamental groups comes to connected manifolds. Our proof heavily follows [1].

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Theorem: Let $M$ be a connected topological manifold, then $\pi_1(M)$ is countable.

Proof: We know that $M$ has a cover ty topological Euclidean balls $\left\{(B_n,\varphi)n)\right\}_{n\in\mathbb{N}}$. Let $B_{n,m}=B_n\cap B_m$. Since our space is second countable we know that $B_{n,m}$ has countably many path components. For each $n\geqslant m$ let $X_{n,m}$ be a collection of representatives from each path component of $B_{n,m}$. Let then, $X=\bigcup_{n\geqslant m} X_{n,m}$–evidently $X$ is countable. If $x,x'\in X$ and $x,x'\in B_n$ then let $p_{x,x'}^n$ to be a path $x\to x'$ in $B_n$. Fix some $x_0\in X$, we will show that $\pi_1(M,x_0)\cong\pi_1(M)$ is countable. Let $G$ be the set of all finite products of paths of the form $p_{x,x'}^n$ as mentioned above. Call a loop $\alpha$ at $x_0$ special if $\alpha\in G$.

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Evidently the set of special loops, being a subset of $G$ (which is clearly countable) is countable. If we can show that every element of $\pi_1(M,x_0)$ is homotopic to a special loop, then we’ll be done. To this end, let $\alpha:[0,1]\to M$ be a loop anchored at $x_0$. Since $[0,1]$ is compact, there exists a finite cover of the set $\{\alpha^{-1}(B_n)\}_{n\in\mathbb{N}}$. Thus, there exists finitely many points $0=a_1<\cdots such that $\alpha^{-1}(B_n)\subseteq[a_i,a_{i+1}]$ for some $n$. Let $\alpha_i$ be the map one obtains by restricting $\alpha$ to $[a_{i},a_{i+1}]$ and reparamaterizing so that it is defined on $[0,1]$ and let $B_{n_i}$ be such that $\alpha_i([0,1])\subseteq B_{n_i}$. By definition, for each $i$ we have that $\alpha(a_i)\in B_{i,i+1}$ and that there exists some point $x_i$ in the same component of $B_{i,i+1}$–we can take $x_1=x_k=x_0$. Let $\beta_i$ be a path $x_i\to \alpha(a_i)$ where we take $\alpha_0=\alpha_k$ to be the constant loop at $x_0$. Note then that

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\begin{aligned}\text{ }[\alpha] &= [\alpha_1]\cdots[\alpha_k]\\ &= ([\beta_0]\cdot[\alpha_1]\cdot[\beta_1]^{-1})\cdot([\beta_1]\cdot[\alpha_2][\beta_2]^{-1})\cdots([\beta_{k-1}]\cdot[\alpha_k]\cdot[\beta_k])\end{aligned}

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Now, each $[\beta_i][\alpha_{i+1}][\beta_{i+1}]^{-1}$ is a path $x_i\to x_{i+1}$ and thus (since $B_{n_i}$ is simply connected) is homotopic to $p_{x_i,x_{i+1}}^{n_i}$. Thus, we see that $[\alpha]$ is a special loop. Since $[\alpha]$ was arbitrary the conclusion follows. $\blacksquare$

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References:

[1] Lee, John M. Introduction to Smooth Manifolds. New York: Springer, 2003. Print.

[2] Lee, John M. Introduction to Topological Manifolds. New York: Springer, 2000. Print.

[3] Milnor, John Willard, and David W. . Weaver. Topology from the Differentiable Viewpoint. Charlottesville (Va.): University of Virginia, 1969. Print.

August 30, 2012 - Posted by | Manifold Theory, Topology