# Some Exercises from Lee’s Introduction to Smooth Manifolds

These are exercises from the first few pages of Lee’s Introduction to Smooth Manifolds. These are pretty elementary from a topology perspective – maybe the middle-to-late part of a semester on point-set topology – but I decided to put them here to (a) remember some of that stuff, and (b) force myself to not delay doing them any longer. As it turns out, I struggled more with these than I should have and even had to consult the internet for a couple thing; in instances of internet borrowing, I provide links.

Note, finally, that if you’re trying to hunt these particular exercises down within Dr. Lee’s book, they’re actually dispersed throughout the text; suffice it to say, I’m not a fan of that particular method of delivery. Nevertheless….

Exercise 1.2
Show that $\mathbb{RP}^n$ is Hausdorff and second countable and is therefore a topological $n$-manifold.
Proof. Note that proving Hausdorffness and second countability will exactly prove the $n$-manifold claim due to the discussion in Example 1.3. Also recall that by Lee’s definition, $\mathbb{RP}^n$ is defined to be the collection of 1-dimensional linear subspaces of $\mathbb{R}^{n+1}$ with the quotient topology determined by $\pi:\mathbb{R}^{n+1}\setminus\{0\}\to\mathbb{RP}^n$ which sends every point $x\in\mathbb{R}^{n+1}$ to the 1-D subspace $[x]=\pi(x)$ spanned by $x$. Perhaps a more intuitively useful definition is that $\mathbb{RP}^n$ is the space obtained from $S^n$ with the identification $x\sim -x$ for all $x\in S^n$. That’s the construction used hereon.

To prove Hausdorffness, suppose $[x]\neq[y]$ are distinct points in $\mathbb{RP}^n$ and consider the subspaces $\pi^{-1}([x])=\{x,-x\}$ and $\pi^{-1}([y])=\{y,-y\}$ in $S^n$. As a subspace of a metric space $\mathbb{R}^{n+1}$, $S^n$ is Hausdorff, whereby there exist open sets $U_1\ni x$ and $V_1\ni y$ in $S^n$ for which $U_1\cap V_1=\varnothing$. Similarly for $-x$ and $-y$, yielding open sets $U_2\ni -x$ and $V_2\ni -y$ which are also disjoint. For notational convenience, let $\iota:x\mapsto -x$ denote the antipodal map, and consider the open sets[1]

$\mathcal{O}_x = \left(\iota(U_1)\cap U_2\right)\cup\left(U_1\cap\iota(U_2)\right)$

and

$\mathcal{O}_y=\left(\iota(V_1)\cap V_2\right)\cup\left(V_1\cap\iota(V_2)\right)$.

Openness and disjointness of the $\mathcal{O}$ sets are obvious, and it’s easily verified that $\iota(\mathcal{O}_\alpha)=\mathcal{O}_\alpha$ for $\alpha\in\{x,y\}$. Hence, $U=\pi(\mathcal{O}_x)$ and $V=\pi(\mathcal{O}_y)$ are disjoint sets which are open (as insured by the fact that $\pi^{-1}(U)=\mathcal{O}_x$ and $\pi^{-1}(V)=\mathcal{O}_y$ are open in $S^n$ along with properties of the quotient topology) and which contain $[x]$ and $[y]$, respectively.

To show second countability, note first that $\mathbb{R}^{n+1}$ has a countable basis $\mathcal{B}=\{B_i\}_{i=1}^\infty$ consisting of balls of rational radii with rational centers. Next, notice that the $n$-sphere $S^n$ also has an induced countable basis $\mathcal{U}$ whose elements are of the form $\mathcal{U}=\{U_i=B_i\cap S^n\}_{i=1}^\infty$. Finally, the claim is that the collection

$\begin{array}{rcl}\mathcal{K}&=&\{\{[x]\in\mathbb{RP}^n\,:\,x\in U_i\}\,:\,U_i\in\mathcal{U}\}\\ &=& \{V\subset\mathbb{RP}^n\,:\,\pi^{-1}(V)=U_i,U_i\in\mathcal{U}\}\end{array}$

is a basis for $\mathbb{RP}^n$[2]. Clearly, $\mathcal{K}$ covers $\mathbb{RP}^n$. Suppose, then, that $U_1,U_2\in\mathcal{U}$, and consider an element $[y]\in V_1\cap V_2$ where $V_i=\{[y]:y\in U_i\}$, $i=1,2$. Because $\mathcal{U}$ is a basis for $S^n$, there exists an element $U_3$ containing $y$ (respectively an element $V_3=\{[y]:y\in U_3\}$ containing $[y]$) for which $y\in U_3\subset U_1\cap U_2$ (respectively, for which $[y]\in V_3\subset V_1\cap V_2$). Thus, as claimed, $\mathcal{K}$ is a basis for $\mathbb{RP}^n$ in one-to-one correspondence with the basis $\mathcal{U}$ of $S^n$ and hence is a countable basis.  $\square$

Exercise 1.3
Show that $\mathbb{RP}^n$ is compact. [Hint: Show that the restriction of $\pi$ to $S^n$ is surjective.]
Proof.
Clearly, $\pi|_{S^n}$ is surjective due to the fact that for given $[y]\in\mathbb{RP}^n$, $\pi(y)=[y]=\pi(-y)$ where $\pm y\in S^n$. It’s also easily shown that $\pi$ is an open map, whereby we have that $\pi$ is a quotient map. One well known property of quotient maps[3] is that if the host space $X$ is compact, so too are all its quotients; in particular, then, because $S^n$ is compact, $\mathbb{RP}^n$ – that is, the image of $X$ under the quotient map $\pi$ – is also compact. Hence, the result.   $\square$

1. I was at first oblivious to the need for unions and intersections. That bit – and therefore the subsequent part – comes from here.

2. The first set description displayed was borrowed from here. The second set description was mine, but I liked the first better so I included it for completeness.

3. To prove it: Let $\mathcal{U}=\{U_\alpha\}$ be an open cover of $\mathbb{RP}^n$. Because $U_\alpha$ is open in $\mathbb{RP}^n$, $\pi^{-1}(U_\alpha)$ is open in $S^n$. Moreover, $\{\pi^{-1}(U_\alpha\}$ is an open cover of $S^n$, and because $S^n$ is compact, there exists a finite subcover $\{\pi^{-1}(U_i)\}_{i=1}^n$ of $S^n$. But then $\{U_i\}_{i=1}^n$ is a finite subcollection of $\mathcal{U}$ which covers $\mathbb{RP}^n$.