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)


\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.]
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.


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