Prove that $S^n/$~$_1$ $\cong RP^n$, where ~$_1$ means identify antipodal points.

Question

I'm self studying Rotman's Algebraic Topology and I've come across this problem.

Prove that $S^n/$~$_1$ $\cong RP^n$, where ~$_1$ means identify antipodal points.

I see that $RP^n = \{[x]: x \in R^{n+1} - \{0\} \text{ and } x$ ~$_2$ $y \text{ iff } x = \lambda y \text{ for some $\lambda \in R-\{0\}$ }\}$ and $S^n/\sim_1=\{x\in R^{n+1} : |x| = 1 \text{ and } x \sim_1 y \text{ iff } x = y \}$.

I tried showing: $$R^{n+1} \cong S^{n+1} - \{\text{North pole}\}$$ $$\Rightarrow R^{n+1} - \{0\} \cong S^{n+1} - \{N\} - \{S\}$$ $$ \Rightarrow (R^{n+1} - \{0\})/\sim_2 \space \cong (S^{n+1} - \{N\} - \{S\}) / \sim_3$$

But from here I couldn't find a way to show $S^n / \sim_1$ is homeomorphic to the RHS.

Anyone have any ideas?

Answer 1

The map $\iota: S^n \hookrightarrow \mathbb{R}^{n+1}\backslash \{0\} $ is continuous. Therefore, the map $\pi \circ \iota: S^n \to \mathbb{R}P^n$ is continuous. It is easily seen to factor through the quotient (since it sends antipodal maps to the same point), and thus induces a continuous map $\widetilde{\pi \circ \iota}: S^n/\sim \to \mathbb{R}P^n$. You can verify that such map is bijective. Since $\mathbb{R}P^n$ is Hausdorff and $S^n/\sim $ is compact (being the image of the quotient map $S^n \to S^n/\sim$), we have that $\widetilde{\pi \circ \iota}$ is a homeomorphism.

Answer 2

You're overthinking this. You are right to start with $RP^n = (\mathbb{R}^{n+1}-\{0\})/\sim_2$. What is $\sim_2$ when restricted to $S^n$?

Answer 3

The idea is to "restrict down." I think both other answers provide a good formalism for this idea, so I will omit it from my answer.

Here is a preliminary idea: suppose that I gave you a set $\{x,y,z\}$, but then imposed the relation $x=y$. In this case, it is enough to consider $\{x,z\}$. We are going to do the same thing.

$\mathbb R P^{n} = \mathbb R^{n+1}/\sim$ where $\vec{x} \sim \lambda \vec{x}$. In particular, notice that every line through the origin passes through $S^n$ in exactly two points (which are antipodal.) Since every point on this line has been identified, it is really enough to consider just these two points on $S^n$, glued together.

You might wonder why we stop here, and you would be right to ask this question. In fact, we can make further identifications, it is clearly then enough to consider only the north hemisphere, and a single point. Then, the only question is what happens at the equator, where each point still has its antipodal point in the hemisphere. In this case, notice that the equator is $S^{n-1}$ and that the rest of the hemisphere is just the interior of $D^n$. This is what allows us to also define $\mathbb R P^n$ as the quotient $D^n$ with $\partial D^n/\sim$, where $\sim$ is the antipodal map.

Answer 4

What every answer is missing is a sexy commutative diagram. Here goes:

Of course, the bottom right corner is $\mathbb R \text P^n := \mathbb R^{n+1}\backslash\{0\}/\sim_2$.

Key

• $f(x) = x$, the simplest map;
• $Q(x)$ is the quotient class of $x$ w.r.t. $\sim_1$, i.e. $Q(x) = \{y \in S^n : y \sim_1 x\}$;
• $P(x)$ is the quotient class of $x$ w.r.t. $\sim_2$; and most importantly,
• $g$ is the map defined on equivalence classes, $g(Q(x)) := P(x)$.

The focus is $g$: it is the map that we want to be well-defined and homeomorphic.

$P$, $Q$ are the quotient maps, whose properties need to be exploited. To make up names for these maps is something of a personal preference, for me it elucidates on the stages below.

This is quite a general treatment, I spell it out in the hope that it elucidates on further quotient-based examples, perhaps the principle of working from upstairs to downstairs.

Why $g$ is well defined

You may well be perplexed by the definition. Why does it make sense? Well, it makes sense provided whenever $Q(x) = Q(y),$ we have $f(P(x)) = f(P(y)).$ Unwrapping this, we want to show whenever $x,y \in S^n$,

$$x \sim_1 y \implies x \sim_2 y.$$

This is something you can happily verify.

Why $g$ is continuous

Firstly, $f$ is continuous. So is $P$, by the definition of the quotient topology (a.k.a. the topology induced by $P$):

$$U\text{ open in }\frac{\mathbb R^{n+1}\backslash\{0\}}{\sim_2} \iff P^{-1}(U) \text{ open in }\mathbb R^{n+1}. \tag{1}$$

(I.e., $P$ is not just continuous, but an open mapping [and the same applies to $Q$].)

So given open $U \in \mathbb R \text P^n$, $f^{-1}P^{-1}(U)$ is open in $S^n$. Now, the fact that $Q$ is an open map (look at $(1)$) means $Q(f^{-1}P^{-1}(U))$ is open. I.e., $g^{-1}(U) = Q(f^{-1}P^{-1}(U))$ is open.

Why $g$ is a bijection.

First, you need to check that $g$ is surjective: whenever you have a class $P(y)$ in $\mathbb R \text P ^{n}$, there is a point $x \in S^{n}$ such that $P(x) = P(y)$, i.e. $x\sim_2 y$. But you can just take $$x = \frac y {\|y\|_2}\phantom{\bigg|}.$$

As for injectivity, this is like the converse of well-definedness. $g$ is injective if, given $x,y$ in $S^n$, $$P(x) = P(y) \implies Q(x) = Q(y).$$

In other words:

$$ x \sim_2 y \implies x \sim_1 y.$$

But this is also pretty simple: $\|x\|_2 = \|y\|_2 = 1$, so if $x\sim_2 y$, i.e. $x = \lambda y$, where $\lambda \in \mathbb R\backslash \{0\}$, then taking norms tells you $|\lambda|=1$, i.e. $\lambda = \pm 1$, so $x \sim_1 y$.

(In my notation, we've shown $g(Q(x)) = g(Q(y)) \implies Q(x) = Q(y)$.)

So altogether, $g$ is bijective.

Why $g$ is a homeomorphism.

You have two options, both mentioned above:

1. You can use the result mentioned a continuous bijection from a compact topological space to another is a homeomorphism. $$S^n/\sim_1$$ is compact because it is the continuous image of the compact space $S^n$;

The fact that $\mathbb R \text P ^n$ is Hausdorff is harder! Given $P(x)$, $P(y)$ with $\|x\|_2= \|y\|_2 = 1$, you can take $$0< \epsilon <\frac 1 2 \min\{\|x-y\|_2, \|x - (-y)\|_2\},$$ in which case for example you can take the double cones

$$ U_x = \left\{r z: r \in \mathbb R \backslash \{0\}, z \in S^n,\ \|z - x\|< \epsilon \right\},\text{ and}\\ U_y = \left\{r z: r \in \mathbb R \backslash \{0\}, z \in S^n,\ \|z - y\|< \epsilon \right\}.$$

And maybe with some sweat, you can show that $U_x$, $U_y$ are open, that they each are unions of $\sim_2$ classes, and that they don't intersect otherwise their normailised points on the sphere would intersect. That altogether means, they project down (via $P$) to open disjoint sets. Alternatively, you can pull out the general quotient-by-cocompact-group action kind of argument to show it's a metric space.

2. You can do it directly (which I prefer after that Hausdorff argument).

We want to show that $g$ is an open mapping.

We need the fact that $\sim_1$ can be viewed as a restriction of $\sim_2$ to the sphere. Being needlessly rigourous, the following is true, since $f$ is an inclusion.

$$P \circ f = P|_{\text{im}(f)} \circ f = P|_{S^n} \circ \hat f = P|_{S^n}$$

where $\hat f:S^n \to S^n$ is $f$ with the range restricted to $S^n$: i.e. the identity map.

Now let $U$ be open in $S^n/\sim_1$. By the definition of quotient topology, $V := Q^{-1}(U)$ is open.

Now, $V \subset S^n$, but we need to enlarge it to an open set in $\mathbb R^{n+1}$: so define the cone:

$$V_2 = \{rv \;|\; r\in \mathbb R\backslash \{0\},\ v \in V\}$$.

This is an open set in $\mathbb R^{n+1}$, perhaps because the map $(x,r)↦ rx$ is a full-rank smooth map $S^n × \mathbb R \backslash \{0\}$, so is an open map; or maybe one can just do something with balls in $S^n$.

Crucially also, $P(V_2)= P(V)$; so we have

$$V_2\text{ open}\stackrel{(1)}\implies P(V_2)\text{ open}\implies P(V)\text{ open} \implies g(U)\text{ open},$$

since $g(U) = P \circ f(V) = P(V) = P(V_2)$ is open, we've showed its an open map, and earned our cup of tea.