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$\newcommand{\RP}{\mathbf RP}$ The real projective plane $\RP^2$ is defined as the quotient space $S^2/\sim$, where $\sim$ identifies the antipodal points of $S^2$.

I want to show that $\RP^2$ is homeomorphic to the quotient space of the closed $2$-disc $D^2$ obtained by identifying the antipodal points on the boundary circle of $D^2$.

This is not at all visually obvious to me.

An analogous problem is showing that $\RP^1=S^1$. Here, the situation is simple because we can think of $S^1$ as sitting inside $\mathbf C$ and consider the function $f:S^1\to S^1$ defined as $f(z)=z^2$ for all $z\in S^1$. The fibres of $f$ are precisely the antipodal points and we get an isomorphism $S^1/\sim \ \cong \ S^1$.

Again, the simpler case is also not visually clear to me.

caffeinemachine
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2 Answers2

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Every point on the upper hemisphere is identified with a point on the lower one. The upper hemisphere is homeomorphic to a disc. What happens at the equator ( the boundary of this disc)?

edit: Let's do a proof. Let $D^2$ be the disc and $D^2/\sim$ the disc with opposite boundary points identified.

For every point $x\in S^2$ there are two points on the line through the origin and $x$. For all points outside the equator, there is a unique point on the upper hemisphere, but on the equator there are two points on the (closed) upper hemisphere. We can't directly define a map $S^2\rightarrow D^2$ this way, however we can map a point to $D^2/\sim$. The problems on the equator disappear when we take the equivalence relation. Thus we have a map

$S^2\rightarrow D^2/\sim$.

By definition of this map it sends antipodal points on $S^2$ to the same points on $D^2/\sim$. Hence it factors to a continuous map $\mathbb{RP}^2\rightarrow D^2/\sim$.

Now you have to show that this map is

  1. injective
  2. surjective

and that $\mathbb{RP}^2$ is compact, and $D^2/\sim$ is Hausdorff. Then you know that the map is a homeomorphism (see https://proofwiki.org/wiki/Continuous_Bijection_from_Compact_to_Hausdorff_is_Homeomorphism).

Thomas Rot
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  • I have seen this to visually motivate the truth of the statement. But I think there is a problem with this. Shouldn't it matter how the points of the upper hemisphere $H^n_+$ are identified with points on the lower hemisphere $H^n_-$? To be more precise, consider any bijection $f:H^n_+\to H^n_-$. Now define a relation $\sim$ on $S^2$ which identifies antipodal points of the equator and identifies $x\in H^n_+$ with $f(x)\in H^n_-$. Is is true that the resulting space is $\mathbf RP^2$ again? – caffeinemachine Aug 10 '15 at 08:02
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    Yes. I was looking for a formal proof. Of course, if some visual insight comes along with it then it is much better. – caffeinemachine Aug 10 '15 at 09:25
  • I cannot see how "Let $U$ be the (closed) upper hemisphere. For every point $x\in S^2$ there is a unique point on the line through the origin and this point that intersects $U$" is true. Since $U$ is closed upper half-sphere, the equator is contained in $U$. If $x\in S^2$ is chosen on the equator, then we get two points which lie on the upper-half sphere as well as on the line joining $x$ and origin, namely $x$ and $-x$. Am I making a mistake? – caffeinemachine Aug 10 '15 at 10:39
  • @caffeinemachine: update – Thomas Rot Aug 10 '15 at 11:05
  • Thanks. I am checking that the map $S^2\to D^2/\sim$ is continuous. It looks promising. – caffeinemachine Aug 10 '15 at 11:45
  • I have posted a follow up question here: http://math.stackexchange.com/questions/1391936/a-map-from-sn-to-dn-sim-is-continuous – caffeinemachine Aug 10 '15 at 13:25
  • I just posted an answer. Please check it out. Thanks. – caffeinemachine Aug 11 '15 at 06:31
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After discussing with a friend, here is a solution:

We think of $D^2$ as the closed upper half-sphere $\bar H^2_+$.

Consider the map $f:S^2 \mapsto \mathbb{RP}^2$ as $f(x)=[x]$, that is, a point in $\bar H^2_+$ is mapped to the corresponding line passing through origin.

Let $g=f|_{\bar H^2_+}$, that is, $g$ is the restriction of $f$ on the closed upper-half sphere.

So we have a continuous map $g$ on $\bar H^2_+$ whose fibres are precisely the one point sets $\{x\}$ whenever $x$ is in the upper-half sphere and $\{x, -x\}$ whenever $x$ is on the equator.

Thus we get a continuous map $\tilde g:\bar H^2_+/\sim\ \mapsto \mathbb{RP}^2$ which is bijective and continuous. This map is in fact a homeomorphism because it is a map from a compact space to a Hausdorff space.

Antimony
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caffeinemachine
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  • this looks good! – Thomas Rot Aug 11 '15 at 07:05
  • How did you show $f$ is continuous? – Aaron Maroja Aug 04 '16 at 15:48
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    @AaronMaroja $\mathbf RP^2$ is usually defined as $S^2/\sim$, where $\sim$ identifies the antipodal points on $S^2$. So $f:S^2\to \mathbf RP^2$ is just the map which takes a point $x\in S^2$ to the equivalence class of $x$ under $\sim$. This is continuous by definition of quotient topology. If you want an explanation as to how this definition of $\mathbf RP^2$ is equivalent to saying that $\mathbf RP^2$ is the "set of all lines in $\mathbf R^2-{\mathbf 0}$", then I can explain that too in a future comment. – caffeinemachine Aug 04 '16 at 16:15