$\newcommand{\RP}{\mathbf RP}$ 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.


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).

  • $\begingroup$ 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? $\endgroup$ – caffeinemachine Aug 10 '15 at 8:02
  • $\begingroup$ Yes. I was looking for a formal proof. Of course, if some visual insight comes along with it then it is much better. $\endgroup$ – caffeinemachine Aug 10 '15 at 9:25
  • $\begingroup$ 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? $\endgroup$ – caffeinemachine Aug 10 '15 at 10:39
  • $\begingroup$ @caffeinemachine: update $\endgroup$ – Thomas Rot Aug 10 '15 at 11:05
  • $\begingroup$ Thanks. I am checking that the map $S^2\to D^2/\sim$ is continuous. It looks promising. $\endgroup$ – caffeinemachine Aug 10 '15 at 11:45

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 $f:S^2\to \RP^2$ 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 $\tilde g:\bar H^2_+/\sim\ \to \RP^2$ 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.

  • $\begingroup$ this looks good! $\endgroup$ – Thomas Rot Aug 11 '15 at 7:05
  • $\begingroup$ How did you show $f$ is continuous? $\endgroup$ – Aaron Maroja Aug 4 '16 at 15:48
  • 1
    $\begingroup$ @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. $\endgroup$ – caffeinemachine Aug 4 '16 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.