# How to construct a map from $\mathbb{ S}^2=\{(a_1,a_2,a_3) | a_1^2+a_2^2+a_3^2=1\}$ to $\mathbb{ RP}^2$?

How would I construct the map? Once constucted, would I be right in saying that there is no Diffeomorphism to map back? As in $\mathbb{RP}^2$ a closed curve would have to have either $2$ points that represent the same $f(x)$ or one point that has $2$ values for $f(x)$? (It's not bijective)

Any advice is appreciated.

-
Easy: just map every point of $\Bbb{S}^2$ to the same point of $\Bbb{RP}^2$. –  Chris Eagle Dec 12 '12 at 11:00
You are correct that the two are not diffeomorphic - one is orientable, and the other is not. It may be possible to find a smooth bijection $\mathbb{S}^2\to\mathbb{RP}^2$ though, it just won't have a smooth inverse. I'm not sure if such a thing exists though. –  Matthew Pressland Dec 12 '12 at 11:17
@MattPressland: A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. –  Chris Eagle Dec 12 '12 at 11:44
@ChrisEagle Ah, indeed. Very nice! –  Matthew Pressland Dec 12 '12 at 11:49
You can construct a map $f\colon S^2 \rightarrow RP^2$ easily. Just choose any point $p$ of $RP^2$ and define $f(x) = p$ for every $x \in S^2$. It is not only continuous, but also smooth. If you mean a bijective smooth map, there is no such thing because they are not homeomorphic. –  Makoto Kato Dec 12 '12 at 12:41

## 1 Answer

There is no global diffeomorphism from $S^2$ to $\mathbb{R}P^2$. This can be seen from the fact that their top de Rham cohomology groups are not isomorphic. Indeed, $H^2(S^2)=\mathbb{R}$ and $H^2(\mathbb{R}P^2)=0$.

As Chris Eagle pointed out, you cannot even have a continuous bijection between them, since their fundamental groups disagree. $\pi(S^2)=0$ while $\pi(\mathbb{R}P^2)=\mathbb{Z}_2$.

That said, there exists local diffeomorphisms between them. The "natural" map from $S^2$ to $\mathbb{R}P^2$ is the projection map $\pi$ with respect to the equivalence relation $x\sim -x \quad \forall x\in S^2\subset \mathbb{R}^3$, in the sense that $\pi$ is the coequalizer of the identity and the antipodal map on $S^2$. This $\pi$ is a local diffeomorphism.

-
I appreciate the responses. Makoto Kato, your map doesn't contain every point of the projective space in the image of f(x), would I be right in saying that a constant say c1*(a1,a2,a3) as because S^2 has magnitude 1 and RP is infinite in that sense? –  The Fanatical Mathematical Dec 12 '12 at 15:32
@TheFanaticalMathematical Of course. But the question does not require a map to be surjective. –  Makoto Kato Dec 12 '12 at 20:30