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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.

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    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.2012-12-12

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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.

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    @TheFanaticalMathematical Of course. But the question does not require a map to be surjective.2012-12-12