Why does there exist a lift from $RP(2)$ to $S^2$? Let $p$ be the quotient map from $S^2$ to $RP(2)$. I am looking for a proof that there exists a map
$g \colon RP(2) \to S^2$
such that
$p(g(x)) = x$ for all $x \in RP(2)$.
 A: There doesn't . As the quotient map is a covering projection and the spaces are Hausdorff and locally path connected, there is a lift if and only if the image of the fundamental group of RP(2) under the identity $Z_2$ is a subgroup of the image of the fundamental group of $S^2$ which is trivial.
A: I guess this is essentially equivalent to Charlie's excellent answer, but perhaps provides an alternative perspective:
Note that this follows from the functoriality of the fundamental group. If $g:\mathbb{RP}^2\to \mathbb{S}^2$ existed with the stated property, then $(p\circ g)_{\ast}$ would be the identity on the fundamental group of $\mathbb{RP}^{2}$ yet $p_{\ast}\circ g_{\ast}$ would be $0$ (since it factors through the fundamental group of $\mathbb{S}^2$, which is trivial). Finally, the identity on the fundamental group of $\mathbb{RP}^{2}$ is non-zero since the fundamental group of $\mathbb{RP}^{2}$ is non-zero. Q.E.D.
Note also that this is, in some sense, analogous to the non-existence of an extension of the identity $i:\mathbb{S}^1\to \mathbb{S}^1$ to a map $\mathbb{D}^2\to \mathbb{S}^1$ (or, at least the proof is).
I hope this helps!
