Covering map of the torus boundary onto itself, lifting from the projective space I'm trying to prove this rather simple fact that the following mapping is a covering and to check it's rank
$p\colon S^1\times S^1\rightarrow S^1\times S^1$; $p(z_1,z_2)=(z_1^2,z_2^3).$
So I think that rank is 6 - $p^{-1}(a,b)$, for every two $z_1$ that would give $a$ after taking a square, there are three $z^2$ that would give $b$ - so there are 6 such points.


*

*$p$ being a surjection is obvious. Being a covering also, but I'm getting stuck trying to give a precise argument.. How to formulate it?

*Does every $f\colon\mathbb{RP}^2\rightarrow S^1\times S^1$ lift with respect to this $p$? I guess that yes, and maybe it doesn't even depend on the domain - projective space. It suffices to take 
$(e^{i\alpha /2},e^{i\beta /3})$, where $(e^{i\alpha},e^{i\beta})$ is the the image of $f$,
as the image of $h\colon\mathbb{RP}^2\rightarrow S^1\times S^1$, and then $ph=f$. I hope it's clear what I want to say. Is that correct?
Thanks
 A: To show that $p$ is a covering, remark that the differential of $p$ at any point is an isomorphism, the inverse function theorem implies that for $x\in S^1\times S^1$, there exists a neighborhood $U_x$ of $x$ such that $f_{\mid U_x}$ the restriction of $U$ is a diffeomorphism onto its image. Since $S^1\times S^1$ is compact, the fibers have finite cardinal, write $f^{-1}(y)=\{x_1,..,x_6\}$, you can find neighborhood $U_{x_i}, i=1,..,6$ and eventual shrink them such that $U_{x_i}\cap U_{x_j}$ is empty if $i\neq j$ such that the restriction of $f$ to $U_{x_i}$ is a diffeomorphism onto its image.
A: For your second question (in general this holds under the  wikipedia hypothesis), a continous map $f : Z \to X$ lifts if and only if $f_*(\pi_1(Z,z)) \subset p_*(\pi_1(Y, y))$ where $p : Y \to X$ is your covering map. 
Here $\pi_1(\mathbb{RP}^2) = \mathbb{Z}/2\mathbb Z$ and $\pi_1(X) = \mathbb Z^2$. Since there is no morphisms $\mathbb{Z}/2\mathbb Z \to \mathbb Z^2$, $f_*(\pi_1(Z,z)) = 0$ in your case so there is always a lift. 
