Checking Existence of Coverings Let T denote the torus. I am currently trying to determine the existence of a few covering maps: $\mathbb{R}^2 \to S^2$, $\mathbb{R}^2 \to T$, $S^2 \to \mathbb{R}^2$, $S^2 \to T$, $T \to S^2$, $T \to \mathbb{R}^2$.
I believe that coverings do not exist for $T \to S^2$ and $T \to \mathbb{R}^2$. I think this is true because this video at 45:07 which states that $\pi_1(X, x)$ is a subgroup of $\pi_1(B, b)$ when X covers B. We know that $\pi_1(\mathbb{R}^2) \simeq \pi_1(\mathbb{R}) \times \pi_1(\mathbb{R}) \simeq 0$, $\pi_1(S^2) = 0$, $\pi_1(T) \simeq \mathbb{Z} \times \mathbb{Z}$. From this, we can see that this is $T$ cannot cover either of the spaces. Additionally, $\mathbb{R}^2$ and $S^2$ cannot cover each other as they are both simply-connected.
Is my reasoning behind this correct?
 A: I think here you have to use this result about covering theory:

*

*Every path connected, locally path connected, semi locally simple connected space admits a universal cover (I.e a simply connected space) up to isomorphism;

The uniqueness of the point $1$ tells us the first and the third map $\mathbb{R}^2 \to S^2$ and $S^2\to \mathbb{R}^2$ cannot exist.
The second map exists, and it’s exactly the universal cover of $T$, defined in the following way. Take the two isometries of $\mathbb{R}^2$ given by $\phi_1(x):=x+e_1$ and $\phi_2(x):=x+e_2$ And the lattice $\Lambda:=\langle \phi_1,\phi_2\rangle $.
Then $\mathbb{R}^2\to \mathbb{R}^2/\Lambda\cong T$ is a universal covering.
Looking at the fundamental domain $[0,1]^2$ of $\mathbb{R}^2/\Lambda$ should be clear that is exactly the torus $T$.
This means that the universal covering of $T$ is $\mathbb{R}^2$.
Thus the fourth map cannot exists because $\mathbb{R}^2$ is the universal covering of $T$ and not $S^2$.
Moreover the fifth map $T\to S^2$ cannot exists  because if you compose with respect the universal cover of $T$, then you have a covering map from $\mathbb{R}^2\to S^2$, that is not possible.
The last map is quite different to prove the non existence. Here you have to take the universal covering map $\mathbb{R}^2$ of $T$ and compose with $T\to \mathbb{R}^2$. Then you have a covering map between simply connected spaces $\mathbb{R}^2 \to \mathbb{R}^2$ and so has to be an isomorphism, so that $T\to \mathbb{R}^2$ is an Iso and it’s not possible.
Another simple reason is that $T$ is compact and so $T\to \mathbb{R}^2$ can’t be surjective, that contradicts the fact that is a covering map.
