I know that $S^2$ cannot cover the torus $\Bbb T^2 \cong \Bbb S^1\times \Bbb S^1$ because $S^2 \not \cong \Bbb R^2$ and both are universal coverings of $\Bbb T^2$
Conversely, is there a covering $p:\Bbb T^2 \to \Bbb S^2$?
Such a covering would induce a trivial injective morphism $p_*:\pi_1(\Bbb T^2) \to \pi_1(\Bbb S^2)$, therefore all loops in $\Bbb T^2$ would be nullhomotopic, but $\Bbb T^2$ is not simply connected.
Is it a correct argument?
Thank you for you help and comments.
I don't know if I fully follow your argument for why $S^2$ cannot cover $T^2$. You mean to say that since $S^2$ is simply connected, it would be a universal cover of $T^2$ if there were a covering map $\rho:S^2\to T^2$, but you know that $\mathbb{R}^2$ is a universal cover of $T^2$, that universal covers are homeomorphic, and $S^2$ is not homeomorphic to $\mathbb{R}^2$?
For the converse, a covering map $\rho:T^2\to S^2$ would not induce an injective homomorphism $\pi_1(T^2)\to\pi_1(S^2)$. That's the end of it; that's why it can't happen. There are no injective group homomorphisms $\mathbb{Z}^2 \to 0$. (Any covering map induces an injective map on fundamental groups by uniqueness of path lifting.)