Prove or disprove any continous map $f$ from $T^2$ to $RP^2$ is null-homotopic.

Question

I tried to solve the following question: Prove or disprove any continous map f from $T^2$ to $RP^2$ is null-homotopic. We know the universal cover of $RP^2$ is $S^2.$ I want to construct a map $g$ from $T^2$ to $S^2.$ Then, I will deduce that since $S^2$ is not contractible, then $g$ is not null-homotopic. How do you construct such a map?

Answer 1

With a little more work your idea works fine to construct a non-null-homotopic map. This is a different idea than the one Amitesh Datta is suggesting in the comments.

Pick a small ball in $T^2$, and collapse everything outside it. This gives a map $f: T^2 \to S^2$ which induces an isomorphism on second homology, and is therefore not null-homotopic. Compose with the projection $S^2 \to \Bbb{RP}^2$.

I claim that the map $pf: T^2 \to \Bbb{RP}^2$ is not null-homotopic. By construction, $pf$ lifts to a map $f: T^2 \to S^2$, so we can lift homotopies (and in particular, null-homotopies) of $pf$ by the homotopy lifting property, which covering spaces satisfy. The endpoint of this lift would be a map $T^2 \to S^2$ whose image is contained in two points (the inverse image of a point in $\Bbb{RP}^2$ under $p$); because $T^2$ is connected, the image is contained in a single point.

Thus we've constructed a null-homotopy of $f$, which is impossible as $f$ induces an isomorphism on second homology - it can't be null-homotopic. So $pf$ was not null-homotopic.