Let $RP$ denote the real projective plane and let $S$ denote the circle. I recently encountered a question of the form:
Prove every map $f:RP \to S\times S$ is null-homotopic. I know that every map $k:RP \to S$ is null-homotopic. Is it valid to argue that $f: RP \to S\times S$ is a product of maps $g,h:RP\to S, x \mapsto (g(x),h(x))$, and since $g$ and $h$ are nullhomotopic, then $f$ is nullhomotopic?
Yes, your argument is correct.
(I'd do it as follows : let $f : \Bbb RP^2 \to S^1 \times S^1$ be your map. Then the induced map on fundamental groups $\pi_1 f : \Bbb Z/2 \to \Bbb Z^2$ is the zero map, as $\Bbb Z^2$ is torsion-free. Thus, by map lifting lemma, you can lift $f$ to the universal cover to get a map $\widetilde{f} : \Bbb RP^2 \to \Bbb R^2$. As $\Bbb R^2$ is contractible, this is nullhomotopic. Pushdown the homotopy to the base space to conclude that $f$ is nullhomotopic)