As in the title, here's the statement of the question:
Let $m>n$ be two natural numbers and let $f: \mathbb{RP}^m \to \mathbb{RP}^n$ be a continuous map. Show that $f_{*}: \tilde{H_{i}}(\mathbb{RP}^m) \to \tilde{H_{i}}(\mathbb{RP}^n)$ are trivial homomorphisms for all $i \in \mathbb{Z}$, where $\tilde{H_{i}}$ denotes the reduced homology.
Here are some hints I've got and ideas I came up with.
Let $p: S^{n} \to \mathbb{RP}^n$ denote the standard two-sheeted covering (I'll also denote the $m$-dimensional covering by $p$). Then, $f \circ p:S^{m} \to \mathbb{RP}^n$ can be lifted to a map $g: S^m \to S^n$, because $S^{m}$ is simply connected. Furthermore, since $p \circ g = f \circ p$, we see that $g(x) = g(-x)$ or $g(x) = -g(-x)$ for all $x \in S^{m}$. Since $S^{m}$ is connected and Hausdorff, it can be proven pretty easily that $g$ is either odd or even. Now, by the Borsuk-Ulam theorem, $g$ must be even.
Now, the idea is to take the commutative diagram involving $f$, $g$ and the two $p$'s and maybe derive the statement form there. Obviously, for $i$ even, $f_{*}$ can only be the trivial homomorphism, so I'm interested in the cases when $i$ is odd.