This is not a homework problem and I'm just doing it for fun.
The problem statement is:
Using the cup product structure, show there is no map $\mathbb{R}P^n \to \mathbb{R}P^m$ inducing a nontrivial map $H^1(\mathbb{R}P^m; \mathbb{Z}_2) \to H^1(\mathbb{R}P^n; \mathbb{Z}_2)$ if $n > m$.
Here is my thought: Assume there is such a map $f$. Using naturally of cup product, we have the square that stacks these two lines together commutes: $$ H^1(\mathbb{R}P^m; \mathbb{Z}_2) \times H^1(\mathbb{R}P^m; \mathbb{Z}_2) \to H^2(\mathbb{R}P^m; \mathbb{Z}_2) $$ and $$ H^1(\mathbb{R}P^n; \mathbb{Z}_2) \times H^1(\mathbb{R}P^n; \mathbb{Z}_2) \to H^2(\mathbb{R}P^n; \mathbb{Z}_2) $$ (Forgive me for not knowing how to load tikzcd packages in MSE).
I want to use the fact that $H^{\ast}(\mathbb{R}P^n; \mathbb{Z}_2) \cong \mathbb{Z}_2[\alpha]/(\alpha^{n + 1})$ and similarly for $m$, $H^{\ast}(\mathbb{R}P^m; \mathbb{Z}_2) \cong \mathbb{Z}_2[\beta]/(\beta^{m + 1})$.
This leads me to discover that on each level $k < n + 1$, $H^k(\mathbb{R}P^n; \mathbb{Z}_2) \cong \mathbb{Z}_2(\alpha^k) \cong \mathbb{Z}_2$. I feel like something related to quadratic forms might be related but I'm a beginner so don't know what I'm talking about.
Thanks!!!
let $m<n$. Then since $H^* (\mathbb R P^{k})=\mathbb Z_2[x]/(x^{n+1})$ for any $k$, we can let $\alpha, \beta$ be generators for the cohomology ring of $H^* (\mathbb R P^{m})$ and $H^* (\mathbb R P^{n})$ respectively, then suppose that there is a nontrivial homomorphism $h^*:H^1(\mathbb RP^m) \to H^1(\mathbb RP^n)$. It follows that $h^*(\alpha)=\beta$.
Then $$0={h}^*(\alpha^{m+1})=h^*(\alpha)^{m+1}=\beta^n \neq 0,$$ a contradiction.