Prove that any continuous map $f:\mathbb{R}P^n\to \mathbb{R}P^m$, $mQuestion: Given a continuous map $f:\mathbb{R}P^n\to \mathbb{R}P^m$, where $m<n$, show that $f$ induces trivial homomorphism $f_\#:\pi_1(\mathbb{R}P^n)\to \pi_1(\mathbb{R}P^m)$.
$\mathbb{R}P^n$ stands for the $n$-dimensional real projective plane, $\pi_1(\mathbb{R}P^n)$ stands for its fundamental group.
My attempt: 
Consider the standard quotients $\pi_1:S^n\to \mathbb{R}P^n$ and $\pi_2:S^m\to \mathbb{R}P^m$, then $f\pi_1:S^n\to \mathbb{R}P^m$ lifts to $\tilde f:S^n\to S^m$, and since for any $x\in S^n$ there is $\pi_2\tilde f(x)=f\pi_1(x)=f\pi_1(-x)=\pi_2\tilde f(-x)$ we see that $\tilde f(x)=\tilde f(-x)$ or $-\tilde f(-x)$ must be true. The sets $B_1=\{x\in S^n\mid \tilde f(x)=\tilde f(-x)\}$ and $B_2=\{x\in S^n\mid \tilde f(x)=-\tilde f(-x)\}$ are both open by a convergence point sequence argument using the fact that the functions are all continuous. Note that $S^n=B_1\cup B_2$, one of $B_1,B_2$ must be empty. If $B_2$ is empty then by the universal property of quotient there exists $g:\mathbb{R}P^n\to S^m$ such that $g\pi_1=\tilde f$, therefore by the surjectivity of $\pi_1$ we see that $g$ is a lifting of $f$ and by the Lifting Criterion we are done. So it remains to show that either $B_2$ must be empty, or to give a lift of $f$ when $B_2$ is not empty. 
I don't know how to proceed. Any hint or solution is appreciated. Thanks in advance.
 A: I don't see how your argument would ever use the fact that $m < n$, so I doubt it can be completed to get your desired result.
The usual way to show this is cohomology rings. I outline the argument and leave you to complete it.
1) Calculate $H^*(\Bbb{RP}^n;\Bbb F_2) = \Bbb F_2[x]/(x^{n+1})$ with $|x| = 1$. (Here $\Bbb F_2$ is the field with two elements, which is isomorphic as a ring to $\Bbb Z/2$; I pass between these two notations as is convenient for the typesetting).
2) Show that the only graded homomorphism $\Bbb F_2[x]/(x^{m+1}) \to \Bbb F_2[x]/(x^{n+1})$ with $m < n$ is the homomorphism sending $x$ to $0$. In particular, your $f$ must be trivial on $\Bbb F_2$-cohomology.
3) Using the identification $H^1(X;\Bbb Z/2) = \text{Hom}(\pi_1 X, \Bbb Z/2)$ and the fact that $\pi_1 \Bbb{RP}^k = \Bbb Z/2$ for all $k > 1$, show using (b) that the map on fundamental groups is zero.
In the edge case $m = 1$, note that the induced map on fundamental groups is a map $\Bbb Z/2 \to \Bbb Z$, which is hence automatically zero. In the even edgier case $m = 0$ we just have $\Bbb{RP}^0 = *$, a point, so its fundamental group is trivial.
A: From the first $\mathbb{Z}/2$ cohomology of $\mathbb{R}P^n$ and the fact $\mathbb{R}P^\infty$ represents $\mathbb{Z}/2$ cohomology, we deduce that the only maps $\mathbb{R}P^n \rightarrow \mathbb{R}P^\infty$ up to homotopy are the inclusion and the constant map.
Suppose we had a map $\mathbb{R}P^n \rightarrow \mathbb{R}P^m$ which is nontrivial on $\pi_1$. Then we could post-compose with the inclusion into $\mathbb{R}P^\infty$ and deduce that this is homotopic to the inclusion  $\mathbb{R}P^n \rightarrow \mathbb{R}P^\infty$ since both factors of the composition are isomorphisms on $\pi_1$, and we've already shown the only map nontrivial on $\pi_1$ has to be the inclusion.
However, the inclusion induces isomorphisms on all homology groups up to dimension $n$. But if $n>m$, the map on $H_n$ then factors through $H_n(\mathbb{R}P^m ; \mathbb{Z}/2)=0$ which is a contradiction since $H_n(\mathbb{R}P^n; \mathbb{Z}/2)=\mathbb{Z}/2$. Hence, the map must be trivial on fundamental group.
A: I find my way. The Borsuk-Ulam Theorem tells that any continuous map $S^n\to \mathbb{R}^n$ must map some pair of antipodal points to a same point, hence we just embed $S^m$ into $\mathbb{R}^n$ using $m<n$ and we see that $B_1$ is not empty, so $B_2$ must be empty and by the universal property of quotient we are done.
Thanks for your answers anyway.
