# Prove that any continuous map $f:\mathbb{R}P^n\to \mathbb{R}P^m$, $m<n$, induces trivial homomorphism between fundamental groups

Question: Given a continuous map $$f:\mathbb{R}P^n\to \mathbb{R}P^m$$, where $$m, show that $$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.

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.

• Thanks for your answer, but I haven't learn cohomology yet...I tried to consider the homology groups with $\mathbb{Z}_2$ coefficients but did not see a clue. Do you have any idea of proving the question using homology groups? May 11 '20 at 13:25
• Uh, I come up with an answer using the Borsuk-Ulam Theorem. Just embedding $S^m$ into $\mathbb{R}^n$ and the theorem tells that $B_1$ is not empty, so $B_2$ must be empty. Thanks anyway. May 11 '20 at 13:46

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.

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 $$m and we see that $$B_1$$ is not empty, so $$B_2$$ must be empty and by the universal property of quotient we are done.