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I have the following problem:

Let $n>m>0$, show that every map $f:\mathbb{RP}^n\to\mathbb{RP}^m$ induces the trivial map on the fundamental groups.

I paste the given solution below: enter image description here Now, this solution looks wrong to me (or at least missing a lot of details). The reason is that the Smith sequence exists only if we take $\mathbb{Z}_2$ as coefficient ring, but we are trying to show this fact with coefficients in $\mathbb{Z}$ (where we have the Hurwics isomorphism, as stated).

Is there a way to save this solution as it is? Or an alternative proof? I tried working one out by myself, but I didn't find anything for the moment. I will update if I find something.

Daniel Robert-Nicoud
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    For $m \ge 2$ you only need coefficients in $\mathbb{Z}_2$ because the fundamental group is $\mathbb{Z}_2$, and you can argue the case $m = 1$ separately. – Qiaochu Yuan Aug 22 '14 at 16:47
  • @QiaochuYuan Ah, I see. Do I have to argue by universal coefficient theorem to show that I have an isomorphism $H_1(\mathbb{RP}^n;\mathbb{Z}_2)\cong\pi_1(\mathbb{RP}^n)$? – Daniel Robert-Nicoud Aug 22 '14 at 16:57

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Suppose it induced a nontrivial map. Then it induces an isomorphism, since both have fundamental group $\mathbb Z_2$. By the Hurewicz theorem, this induces an isomorphism on integral homology (fundamental group is already abelian), and by the universal coefficient theorem, we see that $f^*:H^1( \mathbb RP^m) \to H^1(\mathbb RP^n)$ is an isomorphism, but this is impossible, since $0=f(\alpha^{m+1})=f(\alpha)^{m+1}=\beta^{m+1} \neq 0$, where $\alpha, \beta$ are generators for the cohomology rings $\mathbb Z_2[\alpha]/\alpha^{m}$ and $\mathbb Z_2[\beta]/\beta^{n}$.

Andres Mejia
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