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I've Been trying to prove for many hours now that given a (continuous) map $f:RP^n\to RP^m$ (where $n>m>0$) the induced map $f_* : \pi_1(RP^n)\to \pi_1(RP^m)$ is trivial.

I've seen a few descriptions for real project spaces, the one I have been trying to use is that where describes the space as a quotient of an n-disk with antipodal points on the boundary identified (the description I've seen talks about a closed hemisphere but since they are homeomorphic that's the same) (here's a link where this was explained). so basically this means that $RP^{n+1}$ would look like an n+1-cell glued to $RP^n$ (that is of course if I understand correctly).

Right now I'm trying to show the statement for $n+1$ and $n$ (I think the rest would follow rather simply since one could use the above description inductively) (I've also deduced the case where $m=1$ which is fairly simple since $f_*$ would be a homomorphism between $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}$ and must therefore be trivial) I've tried a few different directions-

1- assuming the statement is false (since the fundamental group of $RP^n$ is $\mathbb{Z}/2\mathbb{Z}$ for $n>1$) that $f_*$ is the identity map which would mean that some loop not homotopic to a point in $RP^{n+1}$ maps to another loop not homotopic to a point. I do not have a grasp of what a non-trivial loop might look like in the real projective plane might "look like".

My intuition seems to be that this means something along the lines of $RP^n$ as a subspace of $RP^{n+1}$ must map into itself for obvious reasons but for some reason this would mean that the disk part of $RP^{n+1}$ would have to "tear" to allow this to happen.

2- I've tried to use relative homology (since the fundamental groups are abelian they are the same as the first homology groups so I wanted to leverage exact sequences to my advantage) but I seem to be doing something wrong. Using relative homology I get the sequence $$H_2(RP^{n+1},RP^{n})\to H_1(RP^n)\to H_1(RP^{n+1})\to H_1(RP^{n+1},RP^{n})$$ where $H_i(RP^{n+1},RP^{n})=H_i(S^{n+1})$ (that is of course if I understand correctly how relative homology works correctly) I'm not sure how to work $f$ into it though. I thought about replacing $H_1(RP^{n+1})$ in some way but I quickly lose exactness.

I'd appreciate some kind of hint or clarification of something I might be misunderstanding

MrP
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    Are you allowed to use cohomology and cup product structure? If yes, then it's very direct. – Kevin.S Jan 26 '21 at 01:40
  • I don't know anything about cohomology and cup product structures, If you could point me towards reference material on the topic I'll go read up. Thanks a lot! – MrP Jan 26 '21 at 01:42
  • That argument uses a corollary of UCT and the cohomology ring of projective space to deduce that the induced map on cohomology is trivial which is dual to the map of homology, then the Hurewicz theorem passes this to the fundamental group. This should be covered in most topology texts. Anyway, since this tool hasn't been introduced so far, I think the induced map of spheres may work, but I'm not sure the details of the proof. – Kevin.S Jan 26 '21 at 02:04
  • That's whats the question says. Is it possible that $f$ described in your question doesn't exist for $n \neq m$? because that could settle it. – MrP Jan 26 '21 at 02:35
  • @Oiler I think the question is correct. One can prove this using the method mentioned in my second comment (under coefficient $\Bbb{Z}_2$). I feel confident about that. The question you mentioned probably have some other conditions. – Kevin.S Jan 26 '21 at 08:30
  • @Oiler have a look at the conditions you need for such a map $f$ to exist... – Tyrone Jan 26 '21 at 14:34
  • @Oiler I think that the phrasing for your comment is a little misleading. There is no odd map $S^{n} \to S^{n-1}$, and hence any odd map $S^n \to S^{m}$ for $m<n$. If there were such a map, then it would induce a nontrivial map on $\pi_1$, so I would be a little careful about this. – Andres Mejia Jan 26 '21 at 21:40

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