Skip to main content
12 events
when toggle format what by license comment
Dec 5, 2021 at 5:44 comment added Arshak Aivazian @AchimKrause Thanks a lot!
Nov 25, 2021 at 12:42 comment added Achim Krause Hatcher, for example. The idea is to obstruct the existence of such a map by looking at the effect on cohomology rings.
Nov 25, 2021 at 4:36 comment added Arshak Aivazian @AchimKrause Still, I fail to understand why there is no continuous mapping from $\mathbb{R}P^3$ to $\mathbb{R}P^2$ inducing an isomorphism of fundamental groups. Can you tell me which textbook to look at?
Nov 10, 2021 at 21:48 comment added Achim Krause Here's a counterexample to that version as well. Write $(BC_p)^n$ for the n-skeleton of the standard CW structure on $BC_p$. Let $p, q$ be two different primes. Then $(BC_p)^2 \times (BC_q)^4$ and $(BC_p)^4 \times (BC_q)^2$ have the same fundamental group, but there's no map either way that induces an isomorphism on it.
Nov 10, 2021 at 21:44 comment added Achim Krause That's a weird thing to expect, don't you think? That there's always a map in at least one direction, but you don't know beforehand in which?
Nov 10, 2021 at 20:34 comment added Arshak Aivazian @HenrikRüping I beg your pardon, of course, I meant there is a mapping in some direction inducing an isomorphism
Nov 10, 2021 at 7:31 comment added HenrikRüping It is also a counterexample to (3). There is no mapping from $\mathbb{R}P^3$ to $\mathbb{R}P^2$ inducing an isomorphism on $\pi_1$.
Nov 10, 2021 at 7:01 comment added Arshak Aivazian Thanks! I have added a 3rd question that matches the title exactly, maybe less trivial.
Nov 10, 2021 at 6:45 vote accept Arshak Aivazian
Nov 10, 2021 at 6:45 comment added Achim Krause You can replace $\mathbb{R}P^\infty$ by $\mathbb{R}P^3$.
Nov 10, 2021 at 6:41 comment added Arshak Aivazian Thanks a lot! I need to think a little about your answer. Are there any counterexamples among finite-dimensional complexes?
Nov 10, 2021 at 6:39 history answered Achim Krause CC BY-SA 4.0