7
$\begingroup$

A pair of continuous mappings $f \colon X \to Y$ and $g \colon Y \to X$ is called $\pi_1$-equivalence if they induce mutually inverse isomorphisms of fundamental groups. Spaces are called $\pi_1$-equivalent if there is $π_1$-equivalence between them.

Let $X, Y$ be CW-complexes

  1. Is it true that if $f \colon X \to Y$ induces an isomorphism of fundamental groups, then $X$ and $Y$ are $π_1$-equivalent?
  2. Is it true that if $\pi_1(X)$ is isomorphic to $\pi_1(Y)$, then $X$ and $Y$ are $\pi_1$-equivalent?

(added later)

  1. Is it true that if $\pi_1(X)$ is isomorphic to $\pi_1(Y)$, then there is of a mapping $f \colon X \to Y$ inducing an isomorphism or there is of a mapping $g \colon Y \to X$ inducing an isomorphism?
$\endgroup$
3
  • $\begingroup$ Do you mean $g:Y\to X$? $\endgroup$
    – Pierre PC
    Nov 10, 2021 at 9:06
  • 1
    $\begingroup$ The area addressing these questions is commonly known as "obstruction theory". As Achim Krause's answer shows, it is a much more subtle problem than this. $\endgroup$ Nov 10, 2021 at 17:06
  • $\begingroup$ @PierrePC Yes, I corrected it, thanks. $\endgroup$ Nov 10, 2021 at 20:30

2 Answers 2

21
$\begingroup$

No and no. For an explicit counterexample to 1. (which is also a counterexample to 2.) take the map $\mathbb{R}P^2\to \mathbb{R}P^{\infty}$.

$\endgroup$
10
  • 4
    $\begingroup$ You can replace $\mathbb{R}P^\infty$ by $\mathbb{R}P^3$. $\endgroup$ Nov 10, 2021 at 6:45
  • 2
    $\begingroup$ 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 $\pi_1$. $\endgroup$ Nov 10, 2021 at 7:31
  • 2
    $\begingroup$ 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? $\endgroup$ Nov 10, 2021 at 21:44
  • 4
    $\begingroup$ 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. $\endgroup$ Nov 10, 2021 at 21:48
  • 1
    $\begingroup$ Hatcher, for example. The idea is to obstruct the existence of such a map by looking at the effect on cohomology rings. $\endgroup$ Nov 25, 2021 at 12:42
5
$\begingroup$

As a consequence of Van Kampen's Lemma, in the special case where $X,Y$ are finite 2-dimensional CW-complexes then the answer is yes to all 3.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.