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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?
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  • $\begingroup$ Do you mean $g:Y\to X$? $\endgroup$
    – Pierre PC
    Commented Nov 10, 2021 at 9:06
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    $\begingroup$ 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. $\endgroup$
    – Denis Nardin
    Commented Nov 10, 2021 at 17:06
  • $\begingroup$ @PierrePC Yes, I corrected it, thanks. $\endgroup$
    – Arshak Aivazian
    Commented Nov 10, 2021 at 20:30

2 Answers 2

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

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

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