# Does the isomorphic of the fundamental groups imply the existence of a mapping inducing an isomorphism?

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?

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?
• Do you mean $g:Y\to X$? Nov 10, 2021 at 9:06
• 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. Nov 10, 2021 at 17:06
• @PierrePC Yes, I corrected it, thanks. Nov 10, 2021 at 20:30

No and no. For an explicit counterexample to 1. (which is also a counterexample to 2.) take the map $$\mathbb{R}P^2\to \mathbb{R}P^{\infty}$$.
• You can replace $\mathbb{R}P^\infty$ by $\mathbb{R}P^3$. Nov 10, 2021 at 6:45
• 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 7:31
• 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 21:48
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.