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Can someone help me find counter examples to the following:

  • $H_1(X) \cong H_1(Y) \implies \pi_1(X) \cong \pi_1(Y)$

  • $\pi_1(X) \cong \pi_1(Y) \implies H_1(X) \cong H_1(Y)$.

I'm aware of examples that show that the implications are false for the homology groups of some dimension, all my examples have isomorphic first homology groups however.

It interests me because I'm aware that the first homology group is isomorphic to the abelianization of $\pi_1$.

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  • $\begingroup$ For the first, you can find a non-abelian group $G$, take its abelianization $G^{ab}$, and then consider $BG$ and $BG^{ab}$: $H_1(BG) = \pi_1(BG)^{ab} = G^{ab} = (G^{ab})^{ab} = \pi_1(BG^{ab})^{ab} = H_1(BG^{ab})$, but $G\not\cong G^{ab}$. For the second, you'll need to look at non-path-connected spaces, as for path connected spaces you have $\pi_1(X)^{ab} = H_1(X)$, so functoriality of abelianization implies $H_1(X)\cong H_1(Y)$ if $X$ and $Y$ are path connected with $\pi_1(X)\cong\pi_1(Y)$. $\endgroup$
    – Stahl
    Nov 7, 2016 at 7:20
  • $\begingroup$ Doesn't the fact that $H_1$ is the abelianization of $\pi_1$ answer your questions? For a counterexample to the first one, find nonisormorphic groups with the same abelianization (for example, the trivial group and a nontrivial simple group) and use those for $\pi_1(X)$ and $\pi_1(Y)$. The second one always holds by that abelianization fact. $\endgroup$
    – Greg Martin
    Nov 7, 2016 at 7:21

3 Answers 3

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For the first one, people have already answered, for example $X = S^1 \vee S^1$ vs. $X = S^1 \times S^1$.

For the second one you need to be careful about base points! In general "the fundamental group of a space" is not well-defined, you need a base point.

For example if you take $X = \{*\} \sqcup S^1$ (the disjoint union of a singleton and a circle) and $Y = \{*\}$ (a singleton), then $\pi_1(X,*) \cong \pi_1(Y,*)$ are both the trivial group, but $H_1(X) = \mathbb{Z}$ while $H_1(Y) = 0$. It is not even true that if for all $x \in X$ and $y \in Y$ you have $\pi_1(X,x) \cong \pi_1(Y,y)$ then $H_1(X) \cong H_1(Y)$, consider for example $X = S^1$ and $Y = S^1 \sqcup S^1$.

However if the two spaces are path-connected, then $H_1(X)$ is the abelianization of $\pi_1(X)$ and $H_1(Y)$ is the abelianization of $\pi_1(Y)$ by Hurewicz's theorem (for any choice of base points), and so if $\pi_1(X) \cong \pi_1(Y)$ then $H_1(X) \cong H_1(Y)$.

Alternatively you can work with groupoids (if you know what that is). If the fundamental groupoid $\pi(X)$ is isomorphic to the fundamental groupoid $\pi(Y)$, then $H_1(X) \cong H_1(Y)$, again by Hurewicz's theorem.

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  • $\begingroup$ Good answer! :) but if the two spaces are path connected the second claim is true, isn't it? $\endgroup$
    – InsideOut
    Nov 7, 2016 at 9:11
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    $\begingroup$ @Gianluca Yes, for path-connected spaces it's true. $\endgroup$
    – Najib Idrissi
    Nov 7, 2016 at 12:17
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For the first implication a counterexample is:

$X=T^2$ and $Y=S^1\vee S^1$, they have non isomorphic fundamental group, but the same first homologous group.

The second is always true, since isomorphic groups have the same (read isomorphic) abelian subgroup generates by commutators.

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For the first question:

An acyclic space $X$ works (see Hatcher example 2.38). You could consider $Y=S^2$.

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