First homology groups are isomorphic but fundamental groups aren't 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 $\pi_1$. 
 A: 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. 
A: 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.
A: For the first question:
An acyclic space $X$ works (see Hatcher example 2.38).  You could consider $Y=S^2$.
