An integral homology circle is a CWcomplex whose integral homology groups are isomorphic to those of the circle.
If $X$ is an integral homology circle with $\pi_1(X)=\mathbb{Z}$, must $X$ be homotopically equivalent to a circle?
An integral homology circle is a CWcomplex whose integral homology groups are isomorphic to those of the circle. If $X$ is an integral homology circle with $\pi_1(X)=\mathbb{Z}$, must $X$ be homotopically equivalent to a circle? 


Example 4.35 of "Algebraic Topology" by Hatcher provides a counterexample. The space is $X=(S^1\vee S^n)\cup e^{n+1}$, where the $(n+1)$cell is attached along the element $2t1\in \pi_n(S^1\vee S^n)\approx \mathbb{Z}[t,t^{1}]$. On the level of cellular homology this has the same effect as attaching the cell along a degree $21=1$ map, so $X$ has the integral homology of $S^1\vee D^n\simeq S^1$. But we have $\pi_{n}(X)\approx \mathbb{Z}[t,t^{1}]/(2t1)\approx \mathbb{Z}[\frac{1}{2}]$, so $X$ is not homotopy equivalent to $S^1$. However if $X$ is an integral homology circle with $\pi_1(X)=\mathbb{Z}$ and $\pi_1(X)$ acts trivially on $\pi_i(X)$ for all $i\geq 2$, then $X$ is homotopy equivalent to a circle (Proposition 4.74 of AT). 

