An integral homology circle is a CW-complex 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?



Integral homology circles
Example 4.35 of "Algebraic Topology" by Hatcher provides a counter-example. The space is $X=(S^1\vee S^n)\cup e^{n+1}$, where the $(n+1)$-cell is attached along the element $2t-1\in \pi\_n(S^1\vee S^n)\approx \mathbb{Z}[t,t^{-1}]$. On the level of cellular homology this has the same effect as attaching the cell along a degree $2-1=1$ map, so $X$ has the integral homology of $S^1\vee D^n\simeq S^1$. But we have $\pi\_{n}(X)\approx \mathbb{Z}[t,t^{-1}]/(2t-1)\approx \mathbb{Z}[\frac{1}{2}]$, 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).


