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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?

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up vote 21 down vote accepted

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).

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Thank you Tom!! – Mark Grant Mar 14 '12 at 18:22

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