I am working on Weibel's K-Book and when defining higher K-Theory for a ring via $BGL(R)^+$, I have encountered a question concerning a homology $n$-sphere.

The statement I want to show is the following:

Let $X$ be a homology $n$-sphere, i.e., a space with $$H_\ast (X) = H_\ast (S^n). $$ Show that there is a homotopy equivalence $S^n \rightarrow X^+$, when $X \rightarrow X^+$ is the $+$-construction on $X$ relative the perfect radical of $\pi_1 (X)$.

For $n \neq 1$ this is not much of a problem, since from

$$0 = H_1 (X) \cong \pi_1 (X) / [\pi_1 (X), \pi_1 (X)]$$

I can conclude that $\pi_1(X)$ is a perfect group and hence $X^+$ is simply connected since I am interested in the $+$-construction relative the perfect radical of $\pi_1 (X)$.

Is there a similar way I get information about the fundamental group of $X^+$ for $n = 1$?

Thank you.

  • $\begingroup$ For $n = 1$ what do you even mean by $X^{+}$? You need to pick a perfect normal subgroup of $\pi_1(X)$ for this to make sense, but a priori $\pi_1(X)$ may not have one. $\endgroup$ – Qiaochu Yuan Apr 3 '15 at 0:04
  • $\begingroup$ I am interested in the +-construction relative the perfect radical of $\pi_1(X)$. In one of the exercises one shows that every group has a unique largest perfect normal subgroup, called perfect radical. $\endgroup$ – Chris Apr 3 '15 at 9:06

A homology $n$-sphere is a smooth closed $n$-dimensional manifold $X$ with $H_*(X) = H_*(S^n)$. In particular, a homology $1$-sphere is a smooth closed one-dimensional manifold, but the only such manifold is $S^1$. As $\pi_1(S^1) = \mathbb{Z}$ is abelian, the perfect radical is $\{0\}$ so $(S^1)^+ = S^1$.

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