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.
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 $(S^1)^+ = S^1$.