I have a corollary stating:
if X is contractible then $H_0(X) = \mathbb{R}$ and $H_n(X) = \{0\}$ for $n>0$.
But $\mathbb{S}^n$ is contractible and $H_n(\mathbb{S}^n) = \mathbb{R}$.
I must be missing something.
Thank you!
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8It’s not contractible! – Ted Shifrin Dec 19 '18 at 22:52
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If contractible means any loop on, say $\mathbb{S}^2$, is homotopic to a point, that is true for $\mathbb{S}^2$ , isn't it ? – Conjecture Dec 19 '18 at 22:55
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5I think you are confusing contractible with simply connected. – positron0802 Dec 19 '18 at 22:57
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Right! Thank you all, I will double check. – Conjecture Dec 19 '18 at 22:58
