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I was reading this post, in the anaswer that used Mayer-Vietoris sequence, Paul separated the Warsaw circle into two groups, one containing the topologist sine curve ($X_2$) and one does not ($X_1$). Later in the Mayer-Vietoris sequence, it seems he concluded $H_n(X_1) \oplus H_n(X_2) = 0$. It is clear that $H_n(X_1) = 0$ for $n > 0$ as it is contractible. However, it raised my doubt about whether $H_n(X_2) = 0$,i.e. the topologist's sine curve.

Thus, my questions is, if a space does not contain nontrivial $S^n$ (and in particular $S^1$) as its subspace, does it have to have trivial homology groups? In particular, if one want to prove $H_n(X_2)=0$ for $n \geq 1$, how might one go about doing that? (Or did he use some other arguments?)

Frank
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The whole secret is that the space $X_2$ deformation retracts to the closed topologist's sine curve $T$. But $T$ has two path components ($L$ and $S$). I mentioned this in my answer, but I didn't explictly say that they are contractible so that we get for $n > 0$ $$H_n(X_2) \approx H_n(T) \approx H_n(L) \oplus H_n(S) = 0 .$$

I edited my answer to the question linked by you.

Paul Frost
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