Proof that $H_1(D^2, S^1) = 0$

Question

I was trying to prove that $H_1(D^2, S^1) = 0$ and I came up with the following proof.

Proof: Since $S^1 \subseteq D^2$ we have the following short exact sequence $$0 \to H_1(S^1) \xrightarrow{i_*} H_1(D^2) \to H_1(D^2, S^1) \to 0$$

where $i_*$ is the induced monomorphism from $H_1(S^1)$ to $H_1(D^2)$ induced by the inclusion map $i ; C_1(S^1) \to C_1(D^2)$.

Now we know that $H_1(D^2) = 0$ thus we obtain the following exact sequence $$0 \to H_1(D^2, S^1) \to 0$$ and so we have $H_1(D^2, S^1) = 0$ as desired. $\square$

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But now I believe that my above proof is errornous since we would also obtain an exact sequence $0 \to H_1(S^1) \to 0$ implying that $H_1(S^1) = 0$ a contradiction since we know by the Hurewicz map that $H_1(S^1) \cong \pi_1(S^1) \cong \mathbb{Z}$.

What have I done wrong in my proof?

Answer 1

We have the long exact sequence

$$ \dots \to H_n(S^1) \to H_n(D^2) \to H_n(D^2,S^1) \to \dots $$

Since $H_n(S^1) = 0$ for $n > 1$, we have that $H_n(D^2,S^1) = 0$ for $n > 2$ and thus we are left with the following exact sequence

$$ 0 \to H_2(D^2,S^1) \to H_1(S^1) \to H_1(D^2) \to H_1(D^2,S^1) \to H_0(S^1) \to H_0(D^2) \to H_0(D^2,S^1) \to 0 $$

Note that only $H_1(D^2)$ is immediately zero, but we don't know about the rest. So you don't have the short exact sequence you claim. A possible simplification is to consider the reduced long exact sequence

$$ \dots \to \tilde{H}_n(S^1) \to \tilde{H}_n(D^2) \to H_n(D^2,S^1) \to \dots $$

from which we have

$$ 0 \to H_2(D^2,S^1) \to \tilde{H}_1(S^1) \to \tilde{H}_1(D^2) \to{H}_1(D^2,S^1) \to 0 $$

and we also get that $H_0(S^1,D^2) = 0$. Now we can use your idea: note that as you say, we have

$$ H_1(D^2,S^1) = 0 $$

but there is no contradiction, since the other part of the sequence is

$$ 0 \to H_2(D^2,S^1) \to \tilde{H}_1(S^1) \to 0 $$

Moreover, we get for free that $H_2(D^2,S^1) \simeq \tilde{H}_1(S^1) \simeq \mathbb{Z}$.

Answer 2

You are assuming that $\iota : C(S^1) \rightarrow C(D^2)$ induces an injection on homology but this in general fails, e.g. the vertical maps in $\require{AMScd}$ \begin{CD} 0 @>>> \mathbb{Z} @>\cdot n>> \mathbb{Z} @>>> 0 @>>> 0 \\ @VVV @V\text{id}VV @V\text{id}VV @VVV @VVV \\ 0 @>>> \mathbb{Z} @>\cdot n>> \mathbb{Z} @>>> \mathbb{Z}/n @>>> 0 \end{CD} make up an injective chain map (between the horzintal complexes) which induces $\mathbb{Z}/n \rightarrow n\mathbb{Z}/n\mathbb{Z} = 0$ on the homology of the middle part. However, you can still use the long exact sequence for the pair $(D^2,S^1)$: $$0 = H_1(D^2) \rightarrow H_1(D^2,S^1) \xrightarrow{\delta} H_0(S^1) \rightarrow H_0(D^2). $$ Since the right map is an isomorphism we have $\delta = 0$ by exactness, hence ker $\delta = H_1(D^2,S^1)$. Again using exactness we have that the left map is surjective, hence $H_1(D^2,S^1)=0$.