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I know this question has been asked several times, but I'm trying to solve it with a different approach that I didn't find online. I'm just wondering if my approach would work.

We look at the quotient map $q: S^1 \times S^1 \to S^2$ by $S^2 \cong S^1 \times S^1 / S^1 \vee S^1$ and want to show that it induces an isomorphism in $H_2$. The approaches I saw use either CW-complexes or homology of good pairs. Unfortunately, I haven't seen homology of good pairs yet, so I'm wondering if we can make the following work:

We have a long exact sequence of the map $q: (S^1 \times S^1, S^1 \vee S^1) \to (S^2, *)$:

$$\begin{array}{c} 0 & \xrightarrow{} & \tilde{H}_2(S^1 \vee S^1)=0 & \xrightarrow{i_*} & \tilde{H}_2(S^1 \times S^1) & \xrightarrow{j_*} & \tilde{H}_2(S^1 \times S^1, S^1 \vee S^1) & \xrightarrow{} &\tilde{H}_1(S^1 \vee S^1) \\ & & \downarrow{q_*} & & \downarrow{q_*} & & \downarrow{q_*} & & \downarrow{q_*} \\ 0 & \xrightarrow{} & \tilde{H}_2(*)=0 & \xrightarrow{g_2} & \tilde{H}_2(S^2) & \xrightarrow{\cong} & \tilde{H}_2(S^2, *) & \xrightarrow{} & \tilde{H}_1(*) = 0 \\ \end{array}$$

Is there a way to deduce from this diagram? Or would one have to go through the CW-complex argument anyway?

M. Wang
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  • Maybe I'm missing something, but don't you get this diagram for any map $q$? The diagram alone won't be able to distinguish between the quotient map and say, a constant map. – Michael Albanese Jan 22 '21 at 13:37
  • @MichaelAlbanese Yes that's true, but maybe the fourth vertical map is a bit easier to understand and could give us some info about $q_*$ (I'm thinking of an application of the Five Lemma after I showed that the fourth vertical map is an isomorphism). But I guess one would need homology of good pairs for that? – M. Wang Jan 22 '21 at 18:34
  • But the fourth map is the zero map, that doesn't depend on $q$. – Michael Albanese Jan 22 '21 at 19:44
  • @MichaelAlbanese Sorry, I meant the third vertical map... – M. Wang Jan 23 '21 at 12:12

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