What is a homology group of $S^2$ with a string attached to it by both ends? (Name it $X$. Drawing in below)
I tried to decompose X as following, but I'm not sure if it's correct.
I'll appreciate any help.
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Saki
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Sure, that is one possible decomposition. Now write down the Mayer-Vietoris sequence and milk it. – Thorgott Jan 15 '21 at 03:05
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Thank you for the comment. I wrote Mayer-Vietoris sequence. Here, can I say that i*:H1(A∩B)→H1(B) is a 0-map? – Saki Jan 15 '21 at 03:36
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When I assumed i* a 0-map, I ended up H_q(X)=Z(q=0,1,2), {0}(otherwise) – Saki Jan 15 '21 at 03:40
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1Yes, $i_*$ is a zero map, since the nontrivial cycle in $A \cap B$ is contractible in $B$. Also, there is another way to do this problem by realizing that your space is homotopy equivalent to $S^2 \vee S^1$; this is because if you draw a line on the sphere connecting your two points, collapsing this line segment to a point is a homotopy equivalence, and the resulting space is $S^2 \vee S^1$. – kamills Jan 15 '21 at 14:45
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Thank you very much. That's really helpful. I think I can proceed now. – Saki Jan 16 '21 at 07:32