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Call $X=Torus-\{D_1,D_2\}$, for $D_1, D_2$ two open disks in the surface of the standard torus. I have to calculate $H_1(X, \partial X)$ for $\partial X$ the union of the two boundary circles of the disks, and I think I've made some progress but not sure about how to finish it up or if what I've done is correct.

I have $(X, \partial X)$ is a good pair, and so I can identify $H_1(X, \partial X)=\tilde{H_1}(X/\partial X)$ reduced homology. I've imagined $X/\partial X$ as joining the boundary of two holes together at a point away from the torus. Then I said after some imagination and stretching:

X with boundary identified ~ Torus wedge $S^1$

And hence calculated: $$H_1(X, \partial X)=\tilde{H_1}(X/\partial X)=\tilde{H_1}(T\vee S^1)=\mathbb{Z}^3.$$ Is this correct? or is there a better way to go about it? And any general tips for stating the generators of such a structure?

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    $\begingroup$ Do you mean $X$ with boundary identified ~ $Torus$ wedge $S^1$? $\endgroup$ May 19, 2017 at 10:00
  • $\begingroup$ Ahh yes thank you, I do! $\endgroup$
    – marineabcd
    May 19, 2017 at 12:23

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What you've done looks good, and the method is what I would have done.

As for the generators, we know that $H_1$ is similar to the fundamental group (that is, it's the abelianisation). The generators of the fundamental group are the normal ones for the torus, and the loop going around the circle. So (admittedly with a bit of handwaving) we get the generators to be the same for homology; take the loops (cycles) from the fundamental group.

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    $\begingroup$ Thank you, that makes sense! $\endgroup$
    – marineabcd
    May 19, 2017 at 12:22

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