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I have the following question concerning excision: consider the torus $T^{2}$ and a disk $D^{2}$ in the torus. Is it possible to say, by excision, that $H_{*}(T^{2}, D^{2}) = H_{*}(T^{2} - D^{2})$? If yes, why? Because I don't see it since $\bar{D^{2}}$ is not contained in $int(D^{2})$. How can I apply excision to calculate $H_{*}(T^{2} - D^{2})$ ?


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how can I apply excision here to calculate for example $H_{*}(T^{2} - D^{2})$ ? – beno Jan 15 '12 at 7:56
Remove a slightly smaller subset $S\subset D^2$, so that the hypothesis of the excision theorem hold and such that the inclusion $T^2-D^2\to T^2-S$ is an deformation retract. – Mariano Suárez-Alvarez Jan 15 '12 at 8:49
what is $R^{2}$? – beno Jan 15 '12 at 8:56
i just got $ H_{n}(T^{2}, D^{2}) = H_{n}(T^{2} - S, D^{2} - S) $, for all $n$. How to go on further? – beno Jan 15 '12 at 9:03
can i do like this: since $(T^{2}-D^{2}, \emptyset) \rightarrow (T^{2}-S, D^{2}-S)$ is a deformation retract, then $H_{n}(T^{2}-S, D^{2}-S) = H_{n}(T^{2}-D^{2})$ ?? – beno Jan 15 '12 at 9:17

First of all, $H_*(T^2,D^2)$ is not isomorphic to $H^*(T^2\setminus D^2)$. By the long exact sequence of the pair $H_*(T^2,D^2)$ it is possible to show that $H_*(T^2,D^2)\cong \tilde{H}_*(T^2)$, where this last notation denotes reduced homology. On the other hand $T^2\setminus D^2$ is homotopy equivalent to the one-point union of two circles. In particular $H_2(T^2\setminus D^2)=0$, but $H_2(T^2,D^2)=\mathbb Z$.

Excision can tell you that $H_*(T^2,D^2)\cong H_*(T^2\setminus int(D^2),\partial D^2)$. This is because we are excising $int(D^2)$. (Recall excision give an isomorphism $H_*(X,Y)\cong H_*(X\setminus U,Y\setminus U)$. Actually, to apply excision, the closure of $U$ needs to be contained in the interior of $Y$, so you need to fuss a little with deformation retracts first.)

(Also, perhaps you meant $H_*(T^2,D^2)\cong \tilde{H}_*(T^2/D^2)$, where $T^2/D^2$ is the quotient space? This is true.)

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how did you use excision when you showed $H_{n}(T^{2}, D^{2}) = H_{n}(T^{2} - D^{2}, \partial D^{2})$, for all $n$ ?? – beno Jan 17 '12 at 6:55
Oops, looks like I missed an 'interior' sign. I'll fix that. In any event, excision lets you cut out the interior of $D^2$ from $T^2$ and $D^2$, leaving behind $T^2\setminus int(D^2)$ and $\partial D^2$. – Grumpy Parsnip Jan 17 '12 at 13:56
@beno He excised a disk a bit smaller that the $D^2$, inside the said $D^2$. Remember that excision does NOT provide a way to compute $H_n(H-B)$; it only tells that in the homology of a pair $(X,A)$, what happens inside $A$ doesn't matter, as long as it's "a bit far" from the border between $A$ and $X-A$. – Samuel T Apr 18 '12 at 9:42

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