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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 $H_{*}(T^{2} - D^{2})$ ?

beno

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    sure. of which pair?2012-01-15

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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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    @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 $X-A$.2012-04-18