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I am calculating the homology group $\#^kT^2=D_k$, and I was able to find using Mayer-Vietoris sequence, with $U=T^2-\{p\}\#^{k-1}T^2$ and $V=T^2\#^{k-1}T^2-\{p\}$, so $U=D_{k-1}$, $V=T^2$ and $U\cap V=S^1$, that $H_n(D_k)=0 $ for all $n\geq 3$. I find also that $H_0(D_k)=\mathbb Z$ because $D_k$ is path connected. So for $n=1,2$ I have in the reduced homology $$ 0\rightarrow H_2(S^1)\rightarrow H_2(D_{k-1})\oplus H_2(T^2)\rightarrow H_2(D_k)\rightarrow H_1(S_1)\rightarrow H_1(D_{k-1})\oplus H_1(T^2)\rightarrow H_1(D_k)\rightarrow H_0(S^1), $$I know that te map $H_1(S^1)\rightarrow H_1(D_{k-1})\oplus H_1(T^2)$ is the zero map, so $H_1(D_k)=\mathbb Z^{2k}$ as wished, but then I got $$ 0\rightarrow \mathbb Z^2\rightarrow H_2(D_k)\rightarrow \mathbb Z\rightarrow 0 $$ that implies $H_2(D_k)=\mathbb Z^3$, but it is supposed to be $\mathbb Z$, what am I missing?

Thanks in advance.

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When you remove a point from the interior of $T^2$ and $D_{k-1}$ you alter their top homology groups. In particular, since they are path connected, the point can be chosen to lie in the open top cell of their CW structure. In this case it is easy to see that the spaces deformation retract onto their 1-skeletons. That is

$T^2-p\simeq S^1\vee S^1$ and $D^{k-1}-p\simeq \bigvee^{2k-2} S^1$

With this it is obvious that $H_2(T^2-p)=0=H_2(D^{k-1}-p)$, so your short exact sequence becomes the isomorphism

$0\rightarrow H_2(D^{k})\xrightarrow{\cong}H_1(S^1)=\mathbb{Z}\rightarrow 0$.

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  • $\begingroup$ I undestand the first part, but how did you use this in the short exact sequence? Because there I have $U$ and $V$, that are not $T^2-p$ and $D_{k-1}-p$ $\endgroup$ – e.turatti Oct 27 '17 at 17:44
  • $\begingroup$ I think it should rather be $U:=(T^2-p)\#^{k-2}T^2\cong D_{k-1}-p$ and $V:=T^2-p$. $\endgroup$ – Tyrone Oct 27 '17 at 19:54
  • $\begingroup$ To be clearer, you are trying to calculate the homology of $D_k=\#^kT^2\cong T^2\#D_{k-1}$. Note that neither $D_{k-1}$, nor $T^2$ are actual subspaces of $D_k$. To get subspace from these spaces you need to cut out a small disk from each of them, and form the gluing over the boundary of the removed disk. Since you can choose the disk that is to be removed to lie in the open top cell of each space, this is the same as removing a point. Hence you have $U:=D_{k-1}-p\cong (T_2-p)#D_{k-2}$, $V:=T^2-p$ (really the points should be disks, but you can treat the problem homotopically). $\endgroup$ – Tyrone Oct 28 '17 at 5:24

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