Homology group of connected sum of torus

Question

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.

Answer 1

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

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