Mayer-Vietoris in reduced homology for a torus. By using the Mayer-Vietoris sequence in reduced homology : 
I have to calculate the homology groups of :
The torus $\mathbb{T}^2 :=[0;1]^2 /\mathcal{T}$ by using the following decomposition
 $X_1 := \mathbb{T}^{2} -\{(1/2;1/2)\}$, $X_2 = ]0,1[^{2}$ et $A=X_1 \cap X_2$.
$\mathcal{T}$ is an equivalence relation meaning that we have identified the boudaries of the square $[0;1]^2$ : $(1,t)\sim (0,t)$ and $(t,1)\sim (t,0)$;$t\in[0;1]$. The point $(1/2;1/2)$ is the center of the square. 
I have already done this work : the rose with 2 petals is a retract by deformation of $X_1$ so 
$\tilde{H}_0(X_1)=\mathbb{Z}$
$\tilde{H}_1(X_1)=\mathbb{Z\oplus\mathbb Z}$ and $\tilde{H}_n(X_1)=0$ for $n\geq 2$
$\tilde{H}_n(X_2)=0$ for all $n$.
But i am stuck here. How can i compute $H_1(\mathbb T^2)$ and $H_2(\mathbb T^2)$ ?
Thanks for any help.
 A: From what you've worked out so far, the key to this problem is computation of the homomorphism
$$\mathbb{Z} \approx H_1(\mathbb{S}^1) \xrightarrow{j_*} H_1(A) \xrightarrow{i_*} H_1(X_1) \approx \mathbb{Z}^2
$$
In this sequence, $i_*$ is the homomorphism induced by the inclusion $i : A \hookrightarrow X_1$, and $j_*$ is the isomorphism induced by a homotopy equivalence $j :\mathbb{S}^1 \to A$. Once you've computed that homomorphism, it will follow that $H_2(\mathbb{T}^2)$ is isomorphic to its kernel. 
Let me give some further hints, without giving the whole thing away.
Your work will be aided by deriving a simple expression for a function $f : \mathbb{S}^1 \to X_1$ which is homotopic to the composition $\mathbb{S}^1 \xrightarrow{i} A \xrightarrow{j} X_1$. 
If you understand how the boundary of a square is glued to a rose $X_1$ to form $\mathbb{T}^2$, then you should be able to use the description of that gluing to guess and to prove a formula for $f$. 
And once you have a formula for $f : S^1 \to X_1$, you should be able to "abelianize" that formula to obtain a formula for the induced function $f_* = j_* \circ i_* : \mathbb{Z} \to \mathbb{Z}^2$.
