Homology group of the Torus using Mayer-Vietoris sequence

Question

I am trying to compute the homology group of the torus $\mathbb T^2$ using Mayer-Vietoris sequence and there is some step I'm struggling with. I know that I should have $$H_k(\mathbb T^2) = \begin{cases} \mathbb Z^2 & \text{if } k = 1,\\ \mathbb Z & \text{if } k = 2,\\ \mathbb 0 & \text{else.} \end{cases}$$ To this end, I consider two disks with the same center embedded in the square $R$ that defines the Torus (just identify the opposite sides two by two), $D_1 \subset D_2 \subset \text{int } R.$ Let us set $$A = \mathbb T^2 \backslash D_1 \quad \text{and} \quad B = D_2.$$ We have that the wedge product of two circles $S^1 \vee S^1$ is a deformation retract of $\mathbb T^2 \backslash D_1 $ and that $S^1$ is a deformation retract of $A \cap B$. The only part of the Mayer-Vietoris sequence which is non-trivial is given by $$0 \to H_2(X) \to H_1(A \cap B) \overset{\varphi}{\to} H_1(A) \oplus H_1(B) \overset{\psi}{\to} H_1(X) \to \tilde{H}_0(A\cap B).$$ By what I said above, it can by rewritten as $$0 \to H_2(\mathbb T^2) \to H_1(S^1) \overset{\varphi}{\to} H_1(S^1 \vee S^1) \oplus H_1(D_1) \overset{\psi}{\to} H_1(\mathbb T^2) \to \tilde{H}_0(S^1),$$ which yields $$0 \to H_2(\mathbb T^2) \to \mathbb Z \overset{\varphi}{\to} \mathbb Z^2 \oplus 0 \overset{\psi}{\to} H_1(\mathbb T^2) \to 0.$$ Now I do not really see how I could conclude. It seems that if $\varphi$ is the zero map everything works, but is that true ?