How can I calculate the homology group of an infinite torus using Mayer-Vietoris? I want to calculate the (simplicial) homology of the following space using Mayer-Vietoris:

I have tried to do it by cutting it along the axis and getting two subspaces homeomorphic to something that looks like an "infinite flute" (a cylinder with infinitely many holes in its surface) such that the intersection is homeomorphic to the 1-dimensional sphere.
But doing this I still have to calculate the homology of the "infinite flute"
Is this the right choice for the subspaces? Or are there any other subspaces which would make the calculations easier?
 A: Here is an answer without Mayer–Vietoris – maybe it's possible to use the technique you describe, but the following approach seems easier to me. Homology commutes with (some) directed colimits. You can e.g. find a precise statement in Hatcher's book Algebraic Topology, Proposition 3.33.
Your space $X$ can naturally be seen as a directed colimit indexed by $\mathbb{N}$ (in fact if you try to define it precisely this is probably how you will do it). Indeed, consider $X_n$ to the be part of your space containing the first $n$ "holes", for example $X_1$ is a torus with a disk removed. More generally $X_n$ is a surface of genus $n$ with a disk removed. It's clear that $X = \bigcup_{n \ge 1} X_n$, and that each compact set of $X$ is contained in some $X_n$. Thus the natural map:
$$\varinjlim H_i(X_n) \to H_i(X)$$
is an isomorphism. All is left to do is compute $H_i(X_n)$ and determine what the maps $H_i(X_n) \to H_i(X_{n+1})$ are, and then see what the colimit is.
As I wrote earlier $X_n$ is a surface $\Sigma_n$ of genus $n$ with a disk removed. Recall that:
$$H_i(\Sigma_n) = \begin{cases}
\mathbb{Z} & i = 0, \\
\mathbb{Z}^{2n} & i = 1, \\
\mathbb{Z} & i = 2,
0 & \text{otherwise}.
\end{cases}$$
Now, using the fact that $X_n$ deformation retracts onto a wedge sum of $2n$ circles (by taking the standard cellular decomposition of $\Sigma_n$ and punching a hole in the middle of the 2-cell), you can compute the homology of $X_n = \Sigma_n \setminus D^2$:
$$H_i(X_n) = \begin{cases}
\mathbb{Z} & i = 0, \\
\mathbb{Z}^{2n} & i = 1, \\
0 & \text{otherwise}.
\end{cases}$$
The inclusion $X_n \to X_{n+1}$ induces an isomorphism on $H_0$, and it is the inclusion $(x_1, \dots, x_{2n}) \mapsto (x_1, \dots, x_{2n}, 0, 0)$ on $H_1$ (you can see this by picking explicit representative of the generating homology classes of $H_1(X_n)$. This finally gives:
$$H_i(X) = \begin{cases}
\mathbb{Z} & i = 0, \\
\bigoplus_{n = 0}^\infty \mathbb{Z} = \mathbb{Z}^{(\infty)} & i = 1, \\
0 & \text{otherwise}.
\end{cases}$$
(Just a warning that will maybe be useless: be careful that $H_1(X)$ is the direct sum of an infinite number of copies of $\mathbb{Z}$, not the direct product. The distinction matters when there is an infinite number of terms. By the universal coefficient theorem, $H^1(X) = \mathbb{Z}^\infty$ is the infinite product though.)
A: Here's a Mayer-Vietoris argument.  Write $X$ as a union $\cup_{i=0}^\infty T_i$ where each $T_i$ is a torus with two boundary components (except $T_0$, which has only one boundary component). Let $A= \sqcup_{i=0}^\infty T_{2i}$ and $B= \sqcup_{i=0}^\infty T_{2i+1}$. Then $A \cap B$ is a countably infinite disjoint union of circles $\sqcup_{i=1}^\infty S^1_i$. Mayer-Vietoris gives you a long exact sequence containing the following terms:
$$0 \to H_2(A) \oplus H_2(B) \to H_2(X) \to \oplus_{i=1}^\infty H_1(S^1_i) \to H_1(A) \oplus H_1(B) \to H_1(X) \to 0.$$
There might be a way to show that $H_2(X)$ vanishes by analyzing the connecting homomorphism. But there's a better approach that mirrors part of Najib Idrissi's answer: Any nonvanishing class in $H_2(X)$ is represented by some cycle $\sigma \in C_2(X)$. Since $\sigma=c_1\sigma_1+ \cdots +c_k \sigma_k$ is a finite linear combination of compact simplices, there is some punctured genus-$g$ surface $\Sigma_g \subset X$ which contains all the simplices $\sigma_1,\ldots,\sigma_k$. So $\sigma \in C_2(X)$ is in the image of the inclusion map $C_2(\Sigma_g) \hookrightarrow C_2(X)$, and thus $[\sigma]\in H_2(X)$ is in the image of the inclusion-induced map $H_2(\Sigma_g) \to H_2(X)$. But it's a standard fact that a punctured compact surface $\Sigma_g$ has $H_2(\Sigma_g)=0$, so $[\sigma] \in H_2(X)$ must equal zero. We conclude that $H_2(X)=0$.
Now the above reduces to a short exact sequence
$$0  \to \oplus_{i=1}^\infty H_1(S^1_i) \to H_1(A) \oplus H_1(B) \to H_1(X) \to 0.$$
We have 
$$H_1(A) \oplus H_1(B) \cong \oplus_{i=1}^\infty H_1(T_i),$$
where $H_1(T_0)\cong \mathbb{Z}^2$ and $H_1(T_i)\cong \mathbb{Z}^3$ for $i\geq 1$. We can view the "third" generator of each $H_1(T_i)$ for $i\geq 1$ as the image of the generator of $H_1(S^1_i)$. This gives
$$H_1(X) \cong \frac{H_1(A)\oplus H_1(B)}{\oplus_{i=1}^\infty H_1(S^1_i)}\cong \oplus_{i=0}^\infty \mathbb{Z}^2.$$
Finally, connectedness gives $H_0(X)\cong \mathbb{Z}$. In summary, we have 
$$H_*(X) \cong \begin{cases} \mathbb{Z} & *=0\\ \oplus_{i=0}^\infty \mathbb{Z} & *=1 \\ 0 & *\geq 2.\end{cases}$$
