Homology of "Intersecting" Spaces It is fairly routine in Topology textbooks to show how to compute the homology groups of spaces that "look like" $X_1 \times X_2$ or $\bigvee X_i$. But how do you compute the homology groups $H_n$ of spaces that merely interest not in any specific way but rather pass through each other. For instance, take $X$ to be the space formed by the two tori passing through each other below. 
My first thought was to create a $\Delta$-complex or CW-complex for the space but it wasn't apparent to me what it should be. Looking through the standard theorems, the only viable approach to me is something like excision but I'm not seeing how this is helping us since I don't see a useful way to decompose the space. How would go about computing $H_n$ in this case? Or perhaps something more simple like a torus passing through a sphere, two spheres passing through each other, or whatever may be less pathological than the case shown here.

 A: This is exactly what Mayer-Vietoris sequences are for.  Suppose $X$ is a space and $U$ and $V$ are open subspaces of $X$ with $U\cup V=X$.  Then there is a long exact sequence of homology groups \begin{align*}\dots & \to H_n(U\cap V)\to H_n(U)\oplus H_n(V)\to H_n(X)
\\
 &\to H_{n-1}(U\cap V)\to H_{n-1}(U)\oplus H_{n-1}(V)\to H_{n-1}(X)\\
&\to\dots
\end{align*}
In this sequence, the map $H_n(U\cap V)\to H_n(U)\oplus H_n(V)$ is given by $x\mapsto (i_*(x),-j_*(x))$ where $i:U\cap V\to U$ and $j:U\cap V\to V$ are the inclusion maps.  The map $H_n(U)\oplus H_n(V)\to H_n(X)$ is given by $(y,z)\mapsto k_*(y)+\ell_*(z)$, where $k:U\to X$ and $\ell:V\to X$ are the inclusion maps.
In your case where $X$ is two tori glued together, you can take $U$ to be a small open neighborhood of the left torus and $V$ to be a small open neighborhood of right torus, so $U$ and $V$ both deformation-retract to the tori and $U\cap V$ deformation-retracts to the intersection of the two tori, which is a disjoint union of two circles.  The inclusion map $U\cap V\to U$ is nullhomotopic (both circles enclose disks in the left torus), but the inclusion map $U\cap V\to V$ sends both generators of $H_1(U\cap V)\cong \mathbb{Z}^2$ to one of the generators of $H_1(V)\cong \mathbb{Z}^2$ (since the two circles in $U\cap V$ are both meridans of the right torus).
We can then compute $H_2(X)$ using the Mayer-Vietoris sequence $$H_2(U\cap V)\to H_2(U)\oplus H_2(V)\to H_2(X)\to H_1(U\cap V)\to H_1(U)\oplus H_1(V)$$ which becomes $$0\to \mathbb{Z}\oplus\mathbb{Z}\to H_2(X)\to \mathbb{Z}^2\to \mathbb{Z}^2\oplus\mathbb{Z}^2.$$  The last map sends both $(1,0)$ and $(0,1)$ to $(0,0,-1,-1)$ by the analysis of the inclusion maps $U\cap V\to U$ and $U\cap V\to V$ of the previous paragraph.  In particular, its kernel is generated by $(1,-1)$ and is isomorphic to $\mathbb{Z}$, so we have a short exact sequence $$0\to\mathbb{Z}\oplus\mathbb{Z}\to H_2(X)\to\mathbb{Z}\to 0$$ which gives $H_2(X)\cong \mathbb{Z}^3$.
I'll leave it as an exercise for you to similarly compute $H_1(X)$ using the Mayer-Vietoris sequence.
A: Mayer-Vietoris could potentially help with intersections (but then you will need the union, too).  For simpler examples, it's usually enough to construct a complex or identify what the intersection is homotopy equivalent to.  Unlike products, wedge sums, or disjoint unions, the intersection's homology is fairly unconstrained. (Exercise: show that for every non-negative $n$ the intersection of two spheres in $\mathbb{R}^3$ can have first homology being $\mathbb{Z}^n$.)
In your particular example, isn't the intersection of the two tori (as drawn) just two circles?  Then $H_0(X)=\mathbb{Z}^2$, $H_1(X)=\mathbb{Z}^2$, and all other homology groups are $0$.

If the question is to find the homology of the union of the tori, then my first attempt would be to find a space which is homotopy equivalent.  For the discs that the second torus cuts out of the first, we can contract each of them to a point, leaving us with a space which is a torus with two spheres attached by their respective north and south poles to the torus.  By drawing out a pole into an arc, and then moving the polar attachment point along the sphere to the torus, one sees that the space is a wedge sum of a torus, two spheres, and two circles.  Hence, $H_0(X)=\mathbb{Z}$, $H_1(X)=\mathbb{Z}^{4}$, and $H_2(X)=\mathbb{Z}^3$.
I can illustrate this if it would be helpful.
