# Question about calculation in Mayer-Vietoris sequence

Suppose I want to to compute the homology group of $$X=S^1 \times (S^1 \vee S^1)$$, which can be seen as two toruses piled together.

Now, by Mayer-Vietoris sequence, I can split $$X$$ into two toruses $$U,V$$, whose intersection is a circle $$S^1$$. Then using the homology groups we already known for torus and circle, we have exact sequence:

$$…\to H_2(U\cap V)\to H_2(U)\oplus H_2(V)\to H_2(X)\to H_1(U\cap V)…\to H_0(X) \to 0$$

which is $$\to H_2(U\cap V)=0\to Z\oplus Z\to H_2(X)\to H_1(U\cap V)=Z\to Z\oplus Z\oplus Z\oplus Z\to H_1(X)\to Z\to Z\oplus Z \to H_0(X)\to 0$$ where $$Z$$ denotes free abelian group.

All we left to do is “filling the blanks”. At first glance, I can’t determine “blanks” explicitly. But if $$H_n(U\cap V)\to H_n(U)\oplus H_n(V)$$ are all injective, we can deduce that $$H_{n+1}(X)\to H_n(U\cap V)$$are zero maps. So $$H_0(X)=Z,H_1(X)=Z\oplus Z\oplus Z, H_2(X)=Z\oplus Z$$ and $$H_n(X)=0$$ for $$n\ge3$$.

I am not sure whether my argument is right and whether all Mayer-Vietoris sequences can be computed in this way? Thank in advance!

Your argument is correct. The injectivity in all degrees can be seen because the intersection circle is a retract of both $$U$$ and $$V$$. In general the map won't be injective so this is not a universal way to compute homology.