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!


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .