# Confusion in using Mayer-Vietoris theorem

For two topological spaces $$A$$ and $$B$$, in order to show that $$H(A \sqcup B) \cong H(A) \oplus H(B)$$ in this question and in general I believe one can use Mayer-Vietoris to obtain the result easily. However, I don't quite understand how it applies in this case - what if $$A \sqcup B$$ is not contained in the interior of $$A$$ and $$B$$?

Also in the link above the OP claims $$A \cap B= \emptyset$$ to conclude that homology group is trivial but $$A \cap B$$ need not be empty or am I missing something?

Would greatly appreciated if someone could shed a light on these.

## 1 Answer

Maybe the notation caused some confusion: $$A \sqcup B$$ is the disjoint union of $$A$$ and $$B$$.

This implies that $$A \cap B = \emptyset$$, which immediately addresses your second point.

Also, this implies that both $$A$$ and $$B$$ are open in $$A \sqcup B$$, which means that the interior of $$A$$ (resp. of $$B$$) in $$A \sqcup B$$ is $$A$$ itself (resp. $$B$$ itself). Hence the interiors of $$A$$ and $$B$$ cover $$A \sqcup B$$, which means that the conditions for Mayer-Vietoris are satisfied.

But I must say that you don't really need Mayer-Vietoris to compute $$H^\star (A \sqcup B)$$ - it's much better to work directly from the definition of homology...

• Arghh it's just that then. Many thanks. Dec 16, 2018 at 13:39