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.
Best 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...