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Let $X=A\cup B\ \ A,B$ are open and $A\cap B$ is contractible. Prove that $H_i(A\cup B)\equiv H_i(A)\oplus H_i(B)$ for $i\geq 2$. I think about using Mayer Vietoris sequence but I don't know how to prove this.
Unfortunately, I was off when my class had this so I don't understand much about Mayer Vietoris sequence and how to apply it to compute homology group. A great help if you could give me a hint.
I think that since $A\cap B$ is contractible which mean $H_i(A\cap B)=0 \ \forall i$ so the sequence will become $H_i(X)\to 0 \to H_i(A)\oplus H_i(B)\to H_{i-1}(X)\to \dots \to H_0(X)\to 0$

Stefan Hamcke
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hung tran
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    The $\forall i$ is wrong, this is the reason why the statement that you are asked to prove only holds for $i\ge2$. So just look at the definition of an exact sequence and figure out what the implications are when there are zeros in it. – Carsten S Jan 11 '14 at 18:14
  • well, my class is for third year so my teacher didn't explain much of the concept, so I know what that sequence is but I don't clearly understand it. Can you explain it more? So for $i\geq 2$ we have $0\to H_i(A)\oplus H_i(B)\to H_i(X)\to 0$ is exact which implies that the map $f:H_i(A)\oplus H_i(B)\to H_i(X)$ is an isomorphism right? So why this does not true when i=0,1 – hung tran Jan 11 '14 at 18:42
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    Right so far. And a space is contractible if it is homotopy equivalent to a point. What are the homology groups of a point? They are not all zero. – Carsten S Jan 11 '14 at 18:46
  • is it $\mathbb{Z}$? So why with $i\geq 2$ the homology group of a point is zeros and this is not true for i=0,1. I don't know much about advanced algebra so studying this class is quite hard for me. – hung tran Jan 11 '14 at 18:58
  • That's it. This was easy, right? – Carsten S Jan 11 '14 at 19:00
  • not for me, hehe. My school concentrate on analysis only, we have algebra faculty too but it for master degree. What I really learn in my algebra class is about definition of group, operator and homomorphism that – hung tran Jan 11 '14 at 19:06

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@CarstenSchultz is right. In this case it is probably more convenient to consider the Mayer-Vietoris sequence with reduced homology groups $$···→\tilde H_n(A ∩ B)→\tilde H_n(A)⊕\tilde H_n(B)→\tilde H_n(X)→\tilde H_{n−1}(A∩B)→···→\tilde H_0(X)→0$$ Then the last part of the sequence reads $$0→\tilde H_0(A)⊕\tilde H_0(B)→\tilde H_0(X)→0$$ Recall that the $\tilde H_0(X)$ is isomorphic to $\Bbb Z^n$ where $n$ is the number of path components minus $1$ (if $X$ is nonempty.) So this tells you how many path components $X$ has.

Stefan Hamcke
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