3
$\begingroup$

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 $H_i(X)\to 0 \to H_i(A)\oplus H_i(B)\to H_{i-1}(X)\to \dots \to H_0(X)\to 0$

$\endgroup$
6
  • 2
    $\begingroup$ 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. $\endgroup$
    – Carsten S
    Commented Jan 11, 2014 at 18:14
  • $\begingroup$ 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 $\endgroup$
    – hung tran
    Commented Jan 11, 2014 at 18:42
  • 1
    $\begingroup$ 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. $\endgroup$
    – Carsten S
    Commented Jan 11, 2014 at 18:46
  • $\begingroup$ 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. $\endgroup$
    – hung tran
    Commented Jan 11, 2014 at 18:58
  • $\begingroup$ That's it. This was easy, right? $\endgroup$
    – Carsten S
    Commented Jan 11, 2014 at 19:00

1 Answer 1

1
$\begingroup$

@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.

$\endgroup$

You must log in to answer this question.

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