1
$\begingroup$

In the Mayer-Vietoris exact sequence $$... \rightarrow H_n(A) \oplus H_n(B) \rightarrow H_n(X) \rightarrow H_{n-1}(A \cap B) \rightarrow H_{n-1}(A) \oplus H_{n-1}(B) \rightarrow...$$ I am confused about exactness at $H_n(X)$.

The map $H_n(A) \oplus H_n(B) \rightarrow H_n(X)$ is just addition: $([\alpha],[\beta]) \mapsto [\alpha] + [\beta].$

The map $H_n(X) \rightarrow H_{n-1}(A \cap B)$ will write (some barycentric subdivision of) a cycle $\gamma$ as $\alpha+\beta$ with $\alpha \in Z_n(A)$ and $\beta \in Z_n(B)$, whose boundaries must be supported on $A \cap B$, and then map $[\gamma]$ to $[\partial(\alpha)] + [\partial(\beta)]$.

So when I do both in a row, I map $([\alpha],[\beta])$ to $[\partial(\alpha)] + [\partial(\beta)]$, and the problem for me is to show that $\partial(\alpha) +\partial(\beta)$ is a boundary of a chain supported on $A \cap B$. I don't know what chain to look at.

Thanks

$\endgroup$
4
  • $\begingroup$ this explains it: en.wikipedia.org/wiki/… $\endgroup$ Commented Apr 17, 2014 at 17:27
  • 1
    $\begingroup$ Isn't this $0$ anyway ? Since $\alpha,\beta$ are cycles. $\endgroup$ Commented Apr 17, 2014 at 17:30
  • $\begingroup$ @StefanHamcke Of course it is! Thank you. I was confused somehow $\endgroup$
    – user142755
    Commented Apr 17, 2014 at 17:40
  • $\begingroup$ You're welcome :-) $\endgroup$ Commented Apr 17, 2014 at 17:41

1 Answer 1

1
$\begingroup$

Recall that $Z_n(A)$ is defined to be the kernel of the map $\partial_n\colon C_n(A)\to C_{n-1}(A)$ and so if $\alpha\in C_n(A)$ then $\partial(\alpha)=0$ and so $[\partial(\alpha)]=0$. Similarly for $\beta\in Z_n(B)$ and so $$[\gamma]\mapsto[\partial(\alpha)]+[\partial(\beta)]=0+0=0$$ hence $[\gamma]$ is in the kernel of the connecting map.

$\endgroup$

You must log in to answer this question.

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