3
$\begingroup$

I'm trying to calculate the homology groups for a simplicial complex $X$, which is a union of subcomplexes $X_1$ and $X_2$ which are both combinatorially equivalent to cones. This is the information I have about them, arranged as a Mayer-Vietoris sequence:

\begin{array}{r|c c c } & H_n(X_1 \cap X_2) & H_n(X_1) \oplus H_n(X_2) & H_n(X) \\ \hline n=5 & 0 & 0 & h_5 \\ n=4 & \mathbb{F} & 0 & h_4 \\ n=3 & 0 & 0 & h_3 \\ n=2 & \mathbb{F}^5 & 0 & h_2 \\ n=1 & 0 & 0 & h_1 \\ n=0 & \mathbb{F} & \mathbb{F} \oplus \mathbb{F} & h_0 \end{array}

I get that an exact sequence $$0 \rightarrow A \rightarrow B \rightarrow 0$$ means the two non-zero elements have to be isomorphic, so $h_5 = \mathbb{F}$ because $$0 \rightarrow h_5 \rightarrow \mathbb{F} \rightarrow 0$$ and similarly to get $h_3 = \mathbb{F}^5$ $$0 \rightarrow h_3 \rightarrow \mathbb{F}^5 \rightarrow 0$$

How do I get the rest?

Does $0 \rightarrow C \rightarrow 0$ mean $C = 0$? That would give me $h_4$ and $h_2 = 0$.

What about $h_1$ and $h_0$?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

As you said, an exact sequence $0 \to M \to N \to 0$ forces $M = N$, since the map $M \to N$ has kernel $\operatorname{im} (0 \to M) = 0$ and image $\ker (N \to 0) = N$. Since all the reduced homology groups $\tilde H_n(X_1) \oplus \tilde H_n(X_2)$ vanish, the Mayer-Vietoris sequence gives isomorphisms $H_n(X) = H_{n-1}(X_1 \cap X_2)$ for each $n$.

$\endgroup$
4
  • 1
    $\begingroup$ You should make explicit that you're switching to the reduced sequence, which is perhaps not an obvious step for a beginner. $\endgroup$ Commented Apr 22, 2015 at 17:54
  • $\begingroup$ Ah, that makes a lot of sense! What I wasn't noticing was the 0 -> C -> 0 is actually 0 -> C -> 0 -> 0. Thanks! $\endgroup$
    – Ollie
    Commented Apr 22, 2015 at 18:02
  • $\begingroup$ @KevinCarlson: Done. $\endgroup$
    – anomaly
    Commented Apr 22, 2015 at 18:14
  • $\begingroup$ @Ollie: Right. More directly, you can also note that exactness at $C$ in the sequence $0 \to C \to 0$ means that the map $C \to 0$ has kernel $\operatorname{im} (0 \to C) = 0$. But $\ker (C \to 0) = C$, so $C = 0$. $\endgroup$
    – anomaly
    Commented Apr 22, 2015 at 18:15

You must log in to answer this question.

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