I need to compute the homology groups under group of integers $ H_k(D; \mathbb Z) $, of the simplicial complex being a triangulation of the following figure:

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I divided it two parts $D = L_1 \cup L_2$ like this:

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Then both parts $L_i$ are homeomorphic to the standard 2-dimentional simplicial complex and hence we know thier homology groups.

$$ H_k(\Delta^2, \mathbb Z) = \begin{cases} \mathbb Z, & k = 0 \\ 0, & k > 0 \end{cases} $$

The intersection $L_1 \cap L_2 = I_1 \cup I_2 \cup I_3$ of the parts is merging of 3 disjoint segments and hence we know their homology groups too.

$$ H_k(I_1 \cup I_2 \cup I_3, \mathbb Z) = \begin{cases} \mathbb Z \oplus \mathbb Z \oplus \mathbb Z, & k = 0 \\ 0, & k > 0 \end{cases} $$

And we can construct the Mayer-Vietoris sequence:

$$ 0 \to H_1(L_1 \cap L_2; \mathbb Z) \to H_1(L_1; \mathbb Z) \oplus H_1(L_2; \mathbb Z) \to H_1(L_1 \cup L_2; \mathbb Z) \to H_0(L_1 \cap L_2; \mathbb Z) \to H_0(L_1; \mathbb Z) \oplus H_0(L_2; \mathbb Z) \to H_0(L_1 \cup L_2; \mathbb Z) \to 0 $$

And substituting computed groups:

$$ 0 \to 0 \to 0 \to H_1(D; \mathbb Z) \to \mathbb Z \oplus \mathbb Z \oplus \mathbb Z \to \mathbb Z \oplus \mathbb Z \to H_0(D; \mathbb Z) \to 0 $$

How can I compute from this groups $ H_0(D, \mathbb Z) $ and $ H_1(D, \mathbb Z) $? Thanks for the help!


From this information alone, you cannot compute the groups, because they could be $$ \mathbb Z \oplus \mathbb Z \oplus \mathbb Z $$ and $$ \mathbb Z \oplus \mathbb Z $$ with the two obvious maps being isomorphisms, or the first could be $\mathbb Z$ and the second be $0$. There are surely other possibilities.

To work out the groups, you'd need to know the actual maps on the various groups.

It's considerably easier to divide the shape with a line running NW-SE rather than NE/SW. Then you get two circles, whose homology groups you know, and the intersection is a line segment, which is contractible. And the map from $H_0$ of the intersection to $H_0$ of either part is an isomorphism (generated by the map taking a 0-simplex to a point in the intersection).

  • $\begingroup$ Oh, yes, it's considerably easier:) Am I right the answer is $ H_k(D; \mathbb Z) = \begin{cases} \mathbb Z, & k = 0 \\ \mathbb Z \oplus \mathbb Z, & k = 1 \\ 0, & k > 1 \end{cases} $ $\endgroup$ – Stanislav Morozov Mar 14 '16 at 14:23
  • $\begingroup$ Yes, that's correct. $\endgroup$ – John Hughes Mar 15 '16 at 3:23

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