I tried to compute the homology groups of 2 disjoint segments $ H_k(I_1 \cup I_2; G) $ and cylinder $ H_k(C, G) $ using the Mayer-Vietoris sequence. It looks like $$ 0 \to H_1(I_1 \cap I_2; G) \to H_1(I_1; G) \oplus H_1(I_2; G) \to H_1(I_1 \cup I_2; G) \to H_0(I_1 \cap I_2; G) \to H_0(I_1; G) \oplus H_0(I_2; G) \to H_0(I_1 \cup I_2; G) \to 0 $$

Given that $ I_1 \cap I_2 = \varnothing $, $ H_1(I; G) = 0 $, $ H_0(I; G) = G $ we obtain: $$ 0 \to 0 \to 0 \oplus 0 \to H_1(I_1 \cup I_2; G) \to 0 \to G \oplus G \to H_0(I_1 \cup I_2; G) \to 0 $$

Am I right?

For cylinder breaking it into 2 rectangles, we analogicaly can obtain the sequence:

$$ 0 \to 0 \oplus 0 \to 0 \oplus 0 \to H_2(C; G) \to 0 \oplus 0 \to 0 \oplus 0 \to H_1(C; G) \to G \oplus G \to G \oplus G \to H_0(C; G) \to 0 $$

The end of both sequence is similar $ G \oplus G \to H_0(T; G) \to 0 $, but as I know $ H_0(C; G) = G $ and $ H_0(I_1 \cup I_2; G) = G \oplus G $

As far as I know, the maps is uniquely defined and should give the same result. Where is my mistake?

Thanks for the help! Don't judge strictly, I'm just starting to learn algebraic topology.


1 Answer 1


Regarding $I_1 \cup I_2$, that is all correct, and then you should go one more step to reach the final conclusion that $H_1(I_1 \cup I_2;G) \approx 0$ and $H_0(I_1 \cup I_2;G) \approx G \oplus G$. I might add here that while this is a good test of your understanding of Mayer-Vietoris, in practice it is much easier to calculate $H_i(I_1 \cup I_2;G)$ using that if $X,Y$ are disjoint closed subspaces then $H_i(X \cup Y;G) \approx H_i(X;G) \oplus H_i(Y;G)$.

The end of the cylinder example is not the same as the end of the previous example. You have ignored the preceding terms of the two examples: $$(1) \qquad 0 \to G \oplus G \to H_0(I_1 \cup I_2;G) \to 0 $$ $$(2) \qquad G \oplus G \to G \oplus G \to H_0(C;G) \to 0 $$ In (2), a general homomorphism $G \oplus G \to G \oplus G$ need not have trivial image (in fact, this homomorphism does not), and so its cokernel, which is $\approx H_0(C;G)$, need not be $\approx G \oplus G$ (in fact, it is not).

  • $\begingroup$ Greate thanks! Could you answer one more question? How can I deduce that $ H_0(C; G) \approx G $ and $ H_1(C; G) \approx G $ exactly? $\endgroup$ Commented Mar 13, 2016 at 16:12
  • $\begingroup$ In your third long exact sequence, where you written have a homomorphism $$G \oplus G \to G \oplus G$$ you need to go back and look at the original expression of that homomorphism in terms of the cylinder broken into two rectangles. Then you need to write down the exact formula for that homomorphism, expressed in terms of various inclusion maps of things into other things. From that formula, compute the kernel and cokernel and that should do it. $\endgroup$
    – Lee Mosher
    Commented Mar 13, 2016 at 17:24

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