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In an algebraic topology course I'm taking we are often asked to compute the homology groups of a space $X = A \cup B$ using the Mayer-Vietoris sequence, and it happens in all of the examples I've seen so far that it's possible to do this without knowing anything about the connecting homomorphisms $\partial_{\ast}$ (say on the level of chains); we only end up needing $H_{\ast}(A), H_{\ast}(B), H_{\ast}(A \cap B)$ and possibly some of the inclusion maps.

My guess is that this is not a typical situation; is there a relatively simple example of a nice space $X$ and nice subspaces $A, B$ such that knowing $H_{\ast}(A), H_{\ast}(B), H_{\ast}(A \cap B)$ is not enough to compute $H_{\ast}(X)$ without knowing the specific form of the connecting homomorphisms? (For maximal relevance to the course $X, A, B, A \cap B$ should be finite simplicial complexes.)

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any situation where the homologies of A,B, A intesrsect B are not trivial I am thinking of S2#S2 for instance –  El Moro May 22 '11 at 19:51
    
@El Moro: some of the examples I'm referring to have that property and you still don't need to know what $\partial_{\ast}$ is in advance. Can you work out an example more specifically? –  Qiaochu Yuan May 22 '11 at 19:52
    
@Yuan I have to admit that I haven't met that case either before although it seems that the delta map comes from snake lemma so you can have its form from there. It is initially a map of chains that goes down to be a map of homologies –  El Moro May 22 '11 at 19:59
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2 Answers 2

up vote 10 down vote accepted

Given a long exact sequence

$$ \cdots \to C_{i+1} \to A_i \to B_i \to C_i \to A_{i-1} \to \cdots $$

let the map $A_i \to B_i$ be denoted $f_i$. Then you have that $C_i$ is an extension

$$ 0 \to coker(f_i) \to C_i \to ker(f_{i-1}) \to 0$$

so up to that extension problem, the maps $f_i$ always determine the $C_i$ groups. So if you want a situation where the group $C_i$ is ambiguous, you could have $ker(f_{i-1}) = \mathbb Z_2$ and $coker(f_i) = \mathbb Z$, that way $C_i$ could be either $\mathbb Z$ or $\mathbb Z \oplus \mathbb Z_2$.

Regardless, the connecting map $\partial_i : C_i \to A_{i-1}$ is determined by this extension problem, and it's easy enough to cook up examples either-way.

So I'm a little confused as to the nature of your question. I guess what I'm saying is that you are in the typical situation, and Grigory's example is also typical in that it's the inclusion map that makes the differences between his examples.

Regarding how useful/useless the MVS is for a typical problem, it really depends on how easily-expressible your space is as a union of spaces you understand (and their intersections). If your space doesn't fit that profile, you've got potentially a lot of work to do. The Serre Spectral Sequence of a Fibration is in a sense something of a souped-up Mayer-Vietoris sequence, and there are plenty of papers where people are happy just computing the $E_3$-page, or determining which page the SS collapses on, or computing a differential. These extension issues tend to be very thorny and consume much literature.

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I phrased the question in what is in hindsight a somewhat confusing way; I think the current edit is closer to what I meant. –  Qiaochu Yuan May 22 '11 at 21:18
    
...and extra +1 for the point "(iterated) M-V = SSS" –  Grigory M May 22 '11 at 21:20
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Yes, the map from $A$ to $B$ determines $C$ up to extension (so if you are over a field, everything is unique), but (1) there is still an extension problem and (2) understanding the maps can be a subtle matter. In particular, in the example that Grigory gave, the inclusions are "the same" in both cases, it's just that the gluing is twisted by a homeomorphism. Put differently, knowing $H_*(A\cap B)\to H_* (A)$ and $H_*(A\cap B)\to H_* (B)$ up to automorphisms does not give $H_*(A\cap B)\to H_* (A)\oplus H_* (A).$ –  Aaron May 22 '11 at 21:24
    
@Aaron: thank you, that is more or less what I wanted to know. –  Qiaochu Yuan May 23 '11 at 7:22
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Knowing only $H(A)$, $H(B)$ and $H(A\cap B)$ is not enough, of course. For example, taking $A=B=S^1\times D^2$ and gluing them by $S^1\times S^1=A\cap B$ one can get either $X_1=S^2\times S^1$ or $X_2=S^3$. This gives two Mayer-Vietoris sequences with identical $H(A)$, $H(B)$ and $H(A\cap B)$ but different H(X).

As for the situation where one also knows inclusion maps, see Ryan Budney's excellent answer.

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This construction can be generalized: take some non-trivial bundle $X\to \Sigma B$ that restricts trivially on B etc. (Example above corresponds to the Hopf fibration.) –  Grigory M May 22 '11 at 20:53
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