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I'm trying to figure out how to actually compute maps in the Mayer-Vietoris sequence as it's quite useless unless you actually know some maps. Hatcher computes the homology for the Klein bottle $K$ on page 151. I still can't seem to understand the logic he uses to determine the map

$H_1(A\cap B)\to H_1(A)\oplus H_1(B)$,

where $A$ and $B$ are the usual Möbius bands. There are at least two steps here:

  1. Finding the actual cycle in $A\cap B$ that generates $H_1(A\cap B)$.

  2. Figuring out where this goes in $H_1(A)$.

I can't seem to be able to visualize what a cycle looks like. For homotopy this is easy, but I've usually just used homology based on the Eilenberg-Steenrod axioms and for MVS problems it actually looks like you need to look at the actual geometry.

I was also wondering if I can actually think about singular $n$-simplices under homotopy equivalences. So if two singular $n$-simplices are homotopic as maps, do they necessarily generate the same element in the homotopy group?

I would appreciate any help regarding ways to visualize this and how people generally think and construct the homomorphisms in the MVS.

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For the computation Hatcher isn't using singular homology, he's using cellular homology. The Mayer-Vietoris sequence is very simple from the perspective of cellular or simplicial (delta complex) homology, and the maps are all easy to see on the chain level. I suggest backing up and reading the section on cellular homology, then run through the proof of Mayer-Vietoris from the perspective of a CW-complex being the union of two subcomplexes. –  Ryan Budney Mar 23 '11 at 1:31
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As Ryan says, this is all straightforward at the simplicial or cellular level. Assume that $A, B$ have a triangulation such that $A \cap B$ is a subcomplex of both (this is possible in most reasonable situations, such as this one). Then $H_1(A \cap B) \to H_1(A) \oplus H_1(B)$ is the direct sum of two inclusion maps - that is, it is the direct sum of the obvious inclusion maps of simplicial $1$-chains (formal sums of edges).

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I guess that this map is uniquely determined by the Eilenberg-Steenrod axioms via repeated cutting up of $A$ and $B$. It might be a valuable exercise to see how this works but I have never tried it. –  Qiaochu Yuan Mar 23 '11 at 1:50
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