Singular Homology, Excisive Couple

Excisive Couple

Let the space S be the union of the two subspaces S1 and S2. Let C be the chains of S. Thus S1 and S2 create subchains in C. Let J be the join of these two subchains at every dimension. The homology of J in C is thus the join of the homologies of S1 and S2 within h(C).

Since the boundary operator keeps J within J, J is its own subchain of C. We may therefore talk about the homology homomorphism from h(J) into h(C) as J embeds into C. If this homomorphism is an isomorphism, then S1 and S2 form an excisive couple.

No, "excisive" isn't in the dictionary, but that's the word they used at Berkeley, so I'm going to run with it.

Interior Cover

Let the interiors of S1 and S2 cover S. Thus S1 and S2 form a covering A of S, and the subchain J is precisely the A-small chains of S. The inclusion of the A-small chains always induces an isomorphism. Thus S1 and S2 form an excisive couple.

Open Intersection

Let S1 and S2 cover S as usual. Let I be the intersection of S1 and S2, where I is a retract of an open set B in S. The interiors of S1 and S2 join with B to cover all of S. Let A be the cover S1, S2, and B. Now the A-small chains, embedded into C, induce a homology isomorphism. Thus h(C) is the join of the images of the three induced homomorphisms from S1, S2, and B.

Map I into B into S, or map I into S; the result is the same. However, since I is a retract of B, the inclusion of I into B is a homology isomorphism. It doesn't matter if we map B into S, or I into S, so map I into S. The join of the three images still produces h(C).

The chains of I are redundant; they are included in the chains of S1 and the chains of S2. The join of these two subchains yields h(C), giving an excisive couple.

Meier Viatora

Let S1 and S2 form an excisive couple for S. To find the homology of S, we only need look at the chains of S1 and S2, as they generate the same homology. That is the definition of an excisive couple.

Fix a dimension n, and look at the simplexes in S1 and S2. Let I be the intersection of S1 and S2, as a topological space and as a group of simplexes. Let W be the direct sum of C(S1) and C(S2). Map I into the first component with coefficient +1, and into the second component with coefficient -1. Clearly this is injective. View the image of I as relations in the free abelian group W. The quotient makes the two copies of I equal. In other words, the quotient is the join J of S1 and S2, and W maps onto J by adding the overlapping sections together. The sequence is exact at every dimension.

Verify that boundary commutes with the maps from I into W, and from W onto J, and turn the short exact sequence into a long exact sequence. The third column is equivalent to h(S). In some cases, the intersection I has a trivial homology, whence h(S) = h(S1)+h(S2). Watch out for the bottom row however, since h0(I) is never trivial. Assuming I is a single path component, let a base point b generate h0. This maps to +b -b in W, which is one of the generators of h0 in that group. The other generator is 0 +b, and that maps to +b in the quotient. In other words, the bottom row is split exact, with reduced homology 1 → 2 → 1.

Relative Homology

A relative version of this theorem holds when S1 contains U1 and S2 contains U2, and U1 and U2 are excisive with union U in S. Fix a dimension n, and build a short exact sequence similar to the one above. This time we're dealing with relative chain groups. The first is S1∩S2/U1∩U2, the second is the direct product S1/U1 + S2/U2, and the third is S1 join S2 mod U1 join U2. Map the simplexes from group one to group two (by inclusion in the first component and negative inclusion in the second component), and from group two to group three (by adding overlapping elements together), just as we did before. With U1∩U2 contained in U1 and U2, contained in their join, the map on cosets, i.e. quotient groups, is well defined. We need to show the sequence is exact.

Each group is generated by the simplexes that lie outside of U. For instance, the first group is generated by the simplexes in S1∩S2, that are not in U. This is a cut-down version of the original group. The earlier reasoning applies. The map from group one to group two is injective, and when you mod out by these relations, you get the third group. The sequence of quotient groups is exact.

Create the long exact sequence. Each column is a relative homology. The first is h(S1∩S2/U1∩U2). The second is the direct product h(S1/U1) + h(S2/U2). The third is the homology of the join of two chains, mod the join of two subchains. We would like to equate this with h(S/U).

The proof uses the 5 lemma - a trick you've seen before. Embed U into S, and write the long exact sequence of homologies. Then embed U1 join U2 into S1 join S2, and write the long exact sequence of homologies. Put the latter sequence above the former, with vertical arrows indicating inclusion. For a given dimension n, place the relative homology in the middle of a 5 term slice of our diagram. Thus the relative homology of S1 join S2 mod U1 join U2 maps into the relative homology of S/U, simply by including the chains of S1 join S2 into the chains of S. The two vertical arrows to the left, and the two vertical arrows to the right, induce homology isomorphisms. Invoke the 5 lemma, and the middle arrow is also an isomorphism. We have indeed produced h(S/U), and that completes the proof.

When S2 has no Homology

For a given dimension n, let hn(S2) = 0. Let various cycles in S1 generate hn(S1), and keep these cycles away from the intersection. Select a linear combination r of these cycles that is nonzero in hn. Suppose r maps to r′, which is 0 in the homology of the join of S1 and S2. In other words, r′ is the boundary of q′. Remember that each simplex of q′ lies in S1 or S2. Pull q′ back to q in S1+S2. If a simplex is in both S1 and S2, let the preimage lie in S2. When restricted to S1, the simplexes of q and q′ are really the same. Their boundaries are the same, namely r = r′. We are left with the simplexes of q that lie in S2. Their boundary maps forward, unchanged, into S2. This boundary has to be 0. Therefore q∩S2 is a cycle. Since the homology of S2 is 0, the cycle is also a boundary, say the boundary of z. Subtract z from q, and the result has r as its boundary. Thus r is actually 0 in the homology. This contradicts the selection of r, Hence the homology of S1 embeds into the homology of S1 join S2.

Review this proof again, and notice that it works just as well when we're talking about relative homology. The portion of q that lies in S2 has its boundary in U, rather than 0, but this is not a problem. Within S2, the boundary lies in U, and in S2, hence in U2. We have a cycle in S2/U2, and that implies a boundary z, and the rest of the proof follows from there.

Examples

As an example, find the homology of Sn by splitting the sphere into its northern and southern hemispheres. Their intersection is the equator, which is the retract of an open band, so we're ok. Set the bottom row aside for the moment, and the homology of the hemisphere is trivial, hence the middle row is 0. The first row contains the homology of the sphere at a lower dimension. Therefore hi(Sn) = hi-1(Sn-1.

The homology of S0 is known, and that gives us a place to start. Moving to S1, the bottom row (reduced) is 2 → 2 → 1, but we need to be a bit more specific about that first homomorphism. This intersection is not one connected space, as described above, so the sequence is not split exact. Instead, the intersection is the two endpoints of the semicircle. These fold into the same homology in either the upper semicircle or the lower semicircle. Let one end point carry through, and represent h0 in both groups, while the difference between these endpoints, which is the other generator in h0(S0), maps to 0. This becomes the image of the previous group, which is h1(S1). Beyond that, the homology of S1 is 0, just like the homology of S0. By induction, the higher spheres have h0(Sn) = hn(Sn) = 1, which agrees with our earlier result.

Next, join two circles at a point b to build the figure 8. The bottom row is 1 → 2 → 1 as usual. This is split exact, so it may as well be 0 as we move up to h1. The left column is 0, since h1(b) = 0. The middle column is 2, from h1(S1)+h1(S1), and the third column must agree. All higher homologies are 0. That completes the figure 8.

Cut a closed disk out of one side of the torus, and call this set S2. Let S1 be the closure of the complement. This is an excisive couple for the torus. Note that the figure 8 is a retract of S1. Since S2 is a disk, its homology is 0. The middle column, based on the figure 8 and the disk, is completely characterized. The intersection of S1 and S2 is a circle, which characterizes the first column. Let's see what this tells us.

The bottom is 1 → 2 → 1 as usual, so that h0(T2) = 1. The first homology has 1 → 2 → ?, from the circle and the figure 8 cross the disk. According to the lemma about S2 having no homology, the figure 8 embeds into the next group. Thus h1(T2) = 2. Moving up one row, we find an isomorphism between the third column, and the first column below. We know it's an isomorphism because the lemma tells us the second column embeds. Therefore h2(T2) = 1. All higher homologies are 0. That completes the homology of the torus, and it is consistent with our earlier results.

Distinct 2-Manifolds

The torus has a different homology from the sphere, hence they are topologically distinct, as you would expect.

The sphere has genus 0, and the torus has genus 1, because it has one hole. An extended torus might have two holes in a row, or more. These are all the compact 2-dimensional manifolds, but I'm only going to prove half the theorem here, the easy half of course. These surfaces have different homologies, and are topologically distinct.

Extend the figure 8 by one more circle, with b as the new point of tangency. We now have three circles in a row. With h1(b) = 0, the first homology of the three circles equals the homology of the figure 8 cross the homology of the circle. This has a rank of 3. All higher homologies are 0.

By induction, h1 of m circles in a row is m, and all higher homologies are 0.

Cut a disk out of the end of a g-hole torus, as we did before. (Here g is the genus of the surface.) The complement retracts to a series of 2g circles. Cross its homology with that of the disk (trivial), to get the homology of the extended torus. Thus h1 = 2g, and the surfaces are distinct.