Singular Homology, Homotopically Equivalent Spaces

Isomorphic Homology

Homeomorphic spaces have isomorphic homology groups. This is intuitive, since we are merely relabeling the points of the space. Or you can be technical and prove the homeomorphism induces a group isomorphism at every dimension, leading to the same homology groups.

Homotopically equivalent spaces also have isomorphic homology groups. This is not intuitive, and it's not easy to prove. Begin by looking at a simplex crossed with the unit interval I. A function on this "cylinder" implements a homotopy between the functions on the two ends. Once the cylinder is partitioned into smaller simplexes, we'll be able to apply the boundary operator and make real progress. For now, let's define, and partition, the standard cylinder.

Partitioning the Cylindrical Box

With I running along dimension n+1, consider the product space δ_n:I. This is a bit like a cylinder whose cross section, at every slice, is δ_n. If n = 2, picture a tall box whose cross section is a triangle. The height of the box is the unit interval from 0 to 1.

This box isn't a simplex, so how can it be cut up into simplexes? This looks like a tricky geometry puzzle, but there is a natural partition. Let e_i:0 be a vertex at the bottom of the box, and let e_i:1 be the corresponding vertex at the top of the box. Start with the simplex determined by e₀:0 and the vertices around the top. Then bring in the simplex bound by e₀:0, e₁:0, and e₁:1 through e_n:1. Then bring in the simplex bound by e₀:0, e₁:0, e₂:0, and e₂:1 through e_n:1. The i^th simplex is determined by e₀:0 through e_i:0 and e_i:1 through e_n:1. The last simplex is framed by the bottom and e_n:1. Proving this is a partition is tricky, but maybe some imagery will help.

Look at the triangular box, which fits conveniently within our 3 dimensional imagination. Draw a line from e₀:0 to e₂:1. This is a diagonal line on one side of the box. A plane is going to rotate about this line, like a page turning. In the beginning the plane is the side of the box. Turn it past e₀:1, until it touches e₁:1. The plane now cuts through the box, and forms the inside face of the first simplex. Then turn the page some more, until the bottom of the plane reaches e₁:0. This face separates the second and third simplexes. Turn the page some more and it once again coincides with the side of the box, having spun 180 degrees.

Higher dimensions are harder to visualize, so let's look at the algebra. It isn't geometrically accurate, but you can (algebraically) arrange the bottom vertices in a circle, from e₀ through e_n. The top vertices form a parallel circle. The i^th simplex repeatse e_i at the bottom and the top. Vertices < i come from the bottom, and vertices > i come from the top. As we move from one simplex to the next, the intersection, i.e. the face they have in common, is e₀ through e_i on the bottom, and e_i+1 through e_n on the top. This is just enough points to form an n dimensional face, which is the intersection of two adjacent simplexes.

To obtain this face from the i^thsimplex, one removes e_i:1. Remember that the vertices of any simplex are ordered, and in this case we are removing the vertex between e_i:0 and e_i+1:1. Moving to the next simplex, we obtain the common face by removing e_i+1:0, which is also the vertex between e_i:0 and e_i+1:1. The boundary of simplex i, and the boundary of simplex i+1, both produce the common face with the same sign, according to the parity of i.

What if we want these internal faces to go away? This can be done by expressing the cylinder as an alternating sum of simplexes. The first simplex is positive, and its internal face, between the first and second simplexes, appears with a coefficient of -1. This is because we remove e₀:1 (the second vertex) from the first simplex to realize this face, and that brings in a coefficient of -1. On the other side, i.e. the second simplex, we are once again removing the second vertex. This brings in a coefficient of -1, but the simplex also has a coefficient of -1, so the internal face disappears. This takes place all the way through the cylinder. The boundary of the alternating sum of these internal simplexes has no internal faces.

What about the outside faces? The first simplex brings in the top, with a coefficient of +1. At the end, we remove e_n:1 from the last simplex, and the bottom face has a coefficient of -1. The walls of the cylinder also appear, in their entirety, though they are cut, by the simplexes, into plus and minus "triangular" regions. Let one of these triangular regions have e_i:0 and e_i:1 in its vertex list for some i in [0,n]. Some other vertex has been removed from the simplex to produce the external face. Each such region appears once, with a sign commensurate with the missing index and the repeated index.

Operator

Build an operator d() as follows. Start with the standard simplex δ_n, and cross this with the unit interval in the next dimension. This is a specific shape in E_j, or in real space if you prefer, which I will call the standard cylinder. Act on the standard cylinder by creating an alternating sum of n+1 simplexes, of dimension n+1, as described above. This is d(δ_n), in E_j, or in any space containing the standard cylinder.

Let S be an arbitrary subspace of E_j, and let v be a singular simplex in S, i.e. a continuous function from δ_n into S. Cross S with the unit interval, giving a topological space S:I. This can always be embedded in an instance of E_j, so there is no trouble. extend v, in the obvious way, across the standard cylinder. The bottom maps to v:0, and the top maps to v:1, with the interior mapping linearly to v:I. Cut the standard cylinder into simplexes, as described above, then apply v. The resulting sum of simplexes in S:I is, by definition, d(v).

Our d() operator is a group homomorphism from G_n(S) into G_n+1(S:I). The boundary operator reduces dimension, but this particular homomorphism increases dimension. It also maps into a different space, S:I, rather than the same space S. However, we may consider S to be a subspace S:0 of S:I, if that serves our purposes.

Notice that d(v) creates simplexes entirely within v cross I, in the larger space S:I. Had we chosen another singular simplex v′, disjoint from v, or sharing only a border with v, d(v′) and d(v) would produce independent simplexes in S:I.

b(d) and d(b)

For notational convenience, use b() for the boundary operator. We have already looked at b(d(δ_n)). This generates +δ_n:1, and -δ_n:0, and a collection of regions that comprise the walls of the box. Conversely, let's look at d(b(δ_n)).

Each face in the boundary of the standard simplex is separate and apart from the others. They share only a lower dimensional border. For example, two edges in our triangle meet at a vertex. If f₁ and f₂ are two such faces, d(f₁) creates simplexes that fill f₁:I, while d(f₂) creates simplexes that fill f₂:i, and these simplexes are independent of each other. In other words, d() fills up each wall of the box in turn.

Now for the magic. Focus on a particular region in the wall of the box. This was characterized earlier as e₀:0 through e_i:0 union e_i:1 through e_n:1, with one of the vertices, other than e_i, missing. The sign depends on the index that is duplicated and the index that is missing. A similar formula holds for d(b(δ_n)). Remove one of the vertices, giving a face f in δ_n, then apply d(), which duplicates one of the remaining vertices at the top and bottom. The same regions appear. If you want to be rigorous, you need more than combinatorics. It isn't enough to reproduce the same set of vertices; the very same function must appear, from δ_n-1 into δ_n:I. Everything is linear, using barycentric coordinates, and b(d) and d(b) do indeed produce exactly the same set of functions into the walls of our cylinder. I'll leave the details to you; I just didn't want you to miss this step.

Since the functions are the same, look at the coefficients. This is an exercise in parity combinatorics, keeping track of what is missing and what is duplicated. Apply the same renumbering argument we saw earlier, when proving boundary² = 0. The sign from d(), or b(), is the parity of the vertex repeated, or the vertex removed, respectively. Let these vertices be i and j. Either i > j or j > i. In one case, bd or db, the higher vertex is repeated or removed first, then the lower. The sign is the parity of i times the parity of j. In the other case, the lower vertex is repeated or removed first, the rest of the vertices are renumbered, and the higher vertex has its parity changed. Thus the sign has been reversed. The signs that result from d(b) and b(d) are opposite. Therefore, the regions in the walls of our cylinder cancel under the formula b(d)+d(b), Leaving only the top minus the bottom.

Compose this with the function v into the space S, and the same relationship holds for the faces and regions of v.

b(d(v)) + d(b(v)) = v:1 - v:0

Homotopic Functions

Let f be a continuous function from S:I into T. In other words, f implements a homotopy on the domain S, into the range T, starting with f restricted to S:0 and ending with f restricted to S:1. Call these functions f₀ and f₁ respectively.

Remember that f₀ and f₁ both implement homology homomorphisms from the homology of S into the homology of T. We would like to relate these homology homomorphisms. In fact, these homomorphisms are the same. The action of f₀ on the homology of S is the same as the action of f₁ on the homology of S. This derives, ultimately, from the fact that f₀ and f₁ are homotopic on S. Here is the proof.

Let v be a singular simplex in S. Thus v:I is a cylinder inside S:I. The homotopy equation tells us b(d(v)) + d(b(v)) = v:1 - v:0. Apply f, and get this.

f(b(d(v))) + f(d(b(v))) = f₁(v) - f₀(v)

Since the boundary operator commutes with f, write the equation this way.

b(f(d(v))) + f(d(b(v))) = f₁(v) - f₀(v)

Let f(d()) be a new function, called q(), that maps simplexes of S into simpliexes of T. This leads to the following equation, on the simplex v, and on the group element represented by v.

b(q(v)) + q(b(v)) = f₁(v) - f₀(v)

At this point I am going to invoke the power of homology theory. Whenever a group homomorphism (such as q) satisfies the above equation, it makes the two chain maps f₀ and f₁ chain homotopic, and the resulting homology homomorphisms are identical.

Homotopically Equivalent

Let S and T be two topological spaces. Let two functions map S into T, and T into S respectively. Let the composition of these functions produce a function from S into itself that is homotopic to the identity map on S. Similarly, let the composition of these two functions, in the reverse order, produce a function from T into itself that is homotopic to the identity map on T. For example, the functions might be a homeomorphism and its inverse, whence the composition of the two maps is trivially homotopic to the identity map.

The two functions (whose composition in either direction is homotopic to the identity map) induce corresponding group homomorphisms. These in turn form a chain equivalence, and that completes the proof. The homology of S equals the homology of T.

This is not a theoretical isomorphism; the map and its inverse are induced by f and g respectively. Apply f to a simplex in S, representing a homology group member, and get a simplex in T, representing the corresponding homology group member.

Another application sets T = S, and g = the identity map. If f is homotopic to the identity map on S, then so are fg and gf. Now S is chain homotopic to S through f, and f produces the identity map on homology. Yes, f applied to a simplex may give a different simplex, but it represents the same homology group member. A generator of the homology, if there is one, maps to another generator of the homology.