Singular Homology, Relative Homology

Relative Homology

In many cases, h_n(S) has less information than π_n(S), as demonstrated by the first homology - so what's the point? For one thing, homology is often easier to calculate than the higher order π groups. Beyond this, homology leads to relative homology, which is a very powerful tool in our arsenal.

Let U be a subspace of S. At every dimension, some of the simplexes of S lie in U, and some do not. When a simplex lies in U, its boundary also lies in U. At each dimension, consider the group of simplexes in S, mod the subgroup of simplexes in U. These quotient maps stack up, and map into one another by the boundary operator. Divide cycles by boundaries, and get the relative homology of S/U.

Continuous Functions Induce Homology Homomorphisms

Let f be a continuous map from S into T, such that f maps U into a subspace V. Review again the page on relative homology, and note that f induces a chain map from the relative simplexes of S/U into the relative simplexes of T/V, and creates a graded homomorphism from the relative homology of S/U into the relative homology of T/V.

This inspires a new category, consisting of topological spaces and subspaces. Morphisms are continuous functions that map the subspace of the domain into the subspace of the range. A functor carries this category into chain complexes, which are the simplexes of each space mod its subspace. Continuous functions induce chain maps from one chain complex into the next. These chain complexes can then be transformed into relative homology groups, whereupon the aforementioned chain maps create relative homology homomorphisms. The composition of continuous functions induces the composition of chain maps and homology homomorphisms, as one would expect from a functor.

Homotopic Functions

Let f₀ and f₁ be homotopic functions from S into T, such that both functions map U into V. Let the homotopy f map U into V at every level. We would like to show that f₀ and f₁ induce the same homomorphisms from the relative homology of S/U into the relative homology of T/V.

We already proved the induced functions on the simplexes of S are chain homotopic. This was done by building a homomorphism d that established the homotopy. These results extend to relative homology, provided d maps U into V. Remember that d takes the standard simplex δ_n and expands it into an alternating sum of n+1 dimensional simplexes in the box δ_n cross I. The same thing takes place inside U, and when we apply f, U cross I maps entirely into V. Thus d, restricted to U, creates higher dimensional simplexes that lie in V. The conditions of the theorem are met, and f₀ and f₁ both induce the same homomorphisms from the relative homology of S/U into the relative homology of T/V.

Homotopically Equivalent

Let S and T be homotopically equivalent spaces via two functions f and g. If h is the homotopy that slides g(f(S)) back to the identity on S, restrict h to U, and g(f(U)) slides onto U. This follows from the restriction of h to U, but, we will make one more assumption. Let the image of U remain inside U throughout the entire homotopy. Also, f(U) lies in V. Similar assumptions are made concerning the homotopy from f(g(T)) into T, restricted to V, and g(V) lying in U. Now the spaces S/U and T/V are homotopically equivalent.

The functions f and g define chain maps between the simplexes of S and the simplexes of T. The same continuous functions, restricted to U and V, produce the same chain maps, restricted to the simplexes of U and V. The composition fg carries the relative homology of S/U into the relative homology of S/U, and this homomorphism is the same as the homotopically equivalent homomorphism produced by the identity map. The composition of the two induced homomorphisms leaves the relative homology unchanged. The same holds in the other direction. Therefore the relative homology of S/U equals the relative homology of T/V.

As a special case, let S be a retract of T, and let U, using the same homotopy, be a retract of V. Now S/U and T/V have the same relative homology.

Continuing the above, let S be a retract of T, such that the homotopy that squashes T down to S keeps S within S at all times. Now S/S is a retract of T/S. The relative homology of S/S is 0, and the same holds for the relative homology of T/S. As an example, the homology of the punctured plane, relative to the unit circle, is 0.

Homotopic Implies Homologous

Let S be a space that is homeomorphic to a ball in n dimensions, and let f be a map from δ_n onto S, such that the boundary of δ_n maps into the boundary of S. This represents an element in the homology of S relative to its boundary. If g is another function satisfying these constraints, and f and g are homotopic, and the homotopy keeps the boundary of δ_n in the boundary of S at all times, then f and g are homologous.

Review the operator d() that partitions the homotopy into simplexes of dimension n+1. Its boundary, denoted b(d), is g - f - d(b). When d is applied to the boundary of δ_n, the result lies in the walls of δ_n cross I. Apply the homotopy, and the resulting n simplexes lie in the boundary of S. These are 0 in the relative homology. Therefore f and g are homologous.

This result can be generalized. Assume f is a linear combination of images of simplexes that partition a domain space U. For instance, U might be a square, cut into two triangles, such that the boundary of the triangles (taken together) forms the boundary of the square, and f implements a simplex from the first triangle into S, and another simplex from the second triangle into S. A homotopy carries f onto g, and keeps the image of the boundary of U in the boundary of S at all times. Apply d to all the simplexes of U in parallel, and then apply the homotopy. Both b and d are group homomorphisms, so b(d) can be computed per simplex and added up. This gives the term d(b), for each simplex. This is d appliec to the sum of the boundaries of the simplexes of U. By assumption, the sum of these boundaries lies in the boundary of U. Apply d, and all faces are in the walls of U cross I. Apply the homotopy, and find simplexes in the boundary of S, which don't matter. This leaves the sum of the simplexes in g, minus the sum of the simplexes in f, which we know better as g-f. Therefore the two group members are homologous.

In some cases we may be able to expand one of the simplexes of U, thus crushing all the others into the wall. Apply g, and only the expanded simplex remains. All the others lie int the boundary of S, and are 0 in the relative homology. Thus a collection of simplexes covering S is often homotopic to, and homologous to, a single simplex covering S.