Homology, Chain Homotopic

Chain Homotopic

If two functions are homotopic, one can be morphed continuously into the other. Two chain maps (which are sort of like functions) are "chain homotopic" if there is an algebraic connection from one to the other. In fact, chain homotopic maps often correspond to homotopic functions in topological spaces. This will become clearer when you explore algebraic topology. For now, let's define chain homotopic, as a purely algebraic relationship.

Let f and g be chain maps taking the chain complex C into E. The chain maps are chain homotopic if there exists a graded homomorphism d from C_i into E_i-1, satisfying the following equation. Here b() is the boundary operator - the internal homomorphisms of the chain complexes.

(g-f)(C_i) = b(d_i(C_i)) + d_i+1(b(C_i))

Notice that every homomorphism d builds a chain map. Start with bd+db and apply b, on the left and on the right, remembering that bb = 0. Either way the result is bdb.

If f = g then f-g is the trivial chain map. Set d = 0 for the coresponding homotopy, hence every chain map is homotopic to itself.

Reverse the sign on the homomorphism d, which commutes with the boundary operator, to prove symmetry. In other words, f is homotopic to g iff g is homotopic to f.

Just as one concatenates homotopies in time, one adds the two homomorphisms to show transitivity. Add d_f, which carries f to g, and d_g, which carries g to h, and the result is (g-f)(C) + (h-g)(C), which is (h-f)(C). Therefore chain homotopic is an equivalence relation. Chaim maps can be clumped together into equivalence classes, based on homotopy.

Chain Contractable

A chain map is chain contractable if it is homotopic to the zero chain map. This is a fancy name for the equivalence class of homotopic functions that has 0 as its canonical representative.

Function Composition

Consider composition, in which f₁ and g₁ take x to y and f₂ and g₂ take y to z. Let d₁ and d₂ be the homomorphisms that implement the respective chain homotopies. Prove g₂(g₁) and f₂(f₁) are homotopic in two steps, exploiting transitivity. First, show g₂(f₁) and g₂(g₁) are homotopic by using the homomorphism g₂(d₁). To go from the first line to the second (in the following), use the fact that g₂ is a chain map that commutes with b.

b(g₂(d₁(x))) + g₂(d₁(b(x))) =
g₂(b(d₁(x))) + g₂(d₁(b(x))) =
g₂(b(d₁(x)) + d₁(b(x))) =
g₂(g₁-f₁)(x) =
(g₂(g₁) - g₂(f₁))(x)

Take g₂(f₁) to f₂(f₁) using d₂(f₁). The algebra is similar. Therefore g₂(g₁) and f₂(f₁) are chain homotopic.

Chain Homotopic Maps Induce the Same Homomorphism on Homology Groups

Since the sum of two maps induces the sum of the two induced homology homomorphisms, it is sufficient to show the homomorphism induced by g-f is 0. Yet g and f are homotopic, so the map is actually b(d)+d(b), where d implements the homotopy.

Start with x in the kernel of C₇ and apply b, then d, giving 0 in E₇. Apply d, then b, giving something in the image of E₆. Hence the entire kernel, and the associated homology group, maps into the image of E₆, which is 0 in the homology group of E₇.

Extending the Above to Relative Homology

Assume f and g are homotopic chain maps acting on a chain C with subchain A. Assume f, g, and the homotopy homomorphism d, when restricted to A, have their images in a subchain B of the range. Like the boundary operator, the d homomorphism can now be applied to cosets of A in C by following coset representatives. Since d(A) lies in B, this is well defined.

Since the homotopy equation holds for elements of C, it holds for coset representatives, and hence for cosets of A, so that the modified homomorphism d defines a chain homotopy connecting the chain maps f and g on the two relative chain complexes. Apply the above, and f and g induce the same homomorphism on the relative homology of C/A.