Singular Homology, The Method of Acyclic Models

The Method of Acyclic Models

In the last section we proved homotopic functions induce chain homotopic maps. In fact we constructed a specific graded homomorphism d that did the trick. However, the construction of d was rather involved. There were lots of details; maybe we didn't get everything right. This section proves the same theorem, without getting into the details of d. Instead, we prove that d exists. After all, existence is all we need. The chain maps become chain homotopic, and they lead to the same homology homomorphisms on S.

Recall that d was defined on the unit simplex, mapping it into a linear combination of higher dimensional simplexes in δ_n cross the unit interval I. This map d was then extended to any simplex v(δ_n) in the space S by applying v to d(δ_n):I. In other words, d and v commute by definition.

This time we are going to define d only for the 0 dimensional simplex. After that, we will prove that d exists, by induction, for each dimension.

The standard simplex in 0 dimensions is a point, and the box, δ₀ cross I, is a segment of length 1. Let d map the point to the entire segment, which happens to be a 1-simplex. Thus d maps every 0-simplex in S to a 1-simplex in S:I.

Have a gander at bd+db. The boundary of a point disappears, taking one of the terms with it. Moving to the other term, apply d and get the segment x:I. The boundary is x:1 - x:0. This is precisely the relationship we expect from bd+db.

Let f implement a homotopy from S:I into T, as was done in the previous page. Let f₀ be f restricted to S:0, and let f₁ be f restricted to S:1. Push everything through f, and d has the desired property, at least in 0 dimensions. That is, bd+db = f₁-f₀.

We're ready for the inductive step. Assume d has been defined for all dimensions below n. Let q = δ_n:0 (the bottom), and let r = δ_n:1 (the top). We are trying to find d such that db = r - q - bd.

If the right side is the boundary of a linear combination of n+1 dimensional simplexes u, then set d(δ_n) = u, and we're done. We only need show the right side is a boundary.

The space δ_n cross I is comvex, hence it has trivial homology beyond h₀. At each level, cycles and boundaries coincide. It is sufficient to prove r-q-bd is a cycle. Apply the boundary operator, and see if the result is 0.

b(r) - b(q) - b(d(b(δ_n)))

The boundary of δ_n lives in n-1 dimensions, where d is already defined. Apply the homotopy relationship at this level.

b(r) - b(q) - { r(b) - q(b) - d(b(b(δ_n))) }

Since boundary² = 0, that term goes away.

b(r) - b(q) - { r(b) - q(b) }

The boundaries of q and r are well understood. In the context of the prior homotopy, q(b) is the boundary of the standard simplex run through q. This is the boundary of q. In the same way, r(b) is the boundary of the standard simplex, pushed up to r, which is the boundary of r. We're really saying boundary commutes with other continuous functions, which is true by definition. Therefore the entire expression drops to 0, and that completes the proof.

Specific homomorphisms d_n have not been constructed for each dimension n, but the method of acyclic models proves these homomorphisms exist, whence homotopic functions induce the same homology homomorphisms.