Singular Homology, Singular Subdivision

Singular Subdivision

The barycentric subdivision of a geometric simplex can be generalized to a singular simplex v in the topological space S. Subdivide δ_n, then map the smaller simplexes into S, using the continuous function v. But it's not that simple. We have to worry about the order of the vertices in the (smaller) simplexes, and their coefficients. We're going to arrange things so that the boundary of the subdivision makes the internal walls go away. This may seem familiar to you - the same thing happened when we constructed a function d() to make homotopic functions chain homotopic.

The subdivision of a point yields the same point back again.

Given the standard 1-simplex [0,1], let w be the center ½. The boundary is 1 - 0, one endpoint minus the other. Connect each simplex in this boundary to w, giving the subdivision [½,1] - [½,0]. Take the boundary of this subdivision and get 1 - 0 back again. The midpoint disappears, and the original boundary returns. In other words, the boundary of the subdivision equals the subdivision of the boundary.

For the 2-simplex, let w be the barycenter, subdivide the boundary (turning three segments into 6), and prepend w to each simplex thereof. This creates the subdivision of the triangle, consisting of 6 smaller triangles. Take the boundary of this subdivision, and focus on one of the six smaller triangles. Remove w, and the original segment returns, with the same sign. The boundary of the subdivision includes the subdivision of the boundary. This is not surprising, since the subdivision was built by adding w to the subdivision of the boundary. But what about the internal walls? When w connects to a midpoint, that midpoint is designed to go away, and in the same fahsion, the incident wall goes away. The lines connecting w to the corners also go away, because we started with the subdivision of the boundary of the simplex. This will become clearer below.

Proceed by induction on the dimension n. Take the boundary of δ_n, and subdivide the resulting linear combination of faces. Join each new simplex to the center w. This creates the subdivision of δ_n.

Take the boundary of the subdivision, and the faces without w are precisely the simplexes in the subdivision of the boundary of δ_n. We only need consider the walls that contain w.

Let y be such a wall. It contains w, and one of the "walls" of the subdivision of one of the faces of δ_n. Call this wall z. Note that y contains n points, w and the points of z. A certain linear combination c of simplexes in the subdivision of δ_n contains y. We are interested in the coefficient on y, in the boundary of c. Pull w out, and we are interested in the coefficient on z, in the boundary of the subdivision of the boundary of δ_n. By induction, subdivision and boundary commute on the faces of δ_n. Thus we are taking the boundary of the boundary of the subdivision of a sum of n-1 simplexes. Since boundary² = 0, the result is 0. The coefficient on z, and on y, is 0, and all the internal walls go away. Therefore subdivision and boundary commute in all dimensions.

Self-Homotopic Chain Map

Let S be a topological space in E_j as usual. Let C be the chain map generated by the simplexes of S at all dimensions, with the boundary operator carrying C_i into C_i-1. The subdivision operator, denoted subd(), is defined on the simplexes of S, which are the generators of C. Thus subd() becomes a homomorphism from C into C. Furthermore, since subd() and bound() commute, subd becomes a valid chain map from C into C, and it induces a map from the homology of C into the homology of C. We will prove this is a homology isomorphism.

For notational convenience, I will subdivide each simplex only once. However, in the following, subd^m would work just as well. In other words, the m^th order subdivision induces an isomorphism on homology.

Remember that homotopic chain maps have the same effect on homology. Thus it is sufficient to prove subd() is chain homotopic to the identity map. This will be demonstrated via the method of acyclic models.

Let d(δ₀) = 0. Passing to S, d(x) = 0 for any point x in S. Note that b(d(x)) + d(b(x)) = 0. On the right side of the equation, subd(x) - x = 0. The homotopy relation holds in zero dimensions. Proceed by induction on n.

b(d(δ_n)) = subd(δ_n) - δ_n - d(b(δ_n))

By induction, d maps δ_n-1 into various higher dimensional simplexes living in δ_n-1. Therefore, the right side is a collection of simplexes in δ_n. The homology of δ_n (beyond h₀) is 0, thus it is sufficient to prove the right side is a cycle. Apply the boundary operator, and remember that d satisfies the homotopy relation at dimension n-1.

b(subd(δ_n)) - b(δ_n) - b(d(b(δ_n)))

b(subd(δ_n)) - b(δ_n) - { subd(b(δ_n)) - b(δ_n) - b(b(d(δ_n))) }

b(subd(δ_n)) - b(δ_n) - { subd(b(δ_n)) - b(δ_n) }

b(subd(δ_n)) - subd(b(δ_n))

Invoke the fact that boundary and subdivision commute, and the expression drops to 0. The right side is a cycle, and a boundary. Thus d implements a homotopy across all dimensions, and the chain map subd^m induces an isomorphism on the homology of S.

Relative Homology

So far, relative homology has been derived from a subspace U of the space S. The simplexes of U form a subchain of the simplexes of S. But there are other ways to build relative homology.

Let C be the chain of simplexes of S, and let E be the subchain subd^m(C). But is E really a subchain? Consider the boundary of subd^m(v) for a simplex v in C. Since boundary and subdivision commute, this is subd^m(b(v)), which is another element of E. The boundary operator keeps E within E, and we have a valid subchain of C. It is reasonable to talk about the homology of E as a stand alone chain, or the homology homomorphism created by embedding E into C, or the relative homology of C/E.

Let g be the operator subd^m, which maps C onto E. Let f be the embedding of E into C. The composition gf is the function described above, a chain map from C into C that induces a homology isomorphism. Since the composition is injective, the homology homomorphisms of g and f are both injective. We already know g is surjective, hence g is an isomorphism. For any space S, the homology of S is the homology of its subdivisions.

Suppose f is not onto. Let y be a simplex of C that represents a homology that is not in the image of E. Let z = g(y), and f(z) has to equal y, at least in its homology. This is a contradiction, hence f is also a homology isomorphism.

The embedding of E into C induces a homology isomorphism, and by the previous theorem, the relative homology of C/E = 0.

Subdividing a Space and its Subspace

Let U be a subspace of S. This creates for chains of simplexes based on S, U, subd^m(S), and subd^m(U). We want to know how the subdivision operator affects the relative homology of S/U.

Build the long exact sequence for the inclusion of U into S. For any dimension n, concentrate on 5 terms in the long exact sequence.

h_n(U) → h_n(S) → h_n(S/U) → h_n-1(U) → h_n-1(S)

A similar 5 term sequence can be built from the subdivisions of U mapping into the subdivisions of S. Place this sequence on top, and draw 5 vertical arrows connecting the top sequence to the bottom. Each vertical arrow is the inclusion of the subchain into the larger chain, or more accurately, the homology homomorphism implied by this embedding.

h_n(U′)	→	h_n(S′)	→	h_n(S′/U′)	→	h_n-1(U′)	→	h_n-1(S′)
↓		↓		↓		↓		↓
h_n(U)	→	h_n(S)	→	h_n(S/U)	→	h_n-1(U)	→	h_n-1(S)

Take a moment to show this is a commutative diagram. This is pretty clear, since homologies are represented by simplexes, and the vertical arrows fix these simplexes in larger chains. Embedding down and over is the same as embedding over and down. I'll leave the details to you.

As shown above, the inclusion of the subdivided chains of S into the chains of S induces a homology isomorphism, and the same for U. Four of the five arrows are isomorphisms, and by the five lemma, the middle arrow is also an isomorphism. Mapping the subdivisions of S/U into S/U induces an isomorphism on relative homology.