4.3 COMPROBACIÓN DE HIPÓTESIS
4.3.4 VERIFICACIÓN DE LA HIPÓTESIS GENERAL
Now we return to the homotopy excision theorem. Its conditions m ≥1 and
n ≥ 1 merely give that π0(C) −→ π0(A) and π0(C) −→ π0(B) are surjective, and any extraneous components ofAorB would not affect the relevant homotopy groups. The conditionm≥2 implies that (X, B) is 1-connected. By the long exact sequence just given, the theorem is equivalent to the following one.
Theorem. Under the hypotheses of the homotopy excision theorem,
πq(X;A, B) = 0 for 2≤q≤m+n−2
and all choices of basepoint ∗ ∈C.
In this form, the conclusion is symmetric inAandBand vacuous ifm+n≤3. Thus our hypotheses m ≥ 2 and n ≥ 1 are the minimal ones under which our strategy can apply.
In order to have some hope of tackling the problem in direct terms, we first reduce it to the case whenAandBare each obtained fromCby attaching a single cell. We may approximate our given excisive triad by a weakly equivalent CW triad. This does not change the triad homotopy groups. More precisely, by our connectivity hypotheses, we may assume thatXis a CW complex that is the union of subcomplexesAandB with intersection C, where (A, C) has no relativeq-cells forq < mand (B, C) has no relativeq-cells forq < n. Since any mapIq −→X has image contained in a finite subcomplex, we may assume thatX has finitely many cells. We may also assume that (A, C) and (B, C) each have at least one cell since otherwise the result holds trivially.
We claim first that, inductively, it suffices to prove the result when (A, C) has exactly one cell. Indeed, suppose thatC⊂A′ ⊂A, whereAis obtained fromA′by attaching a single cell and (A′, C) has one less cell than (A, C). LetX′=A′∪CB. If the result holds for the triads (X′;A′, B) and (X;A, X′), then the result holds for the triad (X;A, B) by application of the five lemma to the following diagram:
πq+1(A, A′) // πq(A′, C) // πq(A, C) // πq(A, A′) // πq−1(A′, C) πq+1(X, X′) //π q(X′, B) //πq(X, B) //πq(X, X′) //πq−1(X′, B). The rows are the exact sequences of the triples (A, A′, C) and (X, X′, B). Note for the caseq= 1 that all pairs in the diagram are 1-connected.
We claim next that, inductively, it suffices to prove the result when (B, C) also has exactly one cell. Indeed, suppose thatC⊂B′ ⊂B, whereB is obtained fromB′ by attaching a single cell and (B′, C) has one less cell than (B, C) and let
X′=A∪CB′. If the result holds for the triads (X′;A, B′) and (X;X′, B), then the result holds for the triad (X;A, B) since the inclusion (A, C)−→(X, B) factors as the composite
88 THE HOMOTOPY EXCISION AND SUSPENSION THEOREMS
Thus we may assume thatA =C∪Dm and B =C∪Dn, where m≥2 and
n≥1, and we fix a basepoint∗ ∈C. Assume given a map of tetrads (Iq;Iq−2× {1} ×I, Iq−1× {1}, Jq−2×I∪Iq−1× {0})
f
(X;A, B,∗),
where 2 ≤q ≤m+n−2. We must prove that f is null homotopic as a map of tetrads. For interior pointsx∈Dmandy∈Dn, we have inclusions of based triads
(A;A, A−x)⊂(X− {y};A, X− {x, y})⊂(X;A, X− {x})⊃(X;A, B).
The first and third of these induce isomorphisms on triad homotopy groups in view of the radial deformation away fromyofX−{y}ontoAand the radial deformation away fromxofX− {x}ontoB. It is trivial to check thatπ∗(A;A, A′) = 0 for any
A′ ⊂A. We shall show that, for well chosen points xandy,f regarded as a map of based triads into (X;A, X− {x}) is homotopic to a mapf′ that has image in (X− {y};A, X− {x, y}). This will imply thatf is null homotopic.
LetDm
1/2⊂DmandDn1/2⊂Dnbe the subdisks of radius 1/2. We can cubically subdivideIq into subcubesIq
αsuch thatf(Iαq) is contained in the interior ofDmif it intersectsDm
1/2andf(Iαq) is contained in the interior ofDn if it intersectsDn1/2. By simplicial approximation,f is homotopic as a map of tetrads to a mapgwhose restriction to the (n−1)-skeleton ofIq with its subdivided cell structure does not coverDn
1/2and whose restriction to the (m−1)-skeleton ofIq does not coverDm1/2. Moreover, we can arrange that the dimension ofg−1(y) is at mostq−nfor a point
y∈Dn
1/2that is not in the image undergof the (n−1)-skeleton ofIq. This is the main point of the proof, and to be completely rigorous about it we would have to digress to introduce a bit of dimension theory. Alternatively, we could use smooth approximation to arrive atgandywith appropriate properties. Since the intuition should be clear, we shall content ourselves with showing how the conclusion of the theorem follows.
Letπ:Iq −→Iq−1 be the projection on the firstq−1 coordinates and letK be the prismπ−1(π(g−1(y))). ThenKcan have dimension at most one more than the dimension ofg−1(y), so that
dimK≤q−n+ 1≤m−1.
Thereforeg(K) cannot cover Dm
1/2. Choose a pointx∈D1/2m such that x6∈g(K). Sinceg(∂Iq−1×I)⊂A, we see thatπ(g−1(x))∪∂Iq−1 andπ(g−1(y)) are disjoint closed subsets ofIq−1. By Uryssohn’s lemma, we may choose a mapv:Iq−1−→I such that
v(π(g−1(x))∪∂Iq−1) = 0 and v(π(g−1(y))) = 1.
Defineh:Iq+1−→Iq by
h(r, s, t) = (r, s−stv(r)) for r∈Iq−1 and s, t∈I.
Then let f′ = g◦h1, where h1(r, s) = h(r, s,1). We claim that f′ is as desired. Observe that
3. PROOF OF THE HOMOTOPY EXCISION THEOREM 89
Moreover,
h(r, s, t) = (r, s) if h(r, s, t)∈g−1(x) sincer∈π(g−1(x)) impliesv(r) = 0 and
h(r, s, t) = (r, s−st) if h(r, s, t)∈g−1(y)
sincer∈π(g−1(y)) implies v(r) = 1. Theng◦his a homotopy of maps of tetrads (Iq;Iq−2× {1} ×I, Iq−1× {1}, Jq−2×I∪Iq−1× {0})
(X;A, X− {x},∗)
CHAPTER 12