The Homotopy Extension Property

This note augments material in Hatcher, Chapter 0.

We assume throughout that A is a closed subspace of X.

The Homotopy Extension Property Not all inclusions A ⊂ X are created equal. One might expect that if A is contractible, the quotient map q: X → X/A should be a homotopy equivalence. This is not always true. We need a definition.

Definition 1 The pair (X, A) has the homotopy extension property (hep) if given any space Y, a homotopy f_{t}: A → Y, and a map g_{0}: X → Y such that f_{0} = g_{0}|A, there exists a homotopy g_{t}: X → Y that starts from the given map g_{0} and extends the homotopy f_{t}, in the sense that f_{t} = g_{t}|A for all t.

We are interested in knowing that such a homotopy exists; we are not interested in seeing one written out explicitly. We now have a theorem, which is a restatement of Proposition 0.17 in Hatcher. We do not repeat the proof here, as we have nothing to add. Note that A/A is just a complicated way of writing the basepoint in X/A.

Theorem 2 If the pair (X, A) has the hep and A is contractible, then the quotient map q: (X, A) → (X/A, A/A) is a homotopy equivalence of pairs.

Lemma 3 The pair (X, A) has the hep if and only if A × I ∪ X × 0 is a retract of X × I.

Proof Suppose r: X × I → A × I ∪ X × 0 is a retraction. A homotopy f_{t} and map g_{0} as above combine to give a map k: A × I ∪ X × 0 → Y. The desired homotopy is g_{t}(x) = k(r(x, t)).

Conversely, take Y = A × I ∪ X × 0, f_{t}(a) = (a, t), and g_{0}(x) = (x, 0). If (X, A) has the hep, the resulting homotopy g_{t}: X → Y can be rewritten as a retraction X × I → Y.

Corollary 4 If (X, A) has the hep, so does (W ×X, W ×A) for any space W.

Proof If r: X × I → A × I ∪ X × 0 is a retraction, so is

id_{W} × r: W × X × I → W × A × I ∪ W × X × 0.

Proposition 5 If (X, A) has the hep, so does the pair (X ×I, A×I ∪ X ×0).

Proof By Lemma 3, we have to show that (A×I ∪X×0)×I ∪(X×I)×0 is a retract of (X ×I) × I. We can rewrite the subspace as A × (I ×I) ∪ X × (0×I ∪ I ×0). We claim there is a homeomorphism of pairs

h: (I ×I, 0×I ∪ I ×0) ≅ (I ×I, I ×0).

This homeomorphism converts the subspace of X × I × I into A × I × I ∪ X × I × 0, which is a retract of X × I × I by Lemma 3, because (X ×I, A×I) has the hep by Corollary 4.

To see that such a homeomorphism exists, we may treat I × I as the cone C(∂I^{2}) on the boundary square. There is an obvious homeomorphism h: ∂I^{2} → ∂I^{2} (which we refrain from writing out explicitly) that satisfies h(P) = P′, h(Q) = Q′, etc.

h : P Q  R S T ≅ P′  Q′  R′ S′ T′ • •  • • •  • • • • •

With obvious modifications, we can also prove the following.

Proposition 6 If (X, A) has the hep, so does the pair (X ×I, A×I ∪ X ×∂I).

Deformation retracts The following substantial result is exactly Corollary 0.20 in Hatcher, but we rearrange the proof somewhat.

Theorem 7 If the pair (X, A) has the hep and the inclusion i: A → X is a homotopy equivalence, then A is a deformation retract of X.

Remark There are (at least) three different variations in the literature on the concept of deformation retract, not always consistently named. We follow Hatcher in requiring a retraction r: X → A that satisfies id_{X} ≃ i^{∘} r rel A. This is the strongest and most useful variation, and the only one we use.

Proof Let g: X → A be a homotopy inverse. A homotopy g|A = g ^{∘} i ≃ id_{A}: A → A extends by the hep to a homotopy g ≃ r: X → A; since r|A = id_{A}, r is a retraction and is still a homotopy inverse to i.

Denote by J the interval [−1, 1] ⊂ ℝ. Let f_{t}: i ^{∘} r ≃ id_{X}: X → X be a homotopy. Define a homotopy g_{u}: A × J → X by g_{u}(a, t) = f_{max(|t|,u)}(a), so that g_{u}(a, t) = a whenever t = ±1 or u = 1. Also, define a map h_{0}: X × J → X by

h_{0}(x, t) = \{ \begin{matrix}f_{|t|}(x) & if −1 ≤ t ≤ 0; \\ f_{t}(i(r(x))) & if 0 ≤ t ≤ 1. \end{matrix}

Note that these agree at t = 0, since i^{∘} r = i^{∘} r^{∘} i^{∘} r. Since (X×J, A×J) has the hep by Corollary 4, the homotopy g_{u} extends to a homotopy h_{u}: X × J → X, i. e. a map H: X × J × I → X. The map h_{0} may be interpreted as a homotopy id_{X} ≃ i^{∘} r. The other three sides of the rectangle J × I yield the desired homotopy id_{X} ≃ i^{∘} r rel A. (The picture is on page 17.)

Proposition 8 If (X, A) has the hep, then A × I ∪ X × 0 is a deformation retract of X × I.

Proof The space X × 0 is an obvious deformation retract of both X × I and A × I ∪ X × 0. Proposition 5 allows us to apply Theorem 7.

Remark There is an alternate direct (but less conceptual) proof. By Lemma 3, there is a retraction r: X × I → A × I ∪ X × 0, which we write in coordinates as r(x, t) = (r_{1}(x, t), r_{2}(x, t)). The fact that r is a retraction is expressed by the equations

r_{1}(a, t) = a, r_{2}(a, t) = t, r_{1}(x, 0) = x, r_{2}(x, 0) = 0,  (9)

where a ∈ A and x ∈ X. We define the deformation retraction s_{u}: X × I → X × I by

s_{u}(x, t) = (r_{1}(x, ut), (1−u)t + ur_{2}(x, t)).

We can check by (9) that s_{0}(x, t) = (x, t), s_{u}(a, t) = (a, t), s_{u}(x, 0) = (x, 0), and s_{1}(x, t) = r(x, t) ∈ A × I ∪ X × 0, as required.

Theorem 10 Suppose that (X, A) has the hep and that f_{t}: A → B is a homotopy. Then the adjunction spaces Y_{0} = B ∪_{f_{0}} X and Y_{1} = B ∪_{f_{1}} X are homotopy equivalent rel B.

Proof We show that Y_{0} and Y_{1} are both deformation retracts of Y = B ∪_{F} (X ×I), where F: A × I → B is the given homotopy. Consider the commutative diagram

\xymatrix{ & X × I \ar[r] & Y \\ X × 0 \ar[r]^{⊂} & A × I ∪ X × 0 \ar[u]_{⊂} \ar[r] & Y_{0} \ar[u]_{⊂} \\ A × 0 \ar[u]_{⊂} \ar[r]^{⊂} & A × I \ar[u]_{⊂} \ar[r]^{F} & B \ar[u]_{⊂}}

The two large rectangles are pushout squares by construction, and so is the lower left square. By the stacking lemma, all the squares are pushout squares. Since A×I∪X×0 is a deformation retract of X × I by Proposition 8, Y_{0} is a deformation retract of Y. Similarly, so is Y_{1}.

Cell complexes We need one basic example of the hep.

Lemma 11 The pair (D^{n}, S^{n−1}) has the hep.

Our proof may look different from the one in the book, but in fact is almost identical.

Proof Suppose given a homotopy f_{t}: S^{n−1} → Y and a map g_{0}: D^{n} → Y such that g_{0}|S^{n−1} = f_{0}. We assemble these to form a map h: D_{2} → Y from the double-size disk D_{2} = {x ∈ ℝ^{n} : ∥x∥ ≤ 2} by setting

h(x) = \{ \begin{matrix}g_{0}(x) & for ∥x∥ ≤ 1; \\ f_{∥x∥−1}(x/∥x∥) & for 1 ≤ ∥x∥ ≤ 2. \end{matrix}

The desired homotopy g_{t}: D^{n} → Y is then just g_{t}(x) = h((1+t)x).

Theorem 12 Any cw-pair (X, A) has the hep.

Proof Given a homotopy f_{t}: A → Y and a map g_{0}: X → Y such that g_{0}|A = f_{0}, we construct a homotopy g_{t}: X → Y by induction on skeletons, g_{t,n}: X^{n} → Y, starting from X^{−1} = ∅. Suppose we have g_{t,n−1} that satisfies g_{0,n−1} = g_{0}|X^{n−1} and g_{t,n−1}|A^{n−1} = f_{t}|A^{n−1}.

By definition, X^{n} may be obtained from X^{n−1} by attaching n-cells, one for each α ∈ Λ, that have characteristic maps Φ_{α}: (D_{α}^{n}, S_{α}^{n−1}) → (X^{n}, X^{n−1}) and attaching maps ϕ_{α} = Φ_{α}|S_{α}^{n−1}, where each (D_{α}^{n}, S_{α}^{n−1}) is a copy of (D^{n}, S^{n−1}). In other words, we have the pushout square

\xymatrix{∐_{α} D_{α}^{n} \ar[r]^{Φ} & X^{n} \\ ∐_{α} S_{α}^{n−1} \ar[u]_{⊂} \ar[r]^{ϕ} & X^{n−1} \ar[u]_{⊂}}

where Φ|D_{α}^{n} = Φ_{α} and ϕ|S_{α}^{n−1} = ϕ_{α}.

We use the pushout property to construct g_{t,n} from g_{t,n−1} and, for each α, a map g_{t,α}: D_{α}^{n} → Y. Commutativity requires g_{t,α}|S_{α}^{n−1} = g_{t,n−1} ^{∘} ϕ_{α}. To make g_{t,n} satisfy g_{0,n} = g_{0}|X^{n} and g_{t,n}|A^{n} = f_{t}|A^{n}, we need g_{0,α} = g_{0} ^{∘} Φ_{α}, and, in case the α-th n-cell lies in A, g_{t,α} = f_{t} ^{∘} Φ_{α}.

So in the latter case, we are forced to take g_{t,α} = f_{t}^{∘}Φ_{α}, which makes the other two conditions automatic, f_{t} ^{∘} Φ_{α}|S_{α}^{n−1} = f_{t} ^{∘} ϕ_{α} = g_{t,n−1} ^{∘} ϕ_{α} and g_{0,α} = f_{0} ^{∘} Φ_{α} = g_{0} ^{∘} Φ_{α}. Otherwise, we have only the maps g_{0,α} = g_{0} ^{∘} Φ_{α}: D_{α}^{n} → Y and g_{t,n−1} ^{∘} ϕ_{α}: S_{α}^{n−1} → Y. These are consistent, as g_{0,n−1} ^{∘} ϕ_{α} = g_{0} ^{∘} ϕ_{α} = g_{0} ^{∘} Φ_{α}|S_{α}^{n−1} = g_{0,α}|S_{α}^{n−1}. As the pair (D_{α}^{n}, S_{α}^{n−1}) has the hep by Lemma 11, we can extend these two maps to g_{t,α} as required. This completes the induction step.

Finally, we define g_{t}: X → Y by g_{t}|X^{n} = g_{t,n} for all n.