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Section 7.2 The Homotopy Extension Property

Definition 7.10.

A pair of spaces \((X, A)\) has the homotopy extension property (HEP) with respect to a space \(Y\) if for every map \(f \colon X \to Y\) and every homotopy \(H \colon A \times I \to Y\) with \(H(-,0) = f|_A\text{,}\) there exists a homotopy \(\widetilde{H} \colon X \times I \to Y\) with \(\widetilde{H}|_{A \times I} = H\) and \(\widetilde{H}(-,0) = f\text{.}\) In other words, every solid diagram
admits a filler \(\widetilde{H}\text{.}\) We say \((X, A)\) has the HEP if it has the HEP with respect to every space \(Y\text{.}\)

Definition 7.11.

A map \(i \colon A \to X\) is a cofibration if the pair \((X, i(A))\) has the HEP. In particular, \((X, A)\) has the HEP if and only if the inclusion \(i \colon A \hookrightarrow X\) is a cofibration.

Remark 7.12.

Recall the adjunction \(\mathrm{Map}(X \times I, Y) \cong \mathrm{Map}(X, Y^I)\text{,}\) where \(Y^I = \mathrm{Map}(I, Y)\) carries the compact-open topology and the bijection sends \(H\) to \(x \mapsto (s \mapsto H(x,s))\text{.}\) Under this adjunction, the HEP for \((X, A)\) is equivalent to: the inclusion \(i \colon A \hookrightarrow X\) has the left lifting property with respect to \(\mathrm{ev}_0 \colon Y^I \to Y\) for every \(Y\text{,}\) i.e. every solid square
admits a lift. The name “cofibration” comes from the fact that this is formally dual to the homotopy lifting property defining fibrations, which we will meet in Chapter 10.

Example 7.13.

The pairs \((X, \varnothing)\) and \((X, X)\) have the HEP for trivial reasons. If \((X, A)\) has the HEP and \(B\) is any space, then \((X \sqcup B, A \sqcup B)\) has the HEP: extend the given homotopy on \(A\) using the HEP and keep it as it is on \(B\text{.}\)
The following reformulation turns the HEP into a concrete question about a single pair of spaces.

Proof.
\((\Rightarrow)\) Write \(L = (X \times \{0\}) \cup (A \times I)\) and apply the HEP to the target \(Y = L\text{,}\) with \(f = i_0 \colon X \to L\text{,}\) \(x \mapsto (x,0)\text{,}\) and \(H = i \times \mathrm{id}_I \colon A \times I \to L\text{,}\) which is nothing but the inclusion of \(A \times I\) as one of the two pieces of \(L\text{.}\) These agree on \(A \times \{0\}\text{,}\) so the HEP provides a filler \(\widetilde{H} \colon X \times I \to L\) with \(\widetilde{H}(x,0) = (x,0)\) for all \(x \in X\) and \(\widetilde{H}(a,t) = (a,t)\) for all \((a,t) \in A \times I\text{.}\) Every point of \(L\) is of one of these two forms, so \(\widetilde{H}\) restricts to the identity on \(L\text{;}\) as its codomain is already \(L\text{,}\) it is a retraction.
\((\Leftarrow)\) Suppose \(A\) is closed in \(X\) and \(r\) is such a retraction. Then any maps \(f \colon X \times \{0\} \to Y\) and \(H \colon A \times I \to Y\) agreeing on \(A \times \{0\}\) assemble into a continuous map \(f \cup H \colon (X \times \{0\}) \cup (A \times I) \to Y\) (here we use that \(A\) is closed, so that \(X \times \{0\}\) and \(A \times I\) are closed subsets and the pasting lemma applies), and \((f \cup H) \circ r\) is the required extension. For the general case, see Hatcher, Proposition A.18.

Exercise 7.15.

Recall that the cone on a space \(X\) is \(CX = (X \times I)\,/\,(X \times \{1\})\text{.}\) Show that the pair \((CX, X)\text{,}\) with \(X\) included via \(x \mapsto [(x,0)]\text{,}\) satisfies the HEP. (Hint: first show that \((X \times I, X \times \{0\})\) has the HEP, by taking a retraction \(\rho\) of \(I \times I\) onto \((I \times \{0\}) \cup (\{0\} \times I)\) and using \(\mathrm{id}_X \times \rho\text{.}\) Then check that a suitable choice of \(\rho\) descends to the quotient \(CX\text{.}\)) Deduce, by a similar argument, that \((X \times I, X \times \partial I)\) has the HEP.

Proof.
The map \(CS^{n-1} \to D^n\text{,}\) \([(x,t)] \mapsto (1-t)x\text{,}\) is a homeomorphism sending the base \(S^{n-1} \times \{0\}\) to the boundary sphere \(S^{n-1} \subset D^n\text{.}\) The result therefore follows from Exercise 7.15 applied to \(X = S^{n-1}\text{.}\) For \(n = 0\) the statement is trivial: \(S^{-1} = \varnothing\) and the pair is \((*, \varnothing)\text{.}\)

First consequences.

The HEP is exactly what is needed to make quotient constructions homotopically well behaved. Here is the basic instance.

Proof.
Let \(h \colon A \times I \to A\) be a contraction of \(A\text{,}\) so \(h_0 = \mathrm{id}_A\) and \(h_1\) is constant. By the HEP applied to \(f = \mathrm{id}_X\) and the homotopy \(A \times I \xrightarrow{h} A \hookrightarrow X\text{,}\) there is a homotopy \(F \colon X \times I \to X\) with \(F_0 = \mathrm{id}_X\) and \(F_t|_A = h_t\text{.}\) Since \(F_t(A) = h_t(A) \subseteq A\) for every \(t\text{,}\) the map \(F_t\) descends to a map \(\bar F_t \colon X/A \to X/A\) with \(\bar F_t \circ q = q \circ F_t\text{,}\) and \(\bar F\) is a homotopy from \(\bar F_0 = \mathrm{id}_{X/A}\) to \(\bar F_1\text{.}\) Moreover \(F_1\) is constant on \(A\text{,}\) so it factors through the quotient: \(F_1 = g \circ q\) for a map \(g \colon X/A \to X\text{.}\) We claim that \(g\) is a homotopy inverse of \(q\text{.}\) On one hand, \(g \circ q = F_1 \simeq F_0 = \mathrm{id}_X\) via \(F\text{.}\) On the other hand, \(q \circ g \circ q = q \circ F_1 = \bar F_1 \circ q\text{,}\) and \(q\) is surjective, so \(q \circ g = \bar F_1 \simeq \bar F_0 = \mathrm{id}_{X/A}\) via \(\bar F\text{.}\)

Remark 7.18.

One might hope that \(X \simeq X/A\) whenever \(A \subseteq X\) is contractible. Surprisingly, this is false in general. Let \(X = S^1\) and let \(A = S^1 \setminus \{(1,0)\}\text{,}\) an open arc, which is certainly contractible. The quotient \(S^1/A\) has two points, \(a = [A]\) and \(b = [(1,0)]\text{,}\) and a subset is open if and only if its preimage in \(S^1\) is open. The preimage of \(\{a\}\) is \(A\text{,}\) which is open, while the preimage of \(\{b\}\) is the point \((1,0)\text{,}\) which is not. So the open sets are \(\varnothing\text{,}\) \(\{a\}\) and \(\{a, b\}\text{:}\) this is the Sierpiński space \(T\text{.}\) It is contractible: the map \(H \colon T \times I \to T\) with \(H(x, t) = x\) for \(t < 1\) and \(H(x, 1) = b\) is continuous, because \(H^{-1}(\{a\}) = \{a\} \times [0, 1)\) is open, and it is a homotopy from \(\mathrm{id}_T\) to the constant map at \(b\text{.}\) Hence \(S^1/A \simeq *\text{,}\) whereas \(S^1\) is not contractible since \(\pi_1(S^1) \cong \mathbb{Z}\text{.}\) So \(S^1 \not\simeq S^1/A\text{,}\) and by Proposition 7.17 the pair \((S^1, A)\) cannot have the HEP. The exact condition needed for \(X \simeq X/A\) is that the inclusion \(A \hookrightarrow X\) is a cofibration.

Exercise 7.19.

Let \((X, A)\) be a pair satisfying the HEP.
  1. Suppose the inclusion \(A \hookrightarrow X\) is a homotopy equivalence. Show that \(A\) is a deformation retract of \(X\text{,}\) i.e. there is a homotopy rel \(A\) from \(\mathrm{id}_X\) to a map with image in \(A\text{.}\)
  2. If \((Y, A)\) is another pair satisfying the HEP and \(f \colon X \to Y\) is a map with \(f|_A = \mathrm{id}_A\) that is a homotopy equivalence, show that \(f\) is a homotopy equivalence relative to \(A\text{,}\) i.e. there is a homotopy inverse \(g\) with \(g|_A = \mathrm{id}_A\) such that the homotopies \(g f \simeq \mathrm{id}_X\) and \(f g \simeq \mathrm{id}_Y\) can be chosen rel \(A\text{.}\) (Part (i) is the special case \(Y = A\text{.}\))
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