Pushouts and Adjunction Spaces

This note augments material in Hatcher, Chapter 0.

Pushouts Given maps i: A → X and f: A → B, we wish to complete the commutative square (a) in a canonical way.

[figure: (a); X; g; Y; A; i; f; B; j; (b); Z; X; g; h; Y; m; A; i; f; B; j; k]  (1)

Definition 2 We call (1)(a) a pushout square if it commutes, j ^{∘} f = g ^{∘} i, and is universal, in the sense that given any space Z and maps h: X → Z and k: B → Z such that k^{∘}f = h^{∘}i, there exists a unique map m: Y → Z that makes diagram (1)(b) commute, m ^{∘} g = h and m ^{∘} j = k. We then call Y a pushout of i and f.

The good news is that uniqueness of pushouts is automatic.

Proposition 3 Given any maps i and f as in (1)(a), the pushout space Y is unique up to canonical homeomorphism.

Proof Suppose Y′, with maps g′: X → Y′ and j′: B → Y′, is another pushout. Take Z = Y′; we find a map m: Y → Y′ such that m ^{∘} g = g′ and m ^{∘} j = j′. By reversing the roles of Y and Y′, we find m′: Y′ → Y such that m′ ^{∘} g′ = g and m′ ^{∘} j′ = j. Then m′ ^{∘} m ^{∘} g = m′ ^{∘} g′ = g, and similarly m′ ^{∘} m ^{∘} j = j. Now take Z = Y, h = g, and k = j. We have two maps, m′ ^{∘} m: Y → Y and id_{Y}: Y → Y, that make diagram (1)(b) commute; by the uniqueness in Definition 2, m′ ^{∘} m = id_{Y}. Similarly, m ^{∘} m′ = id_{Y′}, so that m and m′ are inverse homeomorphisms.

However, existence is not automatic; pushouts must be constructed.

Proposition 4 Let i: A → X and f: A → B be any maps. Then there exists a pushout Y as in Definition 2.

Proof Let ∼ be the smallest equivalence relation on the topological disjoint union X ∐ B that satisfies i(a) ∼ f(a) for all a ∈ A. [It is the intersection of all equivalence relations on X ∐ B that have this property.] We take Y as the quotient space (X ∐ B)/ ∼, with quotient map q: X ∐ B → Y, and set g = q|X and j = q|B.

We have commutativity, since for any a ∈ A, g(i(a)) = q(i(a)) = q(f(a)) = j(f(a)). Given h and k as in diagram (1)(b), we define m̃: X ∐ B → Z by m̃|X = h and m̃|B = k. Then for any a ∈ A, m̃(i(a)) = h(i(a)) = k(f(a)) = m̃(f(a)). It follows that m̃ is constant on each equivalence class and hence factors through the map q: X ∐ B → Y to yield the desired map m: Y → Z. Further, m is unique because it is required to satisfy m ^{∘} q = m̃, with m̃ defined as above.

Corollary 5 A subset V ⊂ Y is open (resp. closed) if and only if g^{−1}(V) is open (resp. closed) in X and j^{−1}(V) is open (resp. closed) in B.

We can stack pushout squares. The proof depends only on the universal property in Definition 2 and is omitted. (Try it!)

Proposition 6 Suppose given a commutative diagram

\xymatrix{X \ar[r] & Y \ar[r] & Z \\ A \ar[u] \ar[r] & B \ar[u] \ar[r] & C \ar[u]}

in which ABXY is a pushout square. Then ACXZ is a pushout square if and only if BCY Z is a pushout square.

Remark The third possible implication fails: if ACXZ and BCY Z are pushout squares, ABXY need not be one. For a simple example, take A = X = Y = C = Z to be a point, and B any space with more than one point.

We can also take products.

Proposition 7 If diagram (1)(a) is a pushout square and W is a locally compact space, then

\xymatrix{X × W \ar[r]^{g×id} & Y × W \\ A × W \ar[u]_{i×id} \ar[r]^{f×id} & B × W \ar[u]_{j×id}}

is another pushout square.

Proof This follows from the standard (but non-trivial) topological result that if q: X ∐ B → Y is a quotient map, so is q × id_{W}: (X ∐ B) × W → Y × W. [This statement is false without some condition on W.]

Corollary 8 “The pushout of homotopies is a homotopy.” Given the pushout square (1)(a) and homotopies h_{t}: X → Z and k_{t}: B → Z such that h_{t} ^{∘} i = k_{t} ^{∘} f for all t, define m_{t}: Y → Z by m_{t} ^{∘} g = h_{t} and m_{t} ^{∘} j = k_{t}; then m_{t} is a homotopy.

Proof We take W = I in Proposition 7.

Adjunction spaces The bad news about pushouts is that Y, being a quotient space, is in general poorly behaved. Even if A, B and X are very nice spaces, Y need not even be Hausdorff. In the construction, it is far from clear what the equivalence classes in X ∐ B are, or what the points of Y really are.

We shall say no more at this level of generality. From now on, we limit attention to the following special case.

Proposition 9 In the pushout square (1)(a), if i is a closed embedding, so is j.

Proof Under this hypothesis, it becomes clear what the equivalence classes in X ∐ B are: they are the singletons {x} for each x ∈ X − A, and the sets i(f^{−1}(b)) ∐{b} for each b ∈ B. Thus as a set, Y is the disjoint union of X − A and B; in particular, j is injective. However, the topology on Y is not the disjoint union topology.

Recall that to prove j is a closed embedding, it is only necessary to show that j(F) is closed in Y whenever F is closed in B. Because j is injective, j^{−1}(j(F)) = F and g^{−1}(j(F)) = i(f^{−1}(F)). By Corollary 5, j(F) is closed.

Henceforth, we simplify notation by assuming that A and B actually are closed subspaces of X and Y, and we (usually) suppress i and j. Commutativity is simply expressed by g|A = f, and we have a map of pairs g: (X, A) → (Y, B). Informally, we obtain Y from B by gluing X to B along the subspace A of X as directed by the map f; we identify each point a ∈ A with its image f(a) ∈ B.

Definition 10 When A ⊂ X is a closed subspace, we call Y an adjunction space and f: A → B the attaching map. We write Y = B ∪_{f} X (or B ⊔_{f} X in Hatcher).

Example If the subspaces A and B of a space X are both open or both closed, then

\xymatrix{A \ar[r]^{⊂} & A ∪ B \\ A ∩ B \ar[u]_{∪} \ar[r]^{⊂} & B \ar[u]_{∪}}

is a pushout square. This is simply a restatement of the standard result that given a function f: A ∪ B → Y, if f|A and f|B are continuous, then f itself is continuous.

In general (though not always), j inherits properties from i and g from f.

Proposition 11 Assume diagram (1)(a) is an adjunction square with A ⊂ X a closed subspace. Then:

(a) If A = X, then B = Y;

(b) g|(X −A): X −A → Y −B is a homeomorphism;

(c) If B and X are T_{1} spaces, so is Y;

(d) If B and X are normal spaces, so is Y;

(e) If F is a closed subspace of X with F ∩ A = ∅, then g|F: F → Y is a closed embedding;

(f) If A is a retract of X, then B is a retract of Y;

(g) If (X, A) satisfies the homotopy extension property, so does (Y, B);

(h) If A is a deformation retract of X, then B is a deformation retract of Y.

Proof By now, (a) is trivial.

In (b), the map is obviously a continuous bijection. To see that it is an open map, take an open set V in X − A; then g(V) ∩ B = ∅ and g^{−1}(g(V)) = V show that g(V) is open. Similarly for (e).

For (c), the equivalence classes in X ∐ B in Proposition 9 are obviously closed. For (d), it is convenient to understand “normal” as not implying T_{1}; then the

Tietze Extension Theorem can be restated as: Y is normal if and only if any map u: G → ℝ from a closed subspace G of Y extends to a map v: Y → ℝ.

Given u, because B is normal, the map u|(B ∩ G) extends to a map v_{B}: B → ℝ. Now we put F = g^{−1}(G) and work in X. The two maps v_{B} ^{∘} f: A → ℝ and u ^{∘} (g|F): F → ℝ agree on A ∩ F and so define a map A ∪ F → ℝ. Because X is normal, this extends to a map v_{X}: X → ℝ. Since v_{X}|A = v_{B} ^{∘} f, we find a map v: Y → ℝ that satisfies v ^{∘} g = v_{X} and v|B = v_{B}. By construction, v extends u.

In (f), suppose r: X → A is a retraction, so that r|A = id_{A}. We define the retraction s: Y → B by s ^{∘} g = f ^{∘} r and s|B = id_{B}.

In (g), suppose given a homotopy k_{t}: B → Z and a map m_{0}: Y → Z such that m_{0}|B = k_{0}. We have a homotopy k_{t} ^{∘} f: A → Z and a map m_{0} ^{∘} g: X → Z such that (m_{0} ^{∘} g)|A = m_{0} ^{∘} g^{∘} i = m_{0} ^{∘} j^{∘} f = k_{0} ^{∘} f; by the hep for (X, A), there is a homotopy h_{t}: X → Z such that h_{0} = m_{0} ^{∘} g and h_{t}|A = k_{t} ^{∘} f. For each t, define m_{t}: Y → Z by m_{t} ^{∘} g = h_{t} and m_{t}|B = k_{t}. By Corollary 8, this is the desired homotopy.

In (h), let d_{t}: X → X be a deformation retraction, so that d_{t}|A = i, d_{0} = id_{X}, and d_{1} = r. We use the homotopy h_{t} = g ^{∘} d_{t}: X → Y and the constant homotopy k_{t} = j: B → Y to construct m_{t}: Y → Y, which is a homotopy by Corollary 8. We see that by uniqueness, m_{0} = id_{Y} and m_{1} = s, the retraction in (f).

Examples of adjunction spaces

1. The quotient space X/A is obtained by taking B to be a one-point space. As a set, its points are those of X − A together with one point corresponding to A.

2. The wedge X ∨ Y of two spaces X and Y with basepoints x_{0} and y_{0} is the quotient space (X ∐ Y)/{x_{0}, y_{0}} obtained from the disjoint union X ∐ Y by identifying the two basepoints. (It is often defined as the subspace X × y_{0} ∪ x_{0} × Y of X × Y; it is easy to construct homeomorphisms between these two definitions.)

More generally, one can form the wedge ⋁_{α} X_{α} of any collection of based spaces (X_{α}, x_{α}) as the quotient space (∐_{α} X_{α})/(∐_{α} x_{α}).

3. The cone CX on X is the quotient (X×I)/(X×0). By (e), it contains a copy of X as the image of X × 1.

4. The suspension SX of X is the pushout of X ×∂I ⊂ X ×I and the projection X × ∂I → ∂I = {0, 1}. It contains a copy of X as the image of X × (1/2).

5. The mapping cylinder M_{f} of f: A → B is obtained by taking X = A × I, with inclusion A ≅ A × 1 ⊂ A × I. By (h), B is a deformation retract of M_{f}. Also, by (e), M_{f} contains a copy of A as the image of A × 0, as well as B.

6. The mapping cone C_{f} of f: A → B is obtained by taking X = CA, with the inclusion A ⊂ CA, or equivalently as M_{f}/A (with the help of Proposition 6).

7. If we take X = D^{n}, the closed unit n-disk in ℝ^{n}, and A = S^{n−1}, its boundary sphere, the resulting space Y = B ∪_{f} D^{n}, commonly written Y = B ∪_{f} e^{n}, is said to be obtained from B by attaching an n-cell, using the attaching map f: S^{n−1} → B. Then g: (D^{n}, S^{n−1}) → (Y, B) is called the characteristic map of the n-cell.

One can attach many n-cells by taking X = ∐_{α} D_{α}^{n} and A = ∐_{α} S_{α}^{n−1}, using attaching maps f_{α}: S_{α}^{n−1} → B, where each D_{α}^{n} is a copy of D^{n}, with boundary S_{α}^{n−1}.

8. The smash product X ∧ Y is the quotient space (X ×Y)/(X ∨ Y).

9. The join X ∗ Y is the pushout of X × ∂I × Y ⊂ X × I × Y and the map X ×∂I ×Y → X ∐ Y formed from the projections X ×0×Y → X and X ×1×Y → Y.