Brown representability for connected pointed animae

Proof of Theorem 3.4.3. If \(F\cong [-,Z]_*\) is representable, the wedge axiom follows from the universal property of coproducts. Mapping a pushout into \(Z\) gives a pullback of mapping animae. A pair of homotopy classes whose restrictions to the intersection agree can be represented together with a homotopy between those restrictions, and therefore lifts to a point of this pullback. This proves the Mayer–Vietoris property.

Conversely, suppose that \(F\) satisfies Brown’s conditions. Apply Proposition 3.4.5 to \(X=\pt \) and the unique element of \(F(\pt )\). We obtain a connected pointed anima \(Z\) and a universal element \(\xi \in F(Z)\). We claim that \[ T_\xi \colon [X,Z]_*\longrightarrow F(X) \] is a bijection for every connected pointed anima \(X\).

For surjectivity, let \(\eta \in F(X)\) and let \(\widetilde \eta \in F(X\vee Z)\) correspond under the wedge axiom to \((\eta ,\xi )\). Applying Proposition 3.4.5 to \((X\vee Z,\widetilde \eta )\) produces a universal element \(\widetilde \xi \in F(\widetilde Z)\) and a map \(X\vee Z\to \widetilde Z\) carrying \(\widetilde \xi \) to \(\widetilde \eta \). Its restriction \(g\colon Z\to \widetilde Z\) induces isomorphisms on all positive homotopy groups because both \(\xi \) and \(\widetilde \xi \) are universal. Since both animae are connected, \(g\) is an isomorphism by Proposition 2.4.22. Composing the restriction \(X\to \widetilde Z\) with an inverse of \(g\) gives a class in \([X,Z]_*\) which maps to \(\eta \).

For injectivity, suppose that \(f,g\colon X\to Z\) satisfy \(f^*\xi =g^*\xi =: \eta \). Form the pushout

where \(\nabla \) is the fold map. The Mayer–Vietoris property gives an element \(\widetilde \eta \in F(\widetilde X)\) restricting to \(\xi \) on \(Z\) and to \(\eta \) on \(X\). Apply Proposition 3.4.5 to \((\widetilde X,\widetilde \eta )\). As in the surjectivity argument, the resulting composite \(Z\to \widetilde X\to \widetilde Z\) is an isomorphism because it carries one universal element to another. The pushout square then shows, after composing with its inverse, that both \(f\) and \(g\) are homotopic to the same map \(X\to Z\). Hence \([f]=[g]\). □