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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Classifying maps are natural up to based homotopy

Statement

Assume AC. Let f:(X,x0)→(Y,y0) be a based map of path-connected CW complexes, let G=π1(X,x0) and H=π1(Y,y0), and put ϕ=f∗:G→H. Let cX:X→BG and cY:Y→BH be based classifying maps inducing the identity on the identified fundamental groups. There is a based map Bϕ:BG→BH inducing ϕ, and

Bϕ∘cX≃cY∘f

by a based homotopy. Consequently the induced maps on singular homology commute.

Facts & Assumptions

Given: Full AC, the based map f, the classifying maps cX,cY, and the CW models in the statement.

[A1]

Full AC, written AC, is The Axiom of Choice.

[F1]

A classifying CW model is connected, has its specified fundamental group, and has zero higher homotopy groups. The classifying space of a discrete group is a K(G,1)

[F2]

Under AC, connected CW models K(G,1) exist for every group and the classifying-space construction is available for the stated groups. Existence and homotopy uniqueness of Eilenberg--Mac Lane spaces

[F3]

Homotopic maps induce the same homomorphism on singular homology. Homotopic maps induce the same map on singular homology

Proof

Given: The data above and [A1].

1.1A1F1F2givenconstruct

Take CW structures with the basepoints as vertices and choose a maximal tree T in the connected one-skeleton of BG, using [A1]. Send all vertices and T to the basepoint of BH. For each oriented edge e outside T, the tree paths to its endpoints turn e into a based loop; send e to a loop representing its image under ϕ. These loops generate π1(BG), and each two-cell attaching loop maps to the identity because ϕ is a homomorphism on G. Thus the map extends over the two-cells. For a cell of dimension k≥3, its attaching sphere extends because πk−1(BH)=0 by [F1]. AC chooses the tree, loop representatives, and cell extensions. The weak topology of the CW structure makes these compatible cellwise maps continuous, giving the required based map Bϕ.

2.1A1F1step 1.1givenconstruct

Write a=Bϕ∘cX and b=cY∘f, so a∗=b∗=ϕ. Choose a maximal tree in X(1) rooted at x0, and let αv be its path from x0 to each vertex v. Choose the vertex track from a(v) to b(v) to be a(αv)−1b(αv), with the track at x0 constant. For an edge e:v→w, the boundary of its homotopy square, conjugated by a(αv), represents a∗[αveαw−1] b∗[αveαw−1]−1=1. Hence every such square fills. For each k-cell with k≥2, the boundary of its homotopy prism is a sphere Sk→BH, which fills because πk(BH)=0 by [F1]. AC chooses the fillings and the CW weak topology assembles them into a continuous homotopy fixed at x0. Thus a≃b by a based homotopy.

3.1F3step 2.1∎

By [F3], the homotopy of step 2.1 gives (Bϕ)∗∘(cX)∗=(cY)∗∘f∗. This is the asserted naturality, including degree two.

Depends on

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Sources