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Equal homology does not imply homotopy equivalence

Statement refuted

Refuted claim: if two spaces have isomorphic singular homology groups in every degree, then they are homotopy equivalent.

Take X:=T2,Y:=(S1∨S1)∨S2. May's torus calculation and Miller's CW-complex calculation give H0(X;Z)≅H0(Y;Z)≅Z,H1(X;Z)≅H1(Y;Z)≅Z2, H2(X;Z)≅H2(Y;Z)≅Z,Hn(X;Z)≅Hn(Y;Z)≅0 for n≥3. Nevertheless X and Y are not homotopy equivalent.

Facts & Assumptions

Given: The spaces X=T2 and Y=(S1∨S1)∨S2.

[L1]

The torus has fundamental group π1(T2)≅Z×Z (π1(T2)≅Z×Z).

[L2]

The wedge of two circles has fundamental group freely generated by two loops (π1(S1∨S1) is the free group on two generators).

[L3]

The sphere S2 is simply connected (Sn is simply connected for every n≥2).

[L4]

A simply connected overlap turns van Kampen into a free product (A simply connected overlap turns the van Kampen pushout into a free product).

[L5]

A deformation retract induces an isomorphism on fundamental groups at every basepoint of the retract (A retract induces an injection on fundamental groups, and a deformation retract induces an isomorphism).

[L6]

Pointed maps induce functorial homomorphisms on fundamental groups, and pointed-homotopic maps induce the same homomorphism (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

[L7]

The wedge is the quotient obtained by identifying only the chosen basepoints (The wedge of a family of pointed spaces).

[A1]

The displayed singular-homology computations hold for the chosen spaces X and Y.

Counterexample

technique · direct
1.1A1

By [A1], the spaces X and Y have isomorphic singular homology groups in every degree, so the refuted claim's hypothesis holds for this pair.

1.2L7givenchooseconstruct

Write W2:=S1∨S1 and let q:W2⊔S2→Y be the quotient map from [L7], with both basepoints identified to y0∈Y. Choose a small open disk D⊆S2 around the sphere basepoint and small open arcs O1,O2 around the wedge point in the two circle summands of W2. Put U:=q(W2⊔D),V:=q((O1∪O2)⊔S2). By the quotient description in [L7], these are open path-connected subsets of Y covering Y. The set U deformation retracts onto W2 by contracting D to y0, the set V deformation retracts onto S2 by contracting the two arcs to y0, and the overlap U∩V=q((O1∪O2)⊔D) deformation retracts onto y0.

1.3L1algebra

By [L1], π1(X,([0],[0]))≅Z2. The group Z2 is abelian, while F2 is not abelian because the reduced words ab and ba are distinct. Hence π1(X,([0],[0]))≇π1(Y,y0).

2.1L2L3L4L5step 1.2

By [L2] and [L5], the deformation retraction in step 1.2 gives π1(U,y0)≅F2. By [L3] and [L5], the corresponding deformation retraction gives π1(V,y0)=1. Since U∩V deformation retracts to the point y0, it is simply connected. Applying [L4] to the open cover Y=U∪V therefore yields π1(Y,y0)≅π1(U,y0)∗π1(V,y0)≅F2.

2.2

If X and Y were homotopy equivalent, choose homotopy inverse maps h:X→Y and k:Y→X, together with homotopies H:k∘h≃id⁡X and K:h∘k≃id⁡Y. Put β(t):=H(([0],[0]),t) and γ(t):=K(y0,t). [L6, step 1.3, construct] For any path ρ from a to b, the assignment [α]↦[ρˉ∗α∗ρ] transports loop classes from π1(X,a) to π1(X,b), and reversing ρ gives the inverse transport because the inserted pairs ρ∗ρˉ and ρˉ∗ρ contract to constant loops. Applying this to β and γ yields isomorphisms β#:π1(X,k(h(([0],[0]))))→∼π1(X,([0],[0])),γ#:π1(Y,h(k(y0)))→∼π1(Y,y0). The maps h:(X,([0],[0]))→(Y,h(([0],[0]))) and k:(Y,y0)→(X,k(y0)) are pointed, so [L6] gives induced homomorphisms on fundamental groups. Precomposing H and K with based loops and then transporting the moving basepoints by β# and γ# shows that these induced homomorphisms are inverse isomorphisms up to the two basepoint transports. Therefore π1(X,([0],[0]))≅π1(Y,y0), contradicting step 1.3.

3.1step 1.1step 2.2∎

Therefore X and Y are not homotopy equivalent even though step 1.1 shows that their singular homology groups agree.

Depends on

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