MATH 6280 - CLASS 15 Contents 1. Long exact sequence on homotopy groups for a pair 1 2. Relationship between fibrations and cofibrations 3 3. Homotopy pushouts 4 These notes are based on • Algebraic Topology from a Homotopical Viewpoint, M. Aguilar, S. Gitler, C. Prieto • A Concise Course in Algebraic Topology, J. Peter May • More Concise Algebraic Topology, J. Peter May and Kate Ponto • Algebraic Topology, A. Hatcher 1. Long exact sequence on homotopy groups for a pair Recall that πn(X; A) = πn(X; A; ∗) = πn−1Pi = [(In; @In;J n); (X; A; ∗)] = [(In; @In); (X; A)] = [(Dn;Sn−1); (X; A)]: It follows immediately from the long exact sequence for homotopy groups of a fiber sequence that there is a long exact sequence: ::: / π2(A) / π2(X) / π2(X; A) / π1(A) / π1(X) / π1(X; A) / π0(A) / π0(X) The maps πn(A) ! πn(X) and πn(X) ! πn(X; A) are the obvious restrictions πn(X; A) ! πn−1(A) 1 2 MATH 6280 - CLASS 15 sends (Dn;Sn−1; ∗) ! (X; A; ∗) to the restriction (Sn−1; ∗) ! (A; ∗). Remark 1.1. For ∗ 2 B ⊆ A ⊆ X we get a long exact sequence for a triple: ::: / π3(X; A) / π2(A; B) / π2(X; B) / π2(X; A) / π1(A; B) / π1(X; B) / π1(X; A) Theorem 1.2. Let B be path-connected. If p : E ! B is a Serre fibration, with fiber F = p−1(∗), then πn(E; F ) ! πnB f p◦f is an isomorphism for n ≥ 1, where the map takes (Dn;Sn−1) −! (E; F ) to (Dn;Sn−1) −−! (B; ∗). In particular, for a Serre fibration, there is a long exact sequence on homotopy groups: ::: / π2B / π1F / π1E / π1B / π0F / π0E / ∗ f Proof. Let (In; @In) −! (B; ∗) be a homotopy class. Then In−1 × f0g ⊂ @In, so we can form the commutative diagram ∗ In−1 × f0g / E : fe p In / B f Since p is a Serre fibration, there is a lift fe : In ! E. Such that pfe = f. Since f(@In) = ∗, we must have fe(@In) ⊂ p−1(∗) = F; so fe :(In; @In) ! (E; F ) and πn(E; F ) ! πnB is surjective. fe Now, suppose that (In; @In;J n) −! (E; F; ∗) is such that pfe ' ∗. ∗ fe : • • ∗ E F • • ∗ MATH 6280 - CLASS 15 3 and ∗ pfe : • • ∗ B ∗ • • ∗ We can find a commutative diagram =∼ In × f0g / J n+1 =∼ In × f0g [ In × f1g [ J n × I =∼ In × I / In × I where the horizontal arrows are homeomorphisms. So, E ! B has the HLP with respect to the inclusion J n+1 ! In × I. Choose a null-homotopy H :(In; @In) × I ! (B; ∗) from pfe to ∗. We let h : J n+1 ! E be fe on In × f0g and ∗ on In × f1g [ J n × I. We have a diagram with a lift h J n+1 / E < H e p In × I / B H One now only needs to check that He is a null-homotopy from fe to the constant map. So, we get a long exact sequence on homotopy groups coming from the long exact sequence for the pair (E; F ): ::: / π2B / π1F / π1E / π1B / π0F / π0E We just need to check surjectivity at the end. Suppose that e is a point in E. Choose a path γ from p(e) to ∗ in B. Choose a lift γe which starts at e. Then p(γe(1)) = ∗ so γe(1) is in the same path component as e and also in the fiber F . Hence, π0F ! π0E is surjective. 2. Relationship between fibrations and cofibrations There is a the following close connection between fibrations a cofibration. 4 MATH 6280 - CLASS 15 Proposition 2.1. • A map A ! X is a cofibration with closed image if and only if it has the left lifting property with respect to all Hurewicz fibrations which are also homotopy equivalences. That is, given any fibration p : E −!' B A / E > i p X / B there is map X ! E making the diagram commute. • A map E ! B is a Hurewicz fibration if and only if it has the right lifting property with respect to all cofibrations with closed image which are also homotopy equivalences. That is, given any cofibrations with closed image i : A −!' X A / E > i p X / B there is map X ! E making the diagram commute. 3. Homotopy pushouts Definition 3.1. The homotopy pushout of g A / Y f X is the double mapping cylinder, which is the pushout of ftg A t A / Y t X i0ti1 A × I / Mf;g Proposition 3.2. If f is a cofibration, then Mf;g ! X [A Y is a homotopy equivalence. Proof. Let R : X × I ! X × f0g [ A × I be the retract which exists since A ! X is a cofibration. Let q : Mf;g ! X [A Y be the quotient and p : X [A Y ! M(f; g) be the composite i1 R[g(A)idY X [A Y −! X × I [A×{1g Y −−−−−−−! Mf;g: MATH 6280 - CLASS 15 5 Let rs : Mf;g ! Mf;g be defined by 8 <y z 2 Y rs(z) = :R(x; (1 − s)t + s) z = (x; t) 2 X × f0g [ A × I: Then, r0(z) = idMf;g and r1(z) = p ◦ q. Conversely, let H : X × I × J ! X × I be the homotopy between idX×I to R. Then let ht : X [A Y × I ! X [A Y be 8 <y z 2 Y hs(z) = :πX (Hs(x; 1)) z 2 X: Then one can verify that h is a homotopy between the identity and q ◦ p. Exercise 3.3. Define coequalizers as pushouts and colimits of diagrams ∗ ! ∗ ! ::: as coequal- izers. Use this to define homotopy coequalizers and homotopy colimits of diagrams ∗ ! ∗ ! :::. Definition 3.4. The homotopy pullback of Y X / Z is the double mapping path space, which is the pull-back of Ef;g X × Y f×g (ev0;ev1) ZI / Z × Z That is, I Ef;g = f(x; α; y) j α(0) = x; α(1) = yg ⊂ X × Z × Y Exercise 3.5. Prove that if f is a fibration, then the natural map X ×Z Y ! Ef;g which sends (x; z; y) to (x; αz; y), for αz the constant path at z, is a homotopy equivalence. Exercise 3.6. Define equalizers as pull-backs and limits of diagrams ::: ! ∗ ! ∗ as equalizers. Use this to define homotopy equalizers and homotopy limits of diagrams ::: ! ∗ ! ∗..