Math 6280 - Class 15

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 ∗ ∈ 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 × {0} ⊂ ∂In, so we can form the commutative diagram ∗ In−1 × {0} / 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 × {0} / J n+1 =∼ In × {0} ∪ In × {1} ∪ 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 × {0} and ∗ on In × {1} ∪ 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 × {0} ∪ 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×{1} Y −−−−−−−→ Mf,g. MATH 6280 - CLASS 15 5

Let rs : Mf,g → Mf,g be deﬁned by  y z ∈ Y rs(z) = R(x, (1 − s)t + s) z = (x, t) ∈ X × {0} ∪ 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  y z ∈ Y hs(z) = πX (Hs(x, 1)) z ∈ 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 coequalizers. Use this to deﬁne homotopy coequalizers and homotopy colimits of diagrams ∗ → ∗ → ....

Deﬁnition 3.4. The homotopy pullback of

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 = {(x, α, y) | α(0) = x, α(1) = y} ⊂ 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 ... → ∗ → ∗.