MATH 6280 - CLASS 10

Contents

1. Replacing f : X → Y by a cofibration 1 2. More details about the based case 2 3. Barratt-Puppe sequence 3

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. Replacing f : X → Y by a cofibration f ∼ Lemma 1.1. If X −→ Y is a cofibration, then Cf → Cf /CX = Y/X is a homotopy equivalence, so f i that the cofiber sequence X −→ Y −→ Cf is, up to homotopy equivalence, the same as the sequence

f X −→ Y → X/Y.

Proof. Indeed, Cf is the pushout f X / Y

  CX / Cf so CX → Cf is a cofibration with CX is contractible 

Proposition 1.2. Any map f : X → Y can be factored as

f X / Y >

i r 1 Mf for i a cofibration and f a weak equivalence. 1 2 MATH 6280 - CLASS 10

Proof. Consider the composite

i1 r X −→ Mf −→ Y where r : Mf = X × I ∪f Y  f(x) p = (x, t) ∈ X × I r(p) = y p = y ∈ Y.

i1 Note that r : Mf → Y is a homotopy equivalence and we have seen X −→ Mf is a cofibration. 

The homotopy cofiber is thus a homotopically well-behaved quotient. Indeed,

Cf = Mf /i1(X). and taking the homotopy cofiber is the process of • replacing f by a cofibration. • taking the quotient of the target by the image. f i Further, note that the composite X −→ Y −→ Cf is null, so this is indeed a homotopical analogue of the cokernel.

2. More details about the based case

For the remainder of the course, we will be more focused on well-pointed spaces, with based maps and based homotopies. We take a moment to make this precise. Recall that a based homotopy is a map

H : X × I → Y which is a based map at each time t. In other words, H(∗ × I) = ∗ so that it factors through the quotient

(X × I)/(∗ × I) = (X × I+)/(∗ × I ∪ X × ∗) = X ∧ I+, where I+ = [0, 1] ∪ {∗} is the interval with a disjoint base point. So, a based homotopy is a map

H : X ∧ I+ → Y

We have seen that the reduced suspension is defined as

ΣX = X ∧ S1 =∼ (X × I)/(X × {0} ∪ X × {1} ∪ ∗ × I)

In a similar way, we can define the reduced cone, mapping cone, etc.

Definition 2.1. • The space X ∧ I+ is called the reduced cylinder on X. MATH 6280 - CLASS 10 3

• The space CX = X × I/(X × 1 ∪ ∗ × I)

is called the reduced cone.

• The based mapping cylinder is Mf = Y ∪f (X ∧ I+).

• For CX the reduced cone, Cf = CX ∪f Y is the reduced mapping cone or cofiber. • A based cofibration is a map f : A → X such that the following extension problem always has a solution:

A / A ∧ I+

  X / X ∧ I+

#  , Y

f Exercise 2.2. (a) A based map X −→ Y is null via a based homotopy if and only if it factors f g through the reduced cone CX and that the composite X −→ Y −→ Z is null via a based point preserving homotopy if and only if there is a factorization through the reduced mapping cone

f X / Y / Cf

g  ~ Z

(b) Verify that the results on cofibrations hold in the based case.

From now on, if we are working with based maps, I will implicitly use the based constructions.

3. Barratt-Puppe sequence

Proposition 3.1. Ci1 → Ci1 /CY is a homotopy equivalence, and hence,

∼ ∼ Ci1 ' Ci1 /CY = Cf /Y = ΣX

Similarly, ∼ ∼ Ci2 ' Ci2 /C(Cf ) = Ci1 /Cf = ΣY.

Proof. CY → Ci1 is a cofibration with contractible image.  4 MATH 6280 - CLASS 10

We not have a sequence

i1 i2 i3 i4 X / Y / Cf / Ci1 / Ci2 / Ci3 / ...

q1 q2 q3 π !  −Σf  −Σi1  ΣX / ΣY / ΣCf / ... and need to prove that the squares commute up to homotopy.

Lemma 3.2. The following diagram is commutative up to homotopy

i3 i4 Ci1 / C2 / C3

q1 q2 q3  −Σf  −Σi  ΣX / ΣY / ΣCf . ∼ ∼ ∼ where q1 : Ci1 → Ci1 /CY = ΣX, q2 : Ci2 → Ci2 /C(Cf ) = ΣY and q3 : Ci3 → Ci3 /C(Ci1 ) = ΣCf are the obvious quotient maps and

−Σf(x, t) = (x, 1 − t).

Proof. We prove that the left hand square commute. The proof for the right hand square is similar. Let

(1) H : Ci1 = (CX ∪f CY ) ∧ I+ → ΣY be given by  (f(x), s(1 − t)) p = (x, t) ∈ CX H(p, s) = (y, (1 − s)t + s) p = (y, t) ∈ CY Then  (f(x), 0) p = (x, t) ∈ CX q2 ◦ i3(p) = H(p, 0) = (y, t) p = (y, t) ∈ CY and  (f(x), 1 − t) p = (x, t) ∈ CX −Σf ◦ q1(p) = H(p, 1) = (y, 1) = ∗ p = (y, t) ∈ CY

 MATH 6280 - CLASS 10 5

Figure 1. A depiction of (1).