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Math 6280 - Class 10 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 > i1 r 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 8 <f(x) p = (x; t) 2 X × I r(p) = :y p = y 2 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 × f0g [ X × f1g [ ∗ × 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 8 <(f(x); s(1 − t)) p = (x; t) 2 CX H(p; s) = :(y; (1 − s)t + s) p = (y; t) 2 CY Then 8 <(f(x); 0) p = (x; t) 2 CX q2 ◦ i3(p) = H(p; 0) = :(y; t) p = (y; t) 2 CY and 8 <(f(x); 1 − t) p = (x; t) 2 CX −Σf ◦ q1(p) = H(p; 1) = :(y; 1) = ∗ p = (y; t) 2 CY MATH 6280 - CLASS 10 5 Figure 1. A depiction of (1)..
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