4. Mapping Cylinders and Chain Homotopies This section is preparation for the next section. Its main purpose is to introduce some new concepts. 1. Chain Homotopies between Chain Maps. Definition. C = (C p ) and C = (C p) be two chain complexes and let ϕ, ψ : C C be two chain maps. Then ϕ and ψ are called chain homotopic (notation: ϕψ) if there is a chain map D p : C p C p+1 so that ϕ ψ = D + D Remark. The chain map D is called a chain homotopy between the two chain maps ϕ and ψ. Definition. A chain map ϕ : C C there is a chain map ψ : C C with is called a chain homotopy equivalence if ϕ ψ = id and ψ ϕ = id. Theorem. Let ϕ : C C be a chain homotopy equivalence. Then ϕ : H C H C is a homology isomorphism. In particular, H q C = H q C. Proof. Let z be a q-cycle of C. Then g (z) f (z) = Dz + D z = Dz + 0 So g (z) and f (z) are in the same homology class. Thus g (z) = f (z). 2. Mapping Cones. Definition. Let ϕ : C D the chain complex given by be a chain map. The mapping cone, cone(ϕ), of ϕ is (cone(ϕ)) m : = C m 1 D m m [ cm 1 ] [ ] := m 1 (c m 1 ) m ( ) ϕ m 1 (c m 1 ) [ ][ ] m 1 0 cm 1 =, for all (c,d) C ϕ m m 1 D m. m
2. Cylinders Notation. Cones are also denoted by: C(ϕ), or by: C ϕ. Theorem. Let ϕ : C D be a chain map. Then the assignment defines a short exact sequence. 0 D m f (cone(ϕ)) m g C m 1 0 d (0, d) (c, d) c Proof. Clearly f is injective and g is surjective. Moreover, {(0,d) d D m } cone(ϕ) equals im(f) as well as ker(g). Thus im(f) = ker(g). Moreover, f(d) = (0,d) = (0, d) = f (d) and g(c,d) = ( c) = c = g( c, d ϕ(c)) = g (c,d) This proves that f and g are chain maps. Corollary. Let ϕ : C D. Then there is a long exact homology sequence... H m C ϕ H m D H m cone(ϕ) H m 1 C ϕ H m 1 D... Proof. Because of the short exact sequence of chain complexes there is a long exact homology sequence 0 D f g cone(ϕ) C 0... H m D f H m cone(ϕ) g H m C H m 1 D f H m 1 cone(ϕ)... δ It remains to show that the connecting homomorphism δ equals ϕ. Let [c] H m C. Then we have the following zig-zag [ϕ(c)] f [ c, 0] [0,ϕ(c)] = [ c,ϕ(c)] g [c] Thus, by the definition of the connecting homomorphism, we have δ[c] = ϕ[c].
4 Cylinders 3 3. Mapping Cylinders. Definition. The mapping cylinder, cyl(ϕ) = D, of ϕ is defined by setting Moreover, define m c m c m 1 D m := C m D m C m 1 := = c m 1 + c m ϕ(c m 1 ) + c m 1 0 1 0 ϕ 0 0 c m c m 1 i m :C m D m = C m D m C m 1, c m (c m,0,0) and j m :D m D m = C m D m C m 1, (0,,0) Notation. Cylinders are also denoted by: M(ϕ), or by: M ϕ. Cylinder Theorem. Let ϕ : C D have is a chain map and set D := cyl(ϕ). Then we (0) (D, ) is a chain complex. (1) i and j are injective chain maps, (2) D and the quotients D /im i and D /im j are free chain complexes, (3) j induces homology isomorphisms in all dimensions (i.e. j is a quasi isomorphism), and (4) the diagram C Φ D i ց ւ j commutes up to chain homotopy. D Proof. Recall the above construction of D. We verify that the pair D = (D p, p) satisfies all conclusions of the theorem.
4. Cylinders ad (0). We first verify that (D, ) is a chain complex. Indeed, (c p,0,0) = ( c p,0,0) = ( c p,0,0) =(0,0,0) (0,d p,0) = (0, d p,0) = (0, d p,0) =(0,0,0) (0,0,c p 1 ) = ( c p 1,Φ(c p 1 ), c p 1 ). = ( ( c p 1,Φ(c p 1 ),0) + (0,0, c p 1 )), = ( c p 1,Φ(c p 1 ),0) + (0,0, c p 1 ) =( c p 1, Φ(c p 1 ),0)+ (+ c p 1, Φ(c p 1 ),+ c p 1 ) =(0,0,0) Thus = 0 and so (D, ) is a chain complex. ad (1). i,j are chain maps, because i(c p ) = (c p,0,0) = ( c p,0,0) = i( c p ) = i (c p ) j(d p ) = (0,d p,0) = (0, d p,0) = j( d p ) = j (d p ), and they are injective, because they are inclusions. ad (2). D p is free since D p = C p D p C p 1 and isince C p 1,C p,d p are free. Moreover, D p/im i = C p D p C p 1 /C p 0 0 and so D /im i and D /im j are free. = 0 D p C p 1 D p/im j = C p D p C p 1 /0 D p 0 = C p 0 C p 1 ad (3) To show that j : D D exact sequence induces an isomorphism in homology, consider the short 0 D j D D /D 0. This short exact sequence induces a long exact homology sequence... H p+1 D j H p+1 D H p+1 (D /D) H p D j H p D H p (D /D)...
It suffices to show that the homology of D /jd vanishes. Now, the chain complex (D /jd, ) is given by 4 Cylinders 5 D p/jd p = C p D p C p 1 /0 D p 0 = C p C p 1 and (c p,c p 1 ) = ( c p c p 1, c p 1 ) since, by definition, (c p,d p,c p 1 ) = ( c p, d p,0) + ( c p 1,Φ(c p 1 ), c p 1 ) = ( c p c p 1, d p + Φ(c p 1 ), c p 1 ). and we get from by forgetting the middle entry. Now, to show that H p (D /D) = 0, it remains to show that every p-cycle of D /D is the boundary of some p + 1-chain of D /D. So let (c p,c p 1 ) Z p (D /D) (0,0) = (c p,c p 1 ) = ( c p c p 1, c p 1 ). c p c p 1 = 0. (c p,c p 1 ) = (c p, c p ) = (0,c p ). (c p,c p 1 ) is the boundary of the chain (0,c p ). H p (D /D) = 0. ad (4). We define a transformation D : C p D p+1 = C p+1 D p+1 C p by setting Then D(c p ) := (0,0,c p ) ( D + D )(c p ) = D(c p ) + D (c p ) = (0,0,c p ) + D( c p ) = ( c p,φ(c p ), c p ) + (0,0, c p ) = ( c p,φ(c p ),0) = (0,Φ(c p ),0) (c p,0,0) = (jφ(c p ) i(c p ) = (jφ i)(c p )
6. Cylinders Thus D + D = j Φ i and so D is a chain homotopy between i and j Φ. This finishes the proof of the theorem. The material of this section is from [Munkres, Elements of Algebraic Topology] and [Weibel, Homological Algebra]