Mapping Cones Cylinders and Chain Triangles

Applying [L1] twice gives $d_{n-1}^{\mathrm{Cone}(f)}{d}_n^{\mathrm{Cone}(f)}(y,x)=(d_{n-1}^D{d}_n^D(y)+d_{n-1}^D{f}_{n-1}(x)-f_{n-2}{d}_{n-1}^C(x),d_{n-2}^C{d}_{n-1}^C(x))\mathrm{.}$

The diagonal terms vanish because $C_{\bullet }$ and $D_{\bullet }$ are chain complexes, and [L2] makes the mixed terms cancel. Therefore the displayed pair is $(0,0)$ for every $(y,x)$, so the cone differential squares to zero.

By [L1], $qj=0$. In degree $n$ the sequence is $0\to D_n\stackrel{y\mapsto (y,0)}{\to }{D}_n\oplus C_{n-1}\stackrel{(y,x)\mapsto x}{\to }{C}_{n-1}\to 0,$ which is exact and split by the section $s_n(x):=(0,x)$. This uses the degreewise biproduct description from [L3] and is exactly the split shape of [L4].

The maps $j$ and $q$ are chain maps by [L1], and step 1.1 proves exactness in every degree. Therefore [L2] makes the sequence a short exact sequence of complexes, with the degreewise splittings already exhibited in step 1.1.

For $f=0$, [L1] gives $d_n^{\mathrm{Cone}(0)}(y,x)=(d_n^D(y),-d_{n-1}^C(x)),$ which is exactly the block-sum differential on $D_n\oplus C[1{]}_n$.

Therefore the identity on the graded object $D\oplus C[1]$ is a chain isomorphism from $\mathrm{Cone}(0)$ to the direct-sum complex furnished by [L2].

Define $h_n:\mathrm{Cone}(1_C{)}_n\to \mathrm{Cone}(1_C{)}_{n+1}$ by $h_n(y,x):=(0,y)\mathrm{.}$ Using [L2], one computes $d_{n+1}{h}_n(y,x)=(y,-d_n^C(y)),h_{n-1}{d}_n(y,x)=(0,d_n^C(y)+x)\mathrm{.}$

Adding the two displayed terms yields $d_{n+1}{h}_n+h_{n-1}{d}_n=1_{\mathrm{Cone}(1_C)}\mathrm{.}$ Thus [L3] makes $h$ a null-homotopy of the identity, and then [L1] shows that $\mathrm{Cone}(1_C)$ is contractible.

Define ${\mathrm{\Phi }}_n:\mathrm{Cone}(f{)}_n\to \mathrm{Cone}(f^{\mathrm{\prime }}{)}_n$ by ${\mathrm{\Phi }}_n(y,x):=(v_n(y),u_{n-1}(x))\mathrm{.}$ Using the commutative square together with [L1] and [L2], one gets $d^{\mathrm{Cone}(f^{\mathrm{\prime }})}\mathrm{\Phi }=\mathrm{\Phi }{d}^{\mathrm{Cone}(f)},$ so $\mathrm{\Phi }$ is a chain map.

Because $u$ and $v$ are chain isomorphisms, their inverses are again chain maps by [L3], and the same block-diagonal formula with $u^{-1}$ and $v^{-1}$ defines the inverse chain map to $\mathrm{\Phi }$. Hence $\mathrm{\Phi }$ is a chain isomorphism.

Define ${\mathrm{\Phi }}_n(y,x):=(b_n(y),a_{n-1}(x))$. Using [L2], the composite $d^{\mathrm{Cone}(g)}\mathrm{\Phi }$ has first component $d^{D^{\mathrm{\prime }}}b(y)+ga(x)\mathrm{.}$

Since $a$ and $b$ are chain maps and [L1] gives $ga=bf$, the same first component is $b(d^D(y)+f(x))$, while the second component is $-a(d^C(x))$. Hence $d^{\mathrm{Cone}(g)}\mathrm{\Phi }=\mathrm{\Phi }{d}^{\mathrm{Cone}(f)}$, so $\mathrm{\Phi }$ is a chain map.

By [L1], every arrow-category morphism $(a,b)$ is sent to the block map $(y,x)\mapsto (b(y),a(x))$ on cones, so objects and morphisms are assigned.

If $(a,b)$ and $(a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ are composable, then the block formula shows $\mathrm{Cone}(a^{\mathrm{\prime }},b^{\mathrm{\prime }})\mathrm{Cone}(a,b)=\mathrm{Cone}(a^{\mathrm{\prime }}a,b^{\mathrm{\prime }}b),$ and identity squares give identity block maps. Thus the assignment preserves composition and identities, hence is a functor.

Define ${\mathrm{\Phi }}_n:\mathrm{Cone}(f{)}_n\to \mathrm{Cone}(g{)}_n$ by ${\mathrm{\Phi }}_n(y,x):=(y+s_{n-1}(x),x)\mathrm{.}$ Using [L2], the equality $g-f=-(f-g)$, and then [L1], one checks $d^{\mathrm{Cone}(g)}\mathrm{\Phi }=\mathrm{\Phi }{d}^{\mathrm{Cone}(f)}\mathrm{.}$

Replacing $s$ by $-s$ gives the inverse block map ${\mathrm{\Psi }}_n(y,x):=(y-s_{n-1}(x),x)\mathrm{.}$ Thus $\mathrm{\Phi }$ is a chain isomorphism between the two mapping cones.

Assume $\mathrm{Cone}(f)$ is acyclic. If $[x]\in H_{n-1}(C)$ satisfies $H_{n-1}(f)([x])=0$, choose $y\in D_n$ with $d_n^D(y)=f_{n-1}(x)$. Then $(-y,x)$ is an $n$-cycle of the cone by [L2], so [L3] gives $(-y,x)=d_{n+1}^{\mathrm{Cone}(f)}(z,w)$ for some $(z,w)$. Hence $x=-d_n^C(w)$, so $H_{n-1}(f)$ is injective. Likewise, if $y\in Z_n(D)$, then $(y,0)$ is an $n$-cycle of the cone, so acyclicity gives $(y,0)=d_{n+1}^{\mathrm{Cone}(f)}(z,w)$. Thus $y=d_{n+1}^D(z)+f_n(w)$, and $H_n(f)$ is surjective.

Conversely, assume $f$ is a quasi-isomorphism. Let $(y,x)$ be an $n$-cycle of $\mathrm{Cone}(f)$. Then $x\in Z_{n-1}(C)$ and $f_{n-1}(x)=-d_n^D(y)$ is a boundary in $D$, so [L1] gives $x=d_n^C(w)$ for some $w\in C_n$. Then $y+f_n(w)$ is an $n$-cycle in $D$, so [L1] again gives $z\in Z_n(C)$ and $v\in D_{n+1}$ with $y+f_n(w)=f_n(z)+d_{n+1}^D(v)\mathrm{.}$ Therefore $(y,x)=d_{n+1}^{\mathrm{Cone}(f)}(v,z-w),$ so every cone cycle is a boundary. By [L3], $\mathrm{Cone}(f)$ is acyclic, and together with [L1] this proves the equivalence.

Suppose first that $f$ has homotopy inverse $g$. Apply [L5] to the arrow comparison from $f:C_{\bullet }\to D_{\bullet }$ to $1_D:D_{\bullet }\to D_{\bullet }$ using $u=f$, $v=1_D$, the trivial homotopy $1_{D}f\simeq 1_{D}f$, and the given homotopy $fg\simeq 1_D$. This gives a chain-homotopy equivalence $\mathrm{\Phi }:\mathrm{Cone}(f)\to \mathrm{Cone}(1_D)\mathrm{.}$ By [L6], the target cone is contractible; the second sentence of [L2] then shows that $\mathrm{Cone}(f)$ is contractible as well.

Conversely, suppose $\mathrm{Cone}(f)$ is contractible. By [L2] and [L3], choose a degree-one endomorphism $H$ of $\mathrm{Cone}(f)$ with $dH+Hd=1_{\mathrm{Cone}(f)}\mathrm{.}$ Write $H_n(y,x)=(a_n(y)+b_{n-1}(x),g_n(y)+u_{n-1}(x)),$ where $a_n:D_n\to D_{n+1}$, $b_{n-1}:C_{n-1}\to D_{n+1}$, $g_n:D_n\to C_n$, and $u_{n-1}:C_{n-1}\to C_n$. Expanding $dH+Hd=1$ with [L4] and comparing the bottom-left, bottom-right, and top-left components gives $g_{n-1}{d}_n^D=d_n^C{g}_n,g_{n-1}{f}_{n-1}-d_n^C{u}_{n-1}-u_{n-2}{d}_{n-1}^C=1_{C_{n-1}},$ $d_{n+1}^D{a}_n+a_{n-1}{d}_n^D+f_n{g}_n=1_{D_n}\mathrm{.}$ Thus $g$ is a chain map, $u$ is a homotopy $gf\simeq 1_C$, and $-a$ is a homotopy $fg\simeq 1_D$. Therefore $g$ is a homotopy inverse of $f$, so [L1] makes $f$ a chain homotopy equivalence.

Applying [L1] twice gives $d_{n-1}^{\mathrm{Cyl}(f)}{d}_n^{\mathrm{Cyl}(f)}(x,y,z)=(d_{n-1}^C{d}_n^C(x)+d_{n-1}^C(z)-d_{n-1}^C(z),d_{n-1}^D{d}_n^D(y)-d_{n-1}^D{f}_{n-1}(z)+f_{n-2}{d}_{n-1}^C(z),d_{n-2}^C{d}_{n-1}^C(z))\mathrm{.}$

The diagonal terms vanish because $C_{\bullet }$ and $D_{\bullet }$ are complexes, and [L2] cancels the mixed terms in the middle coordinate. Hence the displayed triple is zero, so the cylinder differential squares to zero.

By [L1], $pi=f$ and $pj=1_D$. The map $r_n:\mathrm{Cyl}(f{)}_n\to C_n$, $r_n(x,y,z)=x$, satisfies $r_n{i}_n=1_{C_n}$, so $i_n$ is split monic in every degree.

Define $H_n:\mathrm{Cyl}(f{)}_n\to \mathrm{Cyl}(f{)}_{n+1}$ by $H_n(x,y,z):=(0,0,x)\mathrm{.}$ Using the cylinder differential, one computes $dH+Hd=1_{\mathrm{Cyl}(f)}-jp\mathrm{.}$ Thus $j$ is a homotopy inverse for $p$, so [L2] shows that $p$ is a chain-homotopy equivalence; [L3] guarantees all composites involved are chain maps.

By [L1], $pj=1_D$ and $H_n(x,y,z)=(0,0,x)$ is a chain homotopy from $1_{\mathrm{Cyl}(f)}$ to $jp$.

For every $y\in D_n$, one has $j_n(y)=(0,y,0)$, hence $H_n{j}_n(y)=H_n(0,y,0)=0.$ Therefore the homotopy vanishes on the target summand, so [L2] gives the stated strong deformation retract.

Apply [L1] to the given map $f$. This produces a factorization $C_{\bullet }\stackrel{i}{\to }\mathrm{Cyl}(f{)}_{\bullet }\stackrel{p}{\to }{D}_{\bullet }\mathrm{.}$

The same theorem already proves that $i$ is a degreewise split inclusion and that $p$ is a chain-homotopy equivalence. That is exactly the asserted factorization.

In degree $n$, [L1] identifies $i_n$ with the inclusion of the first summand. Therefore [L2] identifies $\mathrm{coker}(i_n)$ with $D_n\oplus C_{n-1}$ via ${\theta }_n([(x,y,z)]):=(y,z)\mathrm{.}$ Under this identification the induced differential is $(y,z)⟼(d_n^D(y)-f_{n-1}(z),-d_{n-1}^C(z)),$ which is the cone differential for the chain map $-f$.

The sign map ${\sigma }_n:D_n\oplus C_{n-1}\to D_n\oplus C_{n-1},{\sigma }_n(y,z):=(-y,z)$ conjugates the differential from step 1.1 to the cone differential of $f$ in [L3]. Hence the cokernel complex is canonically chain-isomorphic to $\mathrm{Cone}(f)$.

By [L2], the homotopy convention gives $gu-vf=dt+td$. Using [L3], direct expansion yields $d^{\mathrm{Cone}(g)}\mathrm{\Phi }(y,x)=(v{d}^{D}y-dt(x)+gu(x),-u{d}^{C}x)=(v{d}^{D}y+vf(x)+t{d}^C(x),-u{d}^{C}x)=\mathrm{\Phi }{d}^{\mathrm{Cone}(f)}(y,x)\mathrm{.}$ Thus $\mathrm{\Phi }$ is a chain map.

Let $j_f,j_g$ and $q_f,q_g$ denote the inclusions and projections in the cone triangles from [L4]. The block formula gives $\mathrm{\Phi }{j}_f=j_{g}v$ and $q_g\mathrm{\Phi }=u[1]q_f$ strictly, while the first square commutes in $K(\mathcal{A})$ because $t:gu\simeq vf$. Hence $(u,v,[\mathrm{\Phi }])$ is a morphism from the cone triangle of $f$ to the cone triangle of $g$ in the homotopy category.

The maps $u$ and $v$ are chain-homotopy equivalences by hypothesis, so [L6] applied to the morphism of cone triangles from step 2.1 shows that $\mathrm{\Phi }$ is a chain-homotopy equivalence. This is the asserted conclusion.

By [L1], the map $\alpha$ is ${\alpha }_n(d,c)=(g_n(d),c)\mathrm{.}$ Writing out the cone differential from [L2] shows that $\mathrm{Cone}(\alpha {)}_n=E_n\oplus C_{n-1}\oplus D_{n-1}\oplus C_{n-2}$. The change of coordinates ${\mathrm{\Theta }}_n(e,c,d,c^{\mathrm{\prime }}):=((e,d+f_{n-1}(c)),(c,c^{\mathrm{\prime }}))$ is a degreewise isomorphism from $\mathrm{Cone}(\alpha {)}_n$ to $\mathrm{Cone}(g{)}_n\oplus \mathrm{Cone}(1_{C[1]}{)}_n$.

A direct substitution into the differential formulas shows that $\mathrm{\Theta }$ is a chain map. Therefore $\mathrm{Cone}(\alpha )$ is chain-isomorphic to $\mathrm{Cone}(g)\oplus \mathrm{Cone}(1_{C[1]})$. The second summand is contractible by [L3], so the direct sum is chain-homotopy equivalent to $\mathrm{Cone}(g)$.

By [L1], the map $f$ is chain-homotopic to $0:C_{\bullet }\to D_{\bullet }$. Applying [L2] gives a chain isomorphism $\mathrm{Cone}(f)\cong \mathrm{Cone}(0)\mathrm{.}$

Using [L3], this identifies $\mathrm{Cone}(f)$ with $D_{\bullet }\oplus C[1{]}_{\bullet }$. Therefore the cone triangle of $f$ is isomorphic in the homotopy category to the displayed split triangle.

By [L1], the map $f$ has a homotopy inverse $g:D_{\bullet }\to C_{\bullet }$.

Passing to homotopy classes as in [L2], the relations $gf\simeq 1_C$ and $fg\simeq 1_D$ become $[g][f]=[1_C],[f][g]=[1_D]\mathrm{.}$ Hence $[f]$ is an isomorphism in the homotopy category.

The two complexes have the same degree-$n$ object $\mathrm{Cone}(f[1]{)}_n=D_{n-1}\oplus C_{n-2}=\mathrm{Cone}(f)[1{]}_n\mathrm{.}$ Define ${\sigma }_{f,n}(y,x):=(-y,x)\mathrm{.}$

By [L1], [L2], and [L3], the differential on $\mathrm{Cone}(f[1])$ is $(-d^D(y)+f(x),d^C(x)),$ while the shifted differential on $\mathrm{Cone}(f)[1]$ is $(-d^D(y)-f(x),d^C(x))\mathrm{.}$ The sign in ${\sigma }_f$ changes the first component exactly enough to intertwine these two formulas, so ${\sigma }_f$ is a chain isomorphism.