Chain Homotopy and the Homotopy Category

Using [L1] twice, the $n$th component of $\mathrm{\partial }(\mathrm{\partial }u)$ is $d_{n+r-1}^D{d}_{n+r}^D{u}_n-(-1{)}^r{d}_{n+r-1}^D{u}_{n-1}{d}_n^C-(-1{)}^{r-1}{d}_{n+r-1}^D{u}_{n-1}{d}_n^C-u_{n-2}{d}_{n-1}^C{d}_n^C\mathrm{.}$

The middle two terms cancel because $-(-1{)}^r-(-1{)}^{r-1}=0$, and the outer two terms vanish because successive differentials in $C$ and $D$ compose to zero. Hence every component of $\mathrm{\partial }(\mathrm{\partial }u)$ is zero, so ${\mathrm{\partial }}^2=0$.

By [L1], the equality $\mathrm{\partial }u=0$ means exactly that $d_n^D{u}_n-u_{n-1}{d}_n^C=0$ for every $n\in \mathbb{Z}$.

Rewriting the equality in step 1.1 gives $d_n^D{u}_n=u_{n-1}{d}_n^C$ for every $n$, and [L2] is precisely this condition. Thus $\mathrm{\partial }u=0$ if and only if $u$ is a chain map.

Reflexivity holds because the zero degree-$1$ family satisfies $f-f=0=d0+0d,$ so [L1] gives $f\simeq f$.

If $s:f\simeq g$, then [L1] gives $f-g=ds+sd$, hence $g-f=d(-s)+(-s)d,$ so $g\simeq f$. If $s:f\simeq g$ and $t:g\simeq h$, then $f-h=(f-g)+(g-h)=d(s+t)+(s+t)d,$ so $f\simeq h$. Therefore the relation is symmetric and transitive as well.

By [L1], we have $f-g=ds+sd$ and $f^{\mathrm{\prime }}-g^{\mathrm{\prime }}=dt+td$. Using [L3], add these equalities to obtain $(f+f^{\mathrm{\prime }})-(g+g^{\mathrm{\prime }})=d(s+t)+(s+t)d,$ so $s+t$ is a homotopy from $f+f^{\mathrm{\prime }}$ to $g+g^{\mathrm{\prime }}$.

Since $u$ and $v$ are chain maps by [L2], their differentials commute in the usual way. Therefore $ufv-ugv=u(f-g)v=u(ds+sd)v=d(usv)+(usv)d,$ so $usv$ is a chain homotopy from $ufv$ to $ugv$.

The zero chain map is null-homotopic via the zero degree-$1$ family, and [L3] guarantees this zero map exists in $\mathrm{Ch}(\mathcal{A})$.

Because $f$ and $g$ are each homotopic to zero by [L1], [L2] shows $f+g\simeq 0+0=0$, $uf\simeq 0$, and $fv\simeq 0$. Hence sums and left or right composition preserve null-homotopy, so these maps form a two-sided additive ideal.

Let $z\in Z_n(C)$ be an $n$-cycle. Since $d_n^{C}z=0$, [L1] gives $(f_n-g_n)(z)=d_{n+1}^D{s}_n(z),$ so $(f_n-g_n)(z)$ is a boundary in degree $n$.

By [L2], $H_n(f)$ and $H_n(g)$ are defined on homology classes of cycles. Step 1.1 shows that every $n$-cycle has images under $f_n$ and $g_n$ differing by a boundary, so those induced homology classes coincide. Hence $H_n(f)=H_n(g)$.

By [L1], the map $f$ is homotopic to $0:C_{\bullet }\to D_{\bullet }$.

Applying [L2] to the homotopy from step 1.1 gives $H_n(f)=H_n(0)$. The map $H_n(0)$ is the zero morphism, so $H_n(f)=0$.

From [L1] and [L2], the homotopies $gf\simeq 1_C$ and $fg\simeq 1_D$ imply $H_n(g)H_n(f)=1_{H_n(C)},H_n(f)H_n(g)=1_{H_n(D)}$ for every $n\in \mathbb{Z}$.

By [L3], the equalities in step 1.1 say exactly that each $H_n(f)$ has inverse $H_n(g)$. Therefore every $H_n(f)$ is an isomorphism, and [L4] makes $f$ a quasi-isomorphism.

By [L1], the unique map $C_{\bullet }\to 0$ is a chain homotopy equivalence.

Then [L2] makes $C_{\bullet }\to 0$ a quasi-isomorphism. Since the zero complex has zero homology in every degree, $H_n(C)=0$ for all $n$, so $C_{\bullet }$ is acyclic.

By [L1], the maps $f-f^{\mathrm{\prime }}$ and $g-g^{\mathrm{\prime }}$ are null-homotopic. Using [L3], $gf-g^{\mathrm{\prime }}{f}^{\mathrm{\prime }}=(g-g^{\mathrm{\prime }})f^{\mathrm{\prime }}+g(f-f^{\mathrm{\prime }})$ is a sum of two null-homotopic maps.

The first summand in step 1.1 is null-homotopic by right whiskering, and the second is null-homotopic by left whiskering; this is exactly [L2] and [L3]. Hence $gf-g^{\mathrm{\prime }}{f}^{\mathrm{\prime }}$ is null-homotopic, so [L1] gives $[gf]=[g^{\mathrm{\prime }}{f}^{\mathrm{\prime }}]$.

By [L3], each hom-set of $\mathrm{Ch}(\mathcal{A})$ is an abelian group. Quotienting by the additive subgroup of null-homotopic maps from [L2] therefore gives an abelian group structure on each ${\mathrm{Hom}}_{K(\mathcal{A})}(C,D)$ from [L1].

The zero complex and the degreewise biproduct complex exist in $\mathrm{Ch}(\mathcal{A})$ by [L3] and [L4]. Because [L2] is a two-sided ideal, the usual injections and projections descend to homotopy classes and still satisfy the biproduct identities in the quotient. Therefore $K(\mathcal{A})$ has a zero object and finite biproducts, so it is additive.

By [L1], the functor $Q$ sends each chain map to its coset modulo null-homotopy. Therefore for parallel maps $f,g$, $Q(f+g)=[f+g]=[f]+[g],Q(0)=[0]=0,$ so $Q$ preserves the additive structure on hom-groups.

The zero object and biproduct objects of $\mathrm{Ch}(\mathcal{A})$ are sent to the same underlying complexes in $K(\mathcal{A})$ because $Q$ is identity on objects. Since the structural maps are sent to their homotopy classes and still satisfy the biproduct identities, $Q$ preserves finite biproducts. Hence $Q$ is additive.

Define ${\bar{H}}_n(C):=H_n(C)$ on objects and ${\bar{H}}_n([f]):=H_n(f)$ on morphisms. This is well defined because [L1] shows that homotopic representatives have the same homology map, and [L4] says those are exactly the equal morphisms in $K(\mathcal{A})$.

Because $H_n$ is additive by [L2] and $Q$ is additive by [L3], the definition in step 1.1 gives an additive functor with $H_n={\bar{H}}_{n}Q$. Uniqueness is immediate from [L4]: every morphism of $K(\mathcal{A})$ is a class $[f]$, so any factorization must send $[f]$ to $H_n(f)$.

By [L2], the equality $[1_C]=0$ means precisely that the identity map and the zero map define the same homotopy class. That is equivalent to $1_C$ being null-homotopic, which [L1] says is exactly contractibility.

In the complex $E_{\bullet }$, multiplication by $2$ is injective, reduction modulo $2$ is surjective, and $\ker (<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2)=2\mathbb{Z}=\mathrm{im}(2),$ so $E_{\bullet }$ is acyclic. If $E_{\bullet }$ were contractible, then step 1.1 would make $[1_E]=0$, equivalently $1_E$ would be null-homotopic. In degree $0$ that would force a section $\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ of the quotient map $\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$, which is impossible. Thus $E_{\bullet }$ is not contractible.

Step 2.1 gives an acyclic complex that is not contractible, so by step 1.1 its identity class is not zero in the homotopy category. Therefore vanishing homology alone does not force an object to be zero there. This does not contradict [L4], which gives only the forward implication contractible $\Rightarrow$ acyclic.

By [L1], the group of $0$-cycles in $\underset{}{\mathrm{Hom}}(C,D{)}_{\bullet }$ is exactly the group of chain maps $C_{\bullet }\to D_{\bullet }$. By [L2], its subgroup of $0$-boundaries is exactly the null-homotopic chain maps.

Therefore $H_0(\underset{}{\mathrm{Hom}}(C,D))=Z_0\mathrm{/}{B}_0$ is the quotient of the chain maps by the null-homotopic ones. By [L3] this is $[C,D{]}_K$, and [L4] identifies that quotient with ${\mathrm{Hom}}_{K(\mathcal{A})}(C,D)$.

By [L1], $d_{n-1}^{C[k]}{d}_n^{C[k]}=(-1{)}^k{d}_{n-k-1}^C(-1{)}^k{d}_{n-k}^C=(-1{)}^{2k}{d}_{n-k-1}^C{d}_{n-k}^C\mathrm{.}$

Since $C_{\bullet }$ is a chain complex, the last composite in step 1.1 is zero. Therefore $d_{n-1}^{C[k]}{d}_n^{C[k]}=0$ for all $n$.

By [L1] and [L2], sending $C$ to $C[k]$ and $f$ to $f[k]$ defines a functor on $\mathrm{Ch}(\mathcal{A})$. The formulas are degreewise, so $[k]$ preserves zero maps and sums, and applying $[-k]$ returns the original complex and map on the nose. Hence $[k]$ is an additive autoequivalence of $\mathrm{Ch}(\mathcal{A})$.

If $f\simeq g$, then [L2] gives $f[k]\simeq g[k]$, so [L3] lets the same formula descend to homotopy classes. Since step 1.1 already gives the inverse $[-k]$ and [L4] provides the additive structure on the quotient, $[k]$ is also an additive autoequivalence of $K(\mathcal{A})$.

Because the differential in [L1] differs from $d_{n-k}^C$ only by the unit $(-1{)}^k$, its kernel and image are the same subobjects. Hence $Z_n(C[k])=Z_{n-k}(C),B_n(C[k])=B_{n-k}(C)\mathrm{.}$

Applying [L2] to the equalities of step 1.1 yields $H_n(C[k])=Z_n(C[k])\mathrm{/}{B}_n(C[k])\cong Z_{n-k}(C)\mathrm{/}{B}_{n-k}(C)=H_{n-k}(C)\mathrm{.}$

If $f:C\to D$ has homotopy inverse $g$, then [L1] shifts the homotopies $gf\simeq 1_C$ and $fg\simeq 1_D$ to homotopies $g[k]f[k]\simeq 1_{C[k]},f[k]g[k]\simeq 1_{D[k]}\mathrm{.}$ The inverse shift $[-k]$ from [L2] shows this construction stays inside the same homotopy-equivalence class of objects. Thus [L7] shows that shift preserves chain homotopy equivalences. By [L6], the special case of a homotopy equivalence $C\simeq 0$ shows that shift also preserves contractible complexes.

Let $f$ be a quasi-isomorphism. By [L3], the map $H_n(f[k])$ identifies with $H_{n-k}(f)$ for every $n$, so $H_n(f[k])$ is an isomorphism whenever $H_{n-k}(f)$ is. Then [L4] makes $f[k]$ a quasi-isomorphism. This is compatible with step 1.1 and [L5], since every shifted homotopy equivalence is again a quasi-isomorphism.

Define $s_{n-1}:C_{n-1}\to C_n$ by $s_{n-1}:={\varphi }_n^{-1}{j}_2{\pi }_1{\varphi }_{n-1},$ where $j_2$ is the second inclusion and ${\pi }_1$ is the first projection. Then $d_n{s}_{n-1}=i_{n-1}{\pi }_1{\varphi }_{n-1}={\varphi }_{n-1}^{-1}{j}_1{\pi }_1{\varphi }_{n-1}$ by the formula for $d_n$ and the compatibility of $i_{n-1}$ with ${\varphi }_{n-1}$.

Likewise $s_{n-2}{d}_{n-1}={\varphi }_{n-1}^{-1}{j}_2{\pi }_2{\varphi }_{n-1},$ so $d_n{s}_{n-1}+s_{n-2}{d}_{n-1}={\varphi }_{n-1}^{-1}(j_1{\pi }_1+j_2{\pi }_2){\varphi }_{n-1}=1_{C_{n-1}}\mathrm{.}$ Thus the identity map is null-homotopic, and [L1] makes $C_{\bullet }$ contractible.

By [L2], each short exact sequence $0\to Z_n(C)\to C_n\stackrel{{\pi }_n}{\to }{Z}_{n-1}(C)\to 0$ is exact, and the section ${\sigma }_n$ splits it. Therefore $C_n\cong Z_n(C)\oplus Z_{n-1}(C)$ for every $n$, with differential equal to projection onto the second summand followed by the cycle inclusion.

Step 1.1 is exactly the compatible decomposition required by [L4], so $C_{\bullet }$ is contractible. As [L5] emphasizes, the contraction comes from the chosen sections rather than from projectivity of the terms alone.

Because $i_n$ splits, each exact sequence $0\to Z_n(C)\stackrel{i_n}{\to }{C}_n\to Z_{n-1}(C)\to 0$ decomposes $C_n$ as $C_n\cong Z_n(C)\oplus Z_{n-1}(C),$ with the differential again equal to projection onto the second factor followed by inclusion into $C_{n-1}$.

Step 1.1 is the same compatible splitting pattern used in [L3], so the same contraction argument applies and $C_{\bullet }$ is contractible. The proof depends on the assumed retractions, as [L4] emphasizes; [L5] explains the formal duality with the epimorphism-side criterion.

Every complex is homotopy equivalent to itself: the identity map is its own homotopy inverse, with the zero homotopies witnessing $1_C{1}_C\simeq 1_C,1_C{1}_C\simeq 1_C\mathrm{.}$ Symmetry is immediate by swapping a map with its chosen homotopy inverse.