Derived Categories

For a roof $(s,f)$, take both refinement legs to be its vertex identity. This gives reflexivity. Interchanging the two refinement legs gives symmetry. These arguments also cover identity roofs and coincident vertices.

For transitivity suppose $(s,f)\sim (t,g)$ via $a,b$ and $(t,g)\sim (u,h)$ via $c,d$. Thus $r=sa=tb\in S$ and $q=tc=ud\in S$, while $fa=gb$ and $gc=hd$. Apply Ore to $r$ and $q$: obtain $v$ in $S$ and $w$ with $rv=qw\in S$.

Now $tbv=tcw$. Cancellation of the postcomposed denominator $t$ supplies $e\in S$ such that $bve=cwe$. Therefore $save=udwe=rve\in S$ and $fave=gbve=gcwe=hdwe$. The legs $ave,dwe$ witness $(s,f)\sim (u,h)$. Only $v,e$, and the displayed composite denominator were asserted to lie in $S$.

Ore supplies $a,b$ of the indicated types and $sa\in S$. To compare any two candidate squares, it is enough to give a common refinement of their output roofs; common refinement is an equivalence relation.

First replace $(s,f)$ by a refinement $(sr,fr)$, where $sr\in S$. Compare squares $fa=tb$ and $fr{a}^{\mathrm{\prime }}=t{b}^{\mathrm{\prime }}$, with $a,a^{\mathrm{\prime }}\in S$. Apply Ore to the denominators $sa$ and $sr{a}^{\mathrm{\prime }}$ to obtain $v\in S,w$ with $sav=sr{a}^{\mathrm{\prime }}w\in S$. Cancel $s$ by a further $e\in S$ to obtain $ave=r{a}^{\mathrm{\prime }}we$, then cancel $t$ by $k\in S$ to obtain $bvek=b^{\mathrm{\prime }}wek$. Thus the two composite roofs have equal numerator and denominator after refinement, with common denominator $savek\in S$. Taking $r=1$ proves independence of the square as well.

Next refine the second roof to $(tr,gr)$, where $tr\in S$. Compare $fa=tb$ and $f{a}^{\mathrm{\prime }}=tr{b}^{\mathrm{\prime }}$. Ore gives $v\in S,w$ with $av=a^{\mathrm{\prime }}w\in S$. Cancellation of $t$ in $tbv=tr{b}^{\mathrm{\prime }}w$ gives $e\in S$ with $bve=r{b}^{\mathrm{\prime }}we$. Hence the composites have equal numerator $gbve=gr{b}^{\mathrm{\prime }}we$ and common denominator $save\in S$. Arbitrary equivalent representatives share a refinement, so these two refinement checks and transitivity prove full representative independence.

For a third roof $(u:T\to Z,h:T\to R)$ choose $fa=tb$ as above and $ge=uk$ with $e\in S$. Ore applied to $b:W\to V$ and $e$ gives $i\in S,j$ with $bi=ej$. Then $gbi=ukj$ and $fai=tej$. The two bracketings can therefore both be computed as $(sai,hkj)$. Independence of square choices proves associativity for all choices.

Composing with an identity roof on the target uses $a=1,b=f$; composing with an identity roof on the source uses $a=s,b=1$. Both recover $(s,f)$ exactly. Thus the operation has both identity laws, including when $s$ itself is an identity.

Composition, associativity and identities are supplied by the preceding lemma. Every roof into a fixed source $X$ refines to one with denominator in the supplied set $S_X$; its possible numerators lie in a set of Hom sets. Taking the quotient of this set by refinement gives a set of arrows from $X$ to $Y$. In a small category the set of all roofs already suffices. This also constructs the empty localization when there are no objects.

Identity-denominator roofs show $Q(gf)=Q(g)Q(f)$ and preservation of identities. For $s:U\to X$ in $S$, the roof $(s,1_U)$ is inverse to $Q(s)$: the products are the identity at $U$ and the roof $(s,s)$, which refines the identity at $X$. Thus $Q$ inverts $S$.

Let $F$ invert $S$. Set $\bar{F}(s,f)=F(f)F(s{)}^{-1}$. For a refinement $sa=tb=r\in S$, both $F(a)=F(s{)}^{-1}F(r)$ and $F(b)=F(t{)}^{-1}F(r)$ are invertible, so $fa=gb$ gives equal values. An Ore equality $fa=tb$ gives $F(t{)}^{-1}F(f)=F(b)F(a{)}^{-1}$, proving preservation of composition. Every roof is $Q(f)Q(s{)}^{-1}$, forcing uniqueness. A natural transformation descends on the same objects: naturality for $s$ implies naturality for $Q(s{)}^{-1}$ and hence for each roof. This proves the stated localization property.

Equality of $(1,f)$ and $(1,g)$ gives refinement legs $a=b=v\in S$ with $fv=gv$; conversely such $v$ is a common refinement. For the dual presentation apply the other Ore axiom to $s:U\to X$ and $f:U\to Y$, obtaining $t:Y\to V$ in $S$ and $h:X\to V$ with $tf=hs$. Then $Q(f)Q(s{)}^{-1}=Q(t{)}^{-1}Q(h)$. This describes right roofs in the same category and requires no second smallness assertion.

In each integer degree $H^n(1_X)=1_{H^n(X)}$, including $H^n(X)=0$. Hence identities induce isomorphisms in every degree.

For quasi-isomorphisms $f:X\to Y,g:Y\to Z$, functoriality gives $H^n(gf)=H^n(g)H^n(f)$, an isomorphism with inverse $H^n(f{)}^{-1}{H}^n(g{)}^{-1}$. This holds for every $n$.

Fix any degree $n$ and write $a=H^n(f)$ and $b=H^n(g)$. Functoriality gives $H^n(gf)=ba$. If $a,b$ are invertible then $ba$ is invertible, also when any cohomology object is zero.

If $a,ba$ are invertible, then $b=(ba)a^{-1}$ is invertible. If $b,ba$ are invertible, then $a=b^{-1}(ba)$ is invertible. These are the other two possible pairs, and the argument holds in every degree.

Cohomology is defined on homotopy classes. Identities, composites, and shifts preserve quasi-isomorphisms, and homotopy equivalences are quasi-isomorphisms. These assertions include the zero complex.

For $s:X\to X^{\mathrm{\prime }}$ a quasi-isomorphism and $f:X\to Y$, take the triangle $X\stackrel{f}{\to }Y\to C\stackrel{h}{\to }X[1]$. Complete $s[1]h:C\to X^{\mathrm{\prime }}[1]$ to a triangle $X^{\mathrm{\prime }}\stackrel{f^{\mathrm{\prime }}}{\to }{Y}^{\mathrm{\prime }}\to C\stackrel{s[1]h}{\to }{X}^{\mathrm{\prime }}[1]$. Rotated TR3 gives $t:Y\to Y^{\mathrm{\prime }}$ with $tf=f^{\mathrm{\prime }}s$ and a morphism of triangles whose other components are $s,1_C$. The two long cohomology sequences show $H^n(t)$ invertible: exactness identifies its kernel and cokernel with zero by the adjacent isomorphisms. Thus $t$ is a quasi-isomorphism, giving the outgoing Ore square. Reversing arrows and rotating gives the incoming Ore square.

If $a=f-g:X\to Y$ satisfies $at=0$ for a quasi-isomorphism $t:Z\to X$, the triangle $Z\stackrel{t}{\to }X\stackrel{d}{\to }C\to Z[1]$ has acyclic $C$. Hom exactness gives $a=i\circ d$ for some $i:C\to Y$. Complete $i$ to $C\stackrel{i}{\to }Y\stackrel{j}{\to }W\to C[1]$. Cohomology exactness makes $j$ a quasi-isomorphism and $ja=jid=0$. Reversing arrows gives the converse cancellation direction.

All constructions used only finitely many shifts, sums and cones. These preserve each of termwise upper, lower, and two-sided boundedness (with possibly changed finite bounds). Thus both Ore and cancellation constructions stay in each bounded homotopy category and establish exactly the multiplicative-system axioms there.

The derived category is the localization at quasi-isomorphisms, so the proposed inverse is the allowed roof $(s,1_X)$. This includes $s=1_0$ at the zero complex.

Composing the roof with $Q(s)$ gives the identity roof of $X$ in one order and $(s,s)$ in the other. The latter refines $(1_Y,1_Y)$ via $s$ and $1_X$, so both products are identities.

Cochain reindexing of homology gives a functor $H^n$ on $K(\mathcal{A})$, and it inverts every quasi-isomorphism by definition. In particular it sends the zero complex to zero.

The localization property therefore gives the unique factorization. On a roof $Q(f)Q(s{)}^{-1}$ functoriality forces the displayed value. No triangulation of $D(\mathcal{A})$ is needed for this assertion.

First $Q(0)$ is a zero object: a roof $X\leftarrow U\to 0$ represents the ordinary zero map, because its numerator is the composite $U\to X\to 0$; dually a right roof out of $0$ is zero. Hence both Hom sets involving $Q(0)$ are singletons. If $X$ is acyclic, $X\to 0$ is a quasi-isomorphism, so $Q(X)\cong Q(0)$.

Conversely, if $Q(X)$ is a zero object it is isomorphic to $Q(0)$. The factored cohomology functors take this isomorphism to $H^n(X)\cong H^n(0)=0$ for every $n$. Thus $X$ is acyclic.

For $(s,f),(t,g):X\to Y$, choose $a\in S,b$ with $sa=tb=r\in S$ and define their sum as $(r,fa+gb)$. If roofs already have denominator $r$, they agree exactly when their numerators agree after some further refinement $e$ with $re\in S$: one implication is immediate; for the other, cancel $r$ between the two refinement legs and then precompose by the resulting denominator.

For two choices of common denominator, refine those denominators once more. Equality of each of the two summands can then be witnessed at that denominator by the previous criterion; applying cancellation successively makes both numerator equalities simultaneous. Their sums are equal there by additivity. This proves choice and representative independence. At a common denominator, associativity, commutativity, zero and negation are precisely the abelian-group laws in $\mathcal{C}(U,Y)$.

Postcomposition by an ordinary arrow is additive since it acts on numerators. Precomposition by an ordinary arrow is additive: use one Ore square with the common denominator of both summands. Composition with $Q(s{)}^{-1}$ is the inverse of the additive composition bijection for $Q(s)$, and is therefore additive. Every localized arrow is a product of ordinary arrows and inverse denominators, proving bilinearity.

The image of $1_0=0$ shows $Q(0)$ is a zero object: every arrow to or from it is zero by bilinearity and the identity law. For $B=X\oplus Y$, the equations $p_X{i}_X=1$, $p_Y{i}_Y=1$, $p_X{i}_Y=p_Y{i}_X=0$ and $i_X{p}_X+i_Y{p}_Y=1_B$ survive under the additive $Q$. They supply unique tuples of incoming and outgoing arrows, hence make $Q(B)$ a biproduct. The empty biproduct is $Q(0)$, and binary ones iterate to finite ones.

Write $\alpha =Q(s{)}^{-1}Q(k)$. The outgoing Ore square for $s,f^{\mathrm{\prime }}$ gives $t:Y^{\mathrm{\prime }}\to Y^{\mathrm{\prime }\mathrm{\prime }}$ in $S$ and $f^{\mathrm{\prime }\mathrm{\prime }}:X^{\mathrm{\prime }\mathrm{\prime }}\to Y^{\mathrm{\prime }\mathrm{\prime }}$ with $f^{\mathrm{\prime }\mathrm{\prime }}s=t{f}^{\mathrm{\prime }}$. Write $\beta =Q(q{)}^{-1}Q(i)$, and apply Ore to $t,q$ to find $r:Y^{\mathrm{\prime }\mathrm{\prime }}\to Y^{(3)}$ in $S$ and $j$ with $rt=jq$. Replace $f^{\mathrm{\prime }\mathrm{\prime }},t$ by $r{f}^{\mathrm{\prime }\mathrm{\prime }},rt$ and put $l=ji$. Then $\beta =Q(t{)}^{-1}Q(l)$ and the right square commutes.

The localized commuting square now gives $Q(lf)=Q(f^{\mathrm{\prime }\mathrm{\prime }}k)$. Dual equality detection supplies $d$ in $S$ with $dlf=d{f}^{\mathrm{\prime }\mathrm{\prime }}k$. Replace $l,f^{\mathrm{\prime }\mathrm{\prime }},t$ by $dl,d{f}^{\mathrm{\prime }\mathrm{\prime }},dt$; both ordinary squares now commute and $dt\in S$. This proves the square assertion, including identity or coincident arrows.

For a composable pair $\alpha :QX\to QY$, $\beta :QY\to QZ$, write $\alpha =Q(s{)}^{-1}Q(f)$ with $s:Y\to Y^{\mathrm{\prime }}$ in $S$. Write $\beta Q(s{)}^{-1}=Q(t{)}^{-1}Q(g)$ with $t:Z\to Z^{\mathrm{\prime }}$ in $S$ and $g:Y^{\mathrm{\prime }}\to Z^{\mathrm{\prime }}$. Thus after the object comparisons $1_X,Q(s),Q(t)$ the pair is $Q(f),Q(g)$ and its composite is $Q(gf)$. The equalities follow from the proved composition law, not from an assumption about arbitrary diagrams.

The additive localization exists. Since a quasi-isomorphism remains one after either shift, shifting both arrows of a roof defines mutually inverse additive shifts. The zero complex and identity triangles descend as well.

For any arrow write $\alpha =Q(s{)}^{-1}Q(f)$ with $s:Y\to Y^{\mathrm{\prime }}$ a quasi-isomorphism. A cone triangle on $f:X\to Y^{\mathrm{\prime }}$ descends and transport along $Q(s)$ completes $\alpha :X\to Y$. Closure under isomorphism is built into the definition. Rotation gives $(Y,Z,X[1],v,w,-u[1])$ because this is the rotation in $K$; shifting a roof preserves the minus sign. This proves TR1 and both directions of TR2.

For TR3 first replace the two triangles by images of triangles on ordinary maps $f,f^{\mathrm{\prime }}$. Apply the square-clearing lemma to the prescribed first two components. It supplies a third ordinary arrow $f^{\mathrm{\prime }\mathrm{\prime }}$ and two commuting squares with maps $(k,l)$ and denominator maps $(s,t)$ into $f^{\mathrm{\prime }\mathrm{\prime }}$. Complete each square to a morphism of cone triangles in $K$ by its TR3, giving third maps $m:C_f\to C_{f^{\mathrm{\prime }\mathrm{\prime }}}$ and $r:C_{f^{\mathrm{\prime }}}\to C_{f^{\mathrm{\prime }\mathrm{\prime }}}$.

For every $n$, take the five consecutive terms $H^n(X^{\mathrm{\prime }})$, $H^n(Y^{\mathrm{\prime }})$, $H^n(C_{f^{\mathrm{\prime }}})$, $H^{n+1}(X^{\mathrm{\prime }})$, $H^{n+1}(Y^{\mathrm{\prime }})$ and their counterparts for $f^{\mathrm{\prime }\mathrm{\prime }}$. Exactness holds in $K$; four vertical maps are isomorphisms because $s,t$ are quasi-isomorphisms. The five lemma makes $H^n(r)$ an isomorphism. Hence $r$ is a quasi-isomorphism before any triangulation of $D$ is used.

Put the third localized component equal to $Q(r{)}^{-1}Q(m)$. The two morphisms of triangles in $K$, with the now invertible comparison $Q(r)$, give all three commuting triangle squares and the shifted first component. Transporting back proves TR3 for the original data.

Clear a composable pair $\alpha :QX\to QY$, $\beta :QY\to QZ$ simultaneously to ordinary $f:X\to Y^{\mathrm{\prime }}$, $g:Y^{\mathrm{\prime }}\to Z^{\mathrm{\prime }}$, using denominator isomorphisms of the objects. This includes zero maps or identity maps. Their composite becomes $gf$.

Apply TR4 in $K(\mathcal{A})$ to $f,g$, using cone triangles with structure maps $i_f,p_f$ and similarly for $g,gf$. It gives $a:C_f\to C_{gf}$ and $b:C_{gf}\to C_g$ such that $a{i}_f=i_{gf}g$ on $Y^{\mathrm{\prime }}$, $p_{gf}a=p_f$, $b{i}_{gf}=i_g$, and $p_{g}b=f[1]p_{gf}$. In particular $C_f\stackrel{a}{\to }{C}_{gf}\stackrel{b}{\to }{C}_g\stackrel{i_f[1]p_g}{\to }{C}_f[1]$ is distinguished. These equations, rather than the cone objects alone, are the octahedral data.

Apply the additive functor $Q$ to the entire octahedron: every face equation survives, including $Q(i_f[1]p_g)=Q(i_f)[1]Q(p_g)$, and its fourth triangle is distinguished by definition. Transport along the object isomorphisms from the clearing step.

If the three triangles specified in TR4 are other completions, TR3 compares each with the constructed completion with identity first two components. Such a comparison has invertible third component: applying either representable Hom and the exact sequences established from TR1–TR3 proves this by the five-term argument, then the Hom criterion for invertibility. Transport the octahedral maps along these triangle isomorphisms. This gives TR4 for the originally prescribed completions with all signs unchanged.

The additive structure, shift, TR1, signed TR2 and TR3 have been established, including identity triangles and zero objects. TR4 has also been established. These are precisely the triangulated-category axioms.

The functor $Q$ is additive, its shift comparison is the identity in the roof model, and it takes every cone triangle to a distinguished triangle by definition. It is therefore exact. By [F4], the cohomology functors are defined on $D(\mathcal{A})$, and on a localized cone triangle their maps are the maps in the cone long exact sequence [F5]. Transporting this sequence by a triangle isomorphism preserves exactness, proving the assertion for every distinguished triangle.

The acyclic subcategory is thick, and a map has acyclic cone exactly when it is a quasi-isomorphism. Thus the indicated quotient inverts precisely the denominators used to construct $D(\mathcal{A})$, with its proved triangulation. Its kernel is precisely the complexes with zero cohomology, including the zero complex.

If exact $F$ kills acyclics, a denominator triangle becomes $F(X)\stackrel{F(s)}{\to }F(Y)\to 0\to F(X)[1]$. Hom exactness implies $\mathrm{Hom}(W,F(s))$ is bijective for every $W$. Taking $W=F(Y)$ supplies a right inverse; injectivity at $W=F(X)$ shows it is also a left inverse. Thus $F(s)$ is invertible and localization gives a unique factorization.

Its additivity follows from the common-denominator sum formula and additivity of $F$. The shift comparison descends by naturality for inverse denominators. Each distinguished triangle is an isomorphic image of a $K$ triangle, so its image under $\bar{F}$ is distinguished because $F$ is exact. Conversely any exact factor through $Q$ takes an acyclic to the image of a zero object, hence to zero.

Let $A$ be acyclic and $f:P\to A$ a chain map; shifting the target will give the same argument for any $A[r]$. Choose an upper bound $b$ for $P$, and set $h^n=0$ for $n>b$. We seek $h^n:P^n\to A^{n-1}$ satisfying $f^n=d_A^{n-1}{h}^n+h^{n+1}{d}_P^n$. The zero complex permits all choices to be zero.

Suppose the equation holds in degrees above $n$. Put $u^n=f^n-h^{n+1}{d}_P^n$. The chain-map equation and the equation at $n+1$ give $d_A^n{u}^n=0$. Acyclicity makes $A^{n-1}\to Z^n(A)$ epic, so projectivity of $P^n$ lifts $u^n$ to $h^n$. This establishes the equation in degree $n$, including the initial degree $b$.

The partial homotopies form a nonempty set of finite sequences of maps (all relevant Hom collections are sets), with an entire extension relation. DC, or the supplied successive lifts, gives the infinite descending homotopy. Thus every $P\to A[r]$ is nullhomotopic for every $r$, which is K-projectivity. The recursion requires an upper bound; no assertion for arbitrary unbounded projectives follows.

Let $A$ be acyclic and $f:A\to I$ a chain map. Choose $a$ with $I^n=0$ for $n<a$ and set $h^n:A^n\to I^{n-1}$ equal to zero for $n\le a$. The equation $f^n=d_I{h}^n+h^{n+1}{d}_A$ holds below $a$. In particular it is valid for zero complexes.