Perfect Complexes and Triangulated Grothendieck Groups

$D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ is essentially small. Every finitely generated projective left $A$-module is a direct summand of a finite free module [F5]; choosing a finite generating family of $P$ gives a surjection $A^n\twoheadrightarrow P$, which splits by [F5], so $P\cong \mathrm{im}(e)$ for an idempotent $e\in M_n(A)$. The idempotent matrices in ${\bigcup }_{n\ge 0}{M}_n(A)$ form a set, so the isomorphism classes of finitely generated projective left $A$-modules form a set; bounded cochain complexes of these modules are finite-support sequences of such modules with differentials, and they therefore also form a set of objects. By [F1] every object of $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ is isomorphic to one of these complexes, so the set of isomorphism classes $\mathrm{Iso}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$ is a set. In the graded case [F6] exhibits each finite graded projective as a degree-zero summand of a finite sum $A\{r_1\}\oplus \cdots \oplus A\{r_n\}$; the finite tuples of shifts and the degree-zero idempotent endomorphisms of their sums form a set, and the same finite-support complex argument applies with [F7].

Shifts preserve bounded finite-projective complexes. If $P$ is a bounded complex of finitely generated projective left $A$-modules, then $(P[1]{)}^n=P^{n+1}$ with differential $-d_P^{n+1}$ [F4], so $P[1]$ is again bounded with finitely generated projective terms, and likewise $P[-1{]}^n=P^{n-1}$ is such a complex. Graded complexes with degree-zero differentials behave identically, since cochain shift changes only cochain degrees and moves the sign of the differential.

The cone of a chain map of bounded finite-projective complexes is again one. For a chain map $f:P\to Q$ of such complexes, [F4] gives $\mathrm{Cone}(f{)}^n=Q^n\oplus P^{n+1}$, which is a finite direct sum of finitely generated projective modules, hence finitely generated projective by [F5]; the support of the cone is contained in the sum of the supports of $P$ and $Q$, hence finite. In the graded case the biproduct is computed degreewise [F7] and a finite direct sum of finite graded projectives is again finite graded projective by [F6].

Every bounded complex $P$ of projective objects is K-projective, with no choice principle needed. Let $E$ be acyclic, $r$ an integer, $f:P\to E[r]$ a chain map, and suppose $P^n=0$ for $n>b$; put $F:=E[r]$, which is acyclic, and set $h^n=0$ for $n>b$. Inductively assume $h^{n+1}:P^{n+1}\to F^n$ satisfies $f^{n+1}=d_F^n{h}^{n+1}+h^{n+2}{d}_P^{n+1}$, and put $u^n:=f^n-h^{n+1}{d}_P^n$. Then $d_F^n{u}^n=d_F^n{f}^n-d_F^n{h}^{n+1}{d}_P^n=f^{n+1}{d}_P^n-(f^{n+1}-h^{n+2}{d}_P^{n+1})d_P^n=0$, so $u^n$ factors through the cycles $Z^n(F)=\ker d_F^n$. Since $H^n(F)=0$, the map $F^{n-1}\to Z^n(F)$ is an epimorphism; by projectivity of $P^n$ the composite $P^n\to Z^n(F)$ lifts to $h^n:P^n\to F^{n-1}$ with $d_F^{n-1}{h}^n=u^n$, which is the homotopy equation in degree $n$. Below the support of $P$ we take $h^n=0$, where both sides vanish. Only finitely many lifts are chosen, one for each degree in the finite support of $P$, so no dependent choice is used and the induction terminates. Hence ${\mathrm{Hom}}_K(P,F)=0$ for every acyclic $F$ and every shift, which is [F2].

Every derived morphism between bounded finite-projective complexes is represented by a chain map, uniquely up to homotopy. A bounded complex of finitely generated projectives has projective terms, so step 1.4 makes it K-projective; the published no-roof proposition [F3] then makes $Q:{\mathrm{Hom}}_K(P,X)\to {\mathrm{Hom}}_D(P,X)$ bijective for every complex $X$, in particular for a bounded finite-projective complex $X$. Surjectivity represents every derived morphism $P\to X$ by a chain map, and injectivity says two chain maps represent the same derived morphism exactly when they are chain homotopic.

$D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ is closed under shifts. Let $X$ be perfect with bounded finite-projective representative $P$, so that $X\cong P$ in $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$. Then $X[1]\cong P[1]$ and $X[-1]\cong P[-1]$; by step 1.2 both $P[1]$ and $P[-1]$ are bounded complexes of finitely generated projectives, so [F1] makes $X[1]$ and $X[-1]$ perfect. In the graded case the same argument uses the graded shift of a bounded complex with degree-zero differentials and finite graded projective terms.

$D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ is closed under cones. Let $X\to Y\to Z\to X[1]$ be a distinguished triangle of $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$ with $X,Y$ perfect. Fix bounded finite-projective representatives $P,Q$ and isomorphisms $u:P\to X$, $v:Q\to Y$ in $D$. The composite $g:=v^{-1}\circ (X\to Y)\circ u:P\to Q$ is a derived morphism between bounded finite-projective complexes, so by step 2.1 it is represented by a chain map $f:P\to Q$ with $Q(f)=g$, that is, $vQ(f)=Q(X\to Y)u$. The cone triangle $P\to Q\to \mathrm{Cone}(f)\to P[1]$ is distinguished by [F4], and its cone is a bounded finite-projective complex by step 1.3. Since $v\circ Q(f)=Q(X\to Y)\circ u$, TR3 [F8] supplies a third component $c:\mathrm{Cone}(f)\to Z$ making $(u,v,c)$ a morphism of triangles; the first two components are isomorphisms, so the triangulated five lemma [F9] makes $c$ an isomorphism. Hence $Z\cong \mathrm{Cone}(f)$, and [F1] makes $Z$ perfect. The graded case is identical, with the graded biproduct and graded projectivity supplied by [F6, F7].

Collecting the results: $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ is a strictly full subcategory by [F1], closed under isomorphism by construction, and closed under shifts and cones by steps 2.2 and 3.1. The distinguished triangles with objects in $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ are those of $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$ among these objects; the axioms TR1–TR4 hold in $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$ [F10] and their completions, being built by shifts and cones from perfect objects, again lie in $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ by steps 2.2 and 3.1, while all morphisms between perfect objects are available because the subcategory is strictly full. Hence $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$, with the inherited translation and triangles, is triangulated, and step 1.1 shows it is essentially small. The graded assertions are proved by the same steps with the graded data.

The identity triangle $X\stackrel{1_X}{\to }X\to 0\to X[1]$ is distinguished by TR1, so its relation $[X]=[X]+[0]$ holds in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(\mathcal{T})$; subtracting $[X]$ gives $[0]=0$.

Let $F:\mathcal{T}\to {\mathcal{T}}^{\mathrm{\prime }}$ be exact. The assignment $\phi ([X]):=[F(X)]$ is a function on $\mathrm{Iso}(\mathcal{T})$: a functor preserves isomorphisms, so isomorphic objects have isomorphic images. If $X\to Y\to Z\to X[1]$ is distinguished, exactness of $F$ makes $F(X)\to F(Y)\to F(Z)\to F(X)[1]$ distinguished, using the specified shift isomorphism, so $\phi$ satisfies $\phi ([Y])=\phi ([X])+\phi ([Z])$. By the free-group and quotient universal properties of [F6], in the pattern recalled in [F7], $\phi$ factors uniquely through a homomorphism $F_{\ast }:K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(\mathcal{T})\to K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}({\mathcal{T}}^{\mathrm{\prime }})$ with $F_{\ast }([X])=[F(X)]$.

By TR2 the signed left rotation $X\to 0\to X[1]\stackrel{-1_{X[1]}}{\to }X[1]$ of the identity triangle is distinguished, so $[0]=[X]+[X[1]]$; by step 1.1, $[X[1]]=-[X]$. Applying this to $X[-1]$, and using that $[-1]$ is a quasi-inverse of $[1]$ so that $(X[-1])[1]\cong X$, gives $[X]=-[X[-1]]$ and hence $[X[-1]]=-[X]$.

For the identity functor $1_{\mathcal{T}}$ the class function $[X]\mapsto [1_{\mathcal{T}}(X)]$ is $[X]\mapsto [X]$, so $(1_{\mathcal{T}}{)}_{\ast }=1$. If $F:\mathcal{T}\to {\mathcal{T}}^{\mathrm{\prime }}$ and $G:{\mathcal{T}}^{\mathrm{\prime }}\to {\mathcal{T}}^{\mathrm{\prime }\mathrm{\prime }}$ are exact, then $G\circ F$ is exact: additivity, the composite natural isomorphism ${\xi }_{F(X)}^G\circ G({\xi }_X^F):GF(X[1])\to GF(X)[1]$, and preservation of distinguished triangles all compose. On every generator, $(G\circ F{)}_{\ast }([X])=[(G\circ F)(X)]=[G(F(X))]=G_{\ast }(F_{\ast }([X]))$, so $(G\circ F{)}_{\ast }=G_{\ast }\circ F_{\ast }$.

Let $\eta :F\Rightarrow G$ be a natural isomorphism between exact functors. Each component ${\eta }_X:F(X)\to G(X)$ is an isomorphism by [F5], so $F(X)$ and $G(X)$ have the same isomorphism class in ${\mathcal{T}}^{\mathrm{\prime }}$ and hence $[F(X)]=[G(X)]$ in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}({\mathcal{T}}^{\mathrm{\prime }})$. The two induced homomorphisms agree on every generator of the free group and therefore are equal. This is an equality of group homomorphisms only; no coherence for a collection of such natural isomorphisms, and no group-action data, is asserted or obtained.

For $n\ge 0$, induction on $n$ using $[X[n]]=[(X[n-1])[1]]=-[X[n-1]]$ gives $[X[n]]=(-1{)}^n[X]$; the case $n=0$ uses $X[0]\cong X$ and $n=1$ is step 2.1. For $n<0$, apply the nonnegative case to $X[n]$: $[X]=[(X[n])[-n]]=(-1{)}^{-n}[X[n]]$, so $[X[n]]=(-1{)}^n[X]$. Every integer $n$ is covered by the two cases.

${\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A)$ is an essentially small additive category, so $\chi$ takes values in the group of [F3]. It is additive because the direct sum of two finitely generated projective modules is finitely generated projective and the zero module is finitely generated projective. It is essentially small: every finitely generated projective $P$ is a direct summand of a finite free module $A^n$ [F8], so $P\cong \mathrm{im}(e)$ for an idempotent matrix $e\in M_n(A)$, and the idempotents in ${\bigcup }_{n\ge 0}{M}_n(A)$ form a set. For a bounded complex $P$, the sum $\chi (P)={\sum }_n(-1{)}^n[P^n]$ is finite by boundedness; it uses the classes of the terms in $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))$. The graded category of finite graded projectives is additive and essentially small by [F10], with the same finiteness of the alternating sum.

Every acyclic bounded complex $P$ of finitely generated projectives has $\chi (P)=0$. Induct on the finite number of degrees in which $P$ is nonzero. Let $b$ be the largest such degree, so $P^b\mathrm{\ne }0$ and $P^{b+1}=0$; acyclicity gives $P^b=\mathrm{im}(d^{b-1})$, and the sequence $0\to Z^{b-1}\to P^{b-1}\stackrel{d^{b-1}}{\to }{P}^b\to 0$, with $Z^{b-1}=\ker d^{b-1}$, splits by projectivity of $P^b$ [F8]; hence $[P^{b-1}]=[Z^{b-1}]+[P^b]$ in the split group and $Z^{b-1}$ is finitely generated projective. Let $T$ be the complex with $T^n=P^n$ for $n\le b-2$, $T^{b-1}=Z^{b-1}$ and $T^n=0$ for $n\ge b$, with the restricted differentials. Then $T$ is a bounded complex of finitely generated projectives with fewer nonzero terms, and it is acyclic: in degrees $n\le b-2$ its cohomology is that of $P$, and $H^{b-1}(T)=Z^{b-1}\mathrm{/}\mathrm{im}(d^{b-2})=H^{b-1}(P)=0$. By induction $\chi (T)=0$, while the split relation gives $\chi (P)-\chi (T)=(-1{)}^{b-1}([P^{b-1}]-[Z^{b-1}])+(-1{)}^b[P^b]=0$. Hence $\chi (P)=0$. The graded case repeats the argument with the degreewise kernel $Z^{b-1}$ and the graded splitting of [F10].

For a chain map $f:P\to Q$ of bounded finite-projective complexes, $\chi (\mathrm{Cone}(f))=\chi (Q)-\chi (P)$. By [F7] the cone has terms $\mathrm{Cone}(f{)}^n=Q^n\oplus P^{n+1}$; the split relations of [F3] give $\chi (\mathrm{Cone}(f))={\sum }_n(-1{)}^n[Q^n]+{\sum }_n(-1{)}^n[P^{n+1}]=\chi (Q)-\chi (P)$. The cone is bounded with finitely generated projective terms by [F2], so the left side is defined.

$\chi$ is unchanged by quasi-isomorphism of bounded finite-projective complexes. If $f:P\to Q$ is a quasi-isomorphism, then $\mathrm{Cone}(f)$ is acyclic [F5], so $\chi (\mathrm{Cone}(f))=0$ by step 1.2 and step 1.3 gives $\chi (Q)=\chi (P)$.

$\chi$ takes the same value on any two bounded finite-projective representatives of one object of $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$. Let $X$ be perfect and let $P,P^{\mathrm{\prime }}$ be bounded finite-projective complexes with isomorphisms $X\cong P$ and $X\cong P^{\mathrm{\prime }}$ in $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$; composing gives an isomorphism $P\to P^{\mathrm{\prime }}$ in the derived category. By [F2] this derived morphism is represented by a chain map $f:P\to P^{\mathrm{\prime }}$, and $f$ is a quasi-isomorphism because its image in $D$ is an isomorphism. Step 2.1 gives $\chi (P)=\chi (P^{\mathrm{\prime }})$, so $\chi (X):=\chi (P)$ is well defined on perfect objects, independently of the chosen representatives.

$\chi$ is unchanged by homotopy equivalence: a homotopy equivalence of bounded finite-projective complexes is a quasi-isomorphism by [F6], so step 2.1 applies. Together with step 3.1 this is the invariance asserted for bounded finite-projective complexes and for the represented perfect object.

$\chi$ is additive on distinguished triangles of perfect objects. Let $X\to Y\to Z\to X[1]$ be distinguished, choose bounded finite-projective representatives $P,Q$ with isomorphisms $u:P\to X$, $v:Q\to Y$, and let $g:=v^{-1}\circ (X\to Y)\circ u$. By [F2] and [F4], $g$ is represented by a chain map $f:P\to Q$, that is, $vQ(f)=Q(X\to Y)u$; the cone triangle $P\to Q\to \mathrm{Cone}(f)\to P[1]$ is distinguished [F7]. TR3 [F9] supplies $c:\mathrm{Cone}(f)\to Z$ making $(u,v,c)$ a morphism of triangles, and the first two components are isomorphisms, so the triangulated five lemma [F9] makes $c$ an isomorphism. Therefore $Z\cong \mathrm{Cone}(f)$ in $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$ and, by step 3.1, $\chi (Z)=\chi (\mathrm{Cone}(f))=\chi (Q)-\chi (P)=\chi (Y)-\chi (X)$, which is the asserted additivity.

Steps 1.1–4.2 establish the definition, quasi-isomorphism and homotopy invariance, independence of representatives, and triangle additivity for complexes of finitely generated projective left modules; the graded assertions use the graded splitting and degreewise biproducts of [F10] at every occurrence of a splitting or a direct sum. No global-dimension hypothesis and no choice principle is used, and the argument nowhere asserts an Euler class for an arbitrary bounded complex of modules.

For finitely generated projective left modules $P,Q$ the degreewise split sequence of complexes $0\to P[0]\to (P\oplus Q)[0]\to Q[0]\to 0$ (the biproduct sequence in degree zero, zero in every other degree) is short exact, and all three complexes are bounded with finitely generated projective terms; by [F2] they are perfect objects of $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$, and [F6] gives a distinguished triangle $P[0]\to (P\oplus Q)[0]\to Q[0]\to P[1]$ of $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$, hence of $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$ by [F2]. Its relation $[(P\oplus Q)[0]]=[P[0]]+[Q[0]]$ holds in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$ by [F1]. Thus the class function $[P]\mapsto [P[0]]$ on $\mathrm{Iso}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))$ is additive on biproducts, and [F3] gives a unique homomorphism ${\iota }_{\ast }:K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))\to K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$ with ${\iota }_{\ast }([P])=[P[0]]$.

By [F4] the assignment $\chi (X):={\sum }_n(-1{)}^n[P^n]$, for any bounded finite-projective complex $P$ representing the perfect object $X$, is a well-defined class function on $\mathrm{Iso}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$ with values in $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))$, is additive on distinguished triangles of perfect objects, and is independent of the representative. Extending $\chi$ over the free abelian group on $\mathrm{Iso}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$ and applying the quotient universal property of [F8] in the pattern of the functor-induced class functions of [F4] and [F1] produces a unique homomorphism $\bar{\chi }:K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))\to K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))$ with $\bar{\chi }([X])={\sum }_n(-1{)}^n[P^n]$.

The composite $\bar{\chi }\circ {\iota }_{\ast }$ is the identity of $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))$: for a finitely generated projective $P$, $\bar{\chi }({\iota }_{\ast }([P]))=\chi (P[0])=[P]$ because the degree-zero stalk complex has the single term $P$ in degree $0$ by [F7], and the two homomorphisms agree on every generator of $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}$ [F3], with ${\iota }_{\ast }$ and $\bar{\chi }$ as constructed in steps 1.1 and 1.2.

The composite ${\iota }_{\ast }\circ \bar{\chi }$ is the identity of $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$. Let $P$ be a bounded finite-projective complex with $P^i=0$ for $i<a$ and $i>b$, representing $X$. For every integer $n$ the inclusion ${\sigma }^{\ge n}P\hookrightarrow {\sigma }^{\ge n-1}P$ of [F7] is a degreewise split short exact sequence of complexes with cokernel the degree-$(n-1)$ stalk complex $P^{n-1}[-(n-1)]$, all terms bounded finite projective; by [F6] and [F2] it gives a distinguished triangle of $D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A)$, so [F1] and [F5] give $[{\sigma }^{\ge n-1}P]=[{\sigma }^{\ge n}P]+(-1{)}^{n-1}[P^{n-1}[0]]$. Since ${\sigma }^{\ge b+1}P=0$ and $[{\sigma }^{\ge b+1}P]=0$ by [F5], summing these relations for $n=a+1,\dots ,b+1$ telescopes to $[{\sigma }^{\ge a}P]={\sum }_{j=a}^b(-1{)}^j[P^j[0]]$; and ${\sigma }^{\ge a}P=P$ by [F7]. Hence $[X]=[P]={\sum }_j(-1{)}^j[P^j[0]]={\sum }_j(-1{)}^j{\iota }_{\ast }([P^j])={\iota }_{\ast }(\bar{\chi }([X]))$, using ${\iota }_{\ast }([P^j])=[P^j[0]]$ from step 1.1 and the definition of $\bar{\chi }$ from step 1.2. The classes $[X]$ generate $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$ [F1], so ${\iota }_{\ast }\circ \bar{\chi }$ is the identity; the zero complex, where the range $a\le j\le b$ is empty, satisfies $[0]=0$ by [F5].

Steps 2.1 and 2.2 exhibit $\bar{\chi }$ as a two-sided inverse of ${\iota }_{\ast }$, so degree-zero inclusion induces the asserted isomorphism $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{Proj}}_{\mathrm{f}\mathrm{g}}(A))\to K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D_{\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{f}}(A))$, with inverse sending the class of an object represented by $P$ to ${\sum }_n(-1{)}^n[P^n]$; no finite global-dimension or Noetherian hypothesis was used. The graded comparison is the same argument run in the abelian category ${\mathrm{GrMod}}_0(A)$ of graded modules with degree-zero maps, where finite graded projectives replace finite projectives, the graded Euler lemma and graded closure clause of [F2, F4] replace their ungraded counterparts, internal shifts $\{1\}$ are left untouched, and the biproduct and brutal-truncation sequences are formed degreewise.