Draft lesson.

Derived perfect duality for twisted module complexes

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

Duality for a complex carries signs which do not occur for a single module. A second distinction appears for sheaves: an internal Hom object is a sheaf complex, whereas a global derived morphism is a morphism in the derived category. We construct the twisted internal Hom, prove its bounded tensor adjunction, and obtain duality from finite local projective models. The evaluation maps, their twists and their coherence are part of the construction. The untwisted local-to-global comparison, used in Exercise 8.4, is proved in Sheaf Ext and the local-to-global spectral sequence, Section 2.

1. Twists, bounds and the two Hom constructions

Let $X$ be a topological space with a sheaf of commutative rings $\mathcal R$. Fix universes containing all coverings, diagrams and Hom sets used below. A line cocycle $L$, on an open cover, consists of invertible $\mathcal R$-modules $L_{ij}$, associative multiplications $L_{ij}\otimes L_{jk}\simeq L_{ik}$, and diagonal units. An $L$-twisted module has transitions $L_{ij}\otimes M_j\simeq M_i$ satisfying that multiplication equation. Write $\mathcal T_L$ for their global abelian category. The neutral category $\mathcal T_1$ is $\operatorname{Mod}(\mathcal R)$.

Tensor and Hom for twisted module stacks, Sections 1–6 proves the products $LB$, inverse $L^{-1}$, cancellation $L^{-1}L\simeq1$, effective module descent, and $\mathcal T_L\times\mathcal T_B\longrightarrow\mathcal T_{LB}, \qquad \underline{\operatorname{Hom}}:\mathcal T_L^{\mathrm{op}}\times\mathcal T_B \longrightarrow\mathcal T_{BL^{-1}}. \tag{1.1}$ These are module-sheaf operations. An underline always means internal Hom. Global morphisms between objects with the same twist are sections of the neutral internal Hom sheaf.

The global categories have enough injectives by Injective objects in locally module stacks, Sections 1 and 5. Restriction to an open preserves injectives: Relative pullback of locally module stacks, Section 2 gives its left adjoint $j_!$, whose stalks are the original module inside the open and zero outside. It is exact. Transposing an extension problem through $j_!\dashv j^*$ proves injectivity after restriction; an exact local equivalence preserves the same property.

Complexes are cohomological, $K[1]^i=K^{i+1}$ with differential $-d_K$. Use the bounded-above flat replacements and tensor comparisons of Flat covers and bounded tensor for locally module stacks, Sections 3–7. The local algebra here is $\mathcal R$, which is flat over itself, so the base-flat hypothesis is satisfied. Tensor of twists identifies the product stack with $\mathcal T_{LB}$. Thus $D^-(\mathcal T_L)\times D^-(\mathcal T_B) \longrightarrow D^-(\mathcal T_{LB}),\quad(P,Q)\longmapsto P\otimes^L Q \tag{1.2}$ exists, with its associative constraints and graded symmetry $x\otimes y\mapsto(-1)^{ij}y\otimes x$.

An object $P$ is uniformly perfect if, on an open covering trivializing its twist, it is isomorphic in the local derived category to a complex of modules locally retracting from finite free modules, with terms zero outside one common finite interval $[a,b]$. The interval concerns its local projective models. It need not be the interval of its cohomology. We work with the full subcategory $\operatorname{Perf}^{\mathrm{unif}}_L\subset D^b(\mathcal T_L)$.

Local perfectness can instead be defined with a different interval on each open. The common interval in the definition above supplies the global bounds needed for a dual in $D^b$. None of our constructions chooses or assumes a global bounded projective model.

2. Constructing derived internal Hom

For $P$ bounded above and $I$ bounded below, set $\mathcal H(P,I)^n= \bigoplus_i\underline{\operatorname{Hom}}(P^i,I^{i+n}),\qquad (\partial f)^i=d_I f^i-(-1)^n f^{i+1}d_P. \tag{2.1}$ If $P^i=0$ for $i>b$ and $I^j=0$ for $j<a$, the possible indices satisfy $a-n\le i\le b$. Each sum is finite, equals the corresponding product, and is zero for $n<a-b$. The differential squares to zero: the two mixed terms cancel, while the remaining terms contain $d_I^2$ and $d_P^2$.

If $P$ has twist $L$ and $I$ has twist $B$, each summand has twist $BL^{-1}$. Its transition applies the inverse transition of $P$, contracts its line with $L^*_{ij}$, and applies the forward transition of $I$, exactly as in (1.1). All line maps have degree zero. Since the transitions of the complexes commute with their differentials, they commute with (2.1). The inverse line multiplication contracts the two successive inverse transitions; the forward $B$ transitions multiply in their original order. This gives the $BL^{-1}$ triple equation degree by degree, including its unit. Hence (2.1) is a global twisted complex, with precomposition and postcomposition on maps.

Lemma 2.1. If $I$ is termwise injective and bounded below, then $\mathcal H(-,I)$ takes a quasi-isomorphism of bounded-above complexes to a quasi-isomorphism. A quasi-isomorphism between bounded-below injective complexes induces the same conclusion in the second input.

Proof. On a trivializing open and every smaller open, $I^j$ is injective by Section 1. The sections of $\underline{\operatorname{Hom}}(-,I^j)$ therefore give an exact contravariant functor on module sheaves: a map from a subobject extends to the larger object. Thus an acyclic source gives exact source-direction complexes of internal Hom sheaves.

Their total is exact as well. Put $r=b-i$ and $s=j-a$, so $r,s\ge0$ and total degree is $a-b+r+s$. A total cycle has a finite diagonal. In the smallest target index $s$ where it has a component, its source-direction differential is zero, since the preceding target component is zero. Exactness in that source direction supplies a preimage. Subtract its total boundary; this removes that target component and adds a component only in the next target index. Repeat. On this fixed diagonal the source index $r$ decreases at each step, so the process terminates. At $r=0$, exactness at the initial source term says a closed component is zero. Sections needed for this argument exist locally, which suffices for exactness of sheaves. Applying it to the acyclic cone of a source quasi-isomorphism proves the first assertion.

An acyclic bounded-below injective complex $E$ is contractible. At its first degree its cycle is zero; successively split $0\to Z^jE\to E^j\to Z^{j+1}E\to0$: the already injective $Z^jE$ makes the inclusion split, and $Z^{j+1}E$ is a summand of the injective $E^j$. These splittings give $E^j\simeq Z^jE\oplus Z^{j+1}E$, with differential carrying the second coordinate to the first coordinate in the next term. Sending that first coordinate back to the preceding second coordinate gives $d h+h d=1$. The cone of the stipulated target map is bounded below, injective term by term, and acyclic, hence contractible. Postcomposition by its contraction contracts $\mathcal H(P,E)$; the mixed terms with $d_P$ cancel. The normalized Hom cone comparison gives the asserted target quasi-isomorphism. $\square$

Choose a bounded-below injective replacement $Q\to I_Q$ and define $R\underline{\operatorname{Hom}}(P,Q)=\mathcal H(P,I_Q) \quad\hbox{in }D^+(\mathcal T_{BL^{-1}}). \tag{2.2}$ This is a bifunctor on $D^-(\mathcal T_L)^{\mathrm{op}}\times D^+(\mathcal T_B)$, not merely an object assignment. Indeed bounded-below injective models give an equivalence with $D^+$: the unique homotopy comparison class realizes each derived arrow, as proved in Derived Hom and bounded resolution pairs, Sections 1, 5 and 6. Homotopies in the target act by postcomposition; in the source, for a degree $n$ Hom element, precompose the source homotopy with multiplier $(-1)^n$. Substitution in (2.1) gives the Hom homotopy equation. Lemma 2.1 then inverts every source denominator. Localization yields the stated bifunctor; comparison classes for three replacements compose by uniqueness. The same normalized cone and shift formulas from that prerequisite apply to internal module maps and retain the opposite-source triangle signs.

Restriction to an open computes the same derived object: it commutes with internal Hom and finite sums, and takes $I_Q$ to another bounded-below injective replacement. This provides canonical restriction comparisons compatible with nested opens and refinements. We constructed a global object first; we have not glued objects of local derived categories.

3. The bounded derived tensor adjunction

Theorem 3.1. For $A\in D^-(\mathcal T_C)$, $P\in D^-(\mathcal T_L)$, and $Q\in D^+(\mathcal T_{CL})$, there is a natural bijection $\operatorname{Hom}_D(A\otimes^L P,Q) \simeq \operatorname{Hom}_D\bigl(A,R\underline{\operatorname{Hom}}(P,Q)\bigr). \tag{3.1}$ The right internal Hom has twist $(CL)L^{-1}\simeq C$. Both groups are in the full derived categories of the indicated twists.

Proof. Replace $P$ by a bounded-above complex $F$ of base-flat twisted modules, and $Q$ by a bounded-below injective complex $I$. Formula (2.2) can use $F$, by Lemma 2.1. Each $\underline{\operatorname{Hom}}(F^i,I^j)$ is injective globally: it is the right adjoint to the exact functor $-\otimes F^i:\mathcal T_C\to\mathcal T_{CL}$. Transpose a map from a subobject, extend it into $I^j$, and transpose back. Each term of $\mathcal H(F,I)$ is a finite sum of these injectives, and its complex is bounded below.

Ordinary currying gives an isomorphism of the global Hom complexes, with formula $q(a\otimes p)=\widehat q(a)(p). \tag{3.2}$ There is no extra sign in this formula. For $a$ of degree $i$ and a Hom map $q$ of degree $n$, the right differential evaluates as $d_Iq(a,p)-(-1)^{i+n}q(a,d_Fp)-(-1)^nq(d_Aa,p).$ This is exactly $d_Iq-(-1)^nq\,d_{A\otimes F}$. The equivariance equation is the underived currying equation from (1.1) in each degree; sums are finite on each contributing diagonal. Currying consequently identifies chain maps and chain homotopies.

The complex $A\otimes F$ computes $A\otimes^L P$, by the single-flat replacement comparison in Section 1. A bounded-below injective target computes morphisms from any bounded-above source by homotopy maps; this includes $I$ and $\mathcal H(F,I)$, by the bounded Hom theorem cited in Section 2. Taking homotopy classes in (3.2) therefore proves (3.1). Precomposition, postcomposition and the replacement augmentations commute with currying. The inverse denominator maps then give naturality for every derived roof. Changes of flat and injective models use those same maps, so their comparison squares and three-choice composition equations commute. $\square$

The corresponding evaluation is the counit represented by (3.2). In particular, for bounded $P$ whenever $R\underline{\operatorname{Hom}}(P,Q)$ belongs to $D^-$, it gives $R\underline{\operatorname{Hom}}(P,Q)\otimes^L P\longrightarrow Q. \tag{3.3}$ This boundedness qualification ensures that the left input lies in the domain just proved.

4. Perfect models and the signed tensor–Hom map

If $K$ is a local projective model for a uniformly perfect $P$, every functor $\underline{\operatorname{Hom}}(K^i,-)$ is exact. For a finite free module it is a finite direct sum of the identity functor; taking a retract preserves exactness. A finite filtration by the degrees of $K$ now shows that $\mathcal H(K,-)$ preserves every target quasi-isomorphism: its successive total quotients are shifted exact Hom complexes. Likewise $K\otimes-$ preserves them, since its terms are flat and the filtration is finite. These are assertions about sheaf complexes, not about exactness of global sections.

Consequently the restriction of (2.2) to this open is computed by $\mathcal H(K,Q)$, for any bounded representative of $Q$. Its dual is $D(K)^p=\underline{\operatorname{Hom}}(K^{-p},\mathcal R), \qquad d_{D(K)}^p(\phi)=(-1)^{p+1}\phi\,d_K^{-p-1}. \tag{4.1}$ Dualizing a finite-free retract gives another finite-free retract. If $K$ is supported in $[a,b]$, its dual is supported in $[-b,-a]$.

Define $D_L(P)=R\underline{\operatorname{Hom}}(P,\mathcal R)$. It is uniformly perfect of inverse twist, by (4.1) and the restriction comparison. The uniform interval ensures that its global cohomology is zero outside $[-b,-a]$, so it belongs to $D^b(\mathcal T_{L^{-1}})$.

Theorem 4.1. For uniformly perfect $P$ and bounded $Q$ of twist $B$, the evaluation-induced morphism $\Psi_{P,Q}:D_L(P)\otimes^L Q \xrightarrow{\sim}R\underline{\operatorname{Hom}}(P,Q) \tag{4.2}$ is a natural isomorphism, with twist $BL^{-1}$.

Proof. Its construction is global. Tensor $D_L(P)\otimes Q$ once more with $P$, move $Q$ past $P$ by the graded symmetry, and evaluate $D_L(P)\otimes P$; take the mate under (3.1). All inputs to that adjunction have the required bounds. On a local model, for $\phi\in D(K)^p$, $y\in Q^q$, it is $\Psi(\phi\otimes y)(x)=(-1)^{pq}\phi(x)y \quad(x\in K^{-p}), \tag{4.3}$ and its other components are zero. The differential equation follows from the two identities $(-1)^{p(q+1)}=(-1)^p(-1)^{pq}$ and $(-1)^{(p+1)q}=(-1)^q(-1)^{pq}$, together with (4.1) and (2.1).

In each degree the unsigned map from a dual tensor to internal Hom is an isomorphism by the finite dual-basis argument of the twisted-module prerequisite. The multiplier in (4.3) is a unit; finite diagonals then give an isomorphism of complexes. Thus the global morphism has an acyclic cone on a covering and hence has zero cohomology cone globally. It is an isomorphism. Its construction by natural tensor, symmetry, evaluation and the natural adjunction proves naturality for all derived maps. This argument uses locality to test invertibility, not to infer equality of global derived morphisms. $\square$

For later use put $\chi_{P,Q}=\Psi_{P,Q}\circ c_{Q,D_L(P)}$. It has source $Q\otimes^L D_L(P)$. The two signs cancel locally, giving $\chi(y\otimes\phi)(x)=\phi(x)y$. Globally, by its defining mate equation, its adjoint evaluation is exactly $1_Q\otimes\operatorname{ev}_P$.

5. Evaluation, coevaluation and both global triangles

Write $D=D_L(P)$. Evaluation has neutral target, $\operatorname{ev}_P:D\otimes^L P\longrightarrow\mathcal R. \tag{5.1}$ Theorem 4.1 identifies the right adjoint in (3.1), on bounded objects, with $Q\mapsto Q\otimes^L D$. Tensor by $P$ and by $D$ preserve $D^b$: their local finite flat models have a common finite interval. Their counit at $Q$ is $1_Q\otimes\operatorname{ev}_P$, by the final mate equation in Section 4.

Define $\eta_P:\mathcal R\to P\otimes^L D$ to be the unit of this adjunction at $\mathcal R$, or equivalently the map corresponding to $1_P$. On a local model, choose a finite dual basis $(e_{i,r},e^*_{i,r})$ for each $K^i$. Then $\eta_K(1)=\sum_{i,r}e_{i,r}\otimes e^*_{i,r}, \qquad \operatorname{ev}_K(\phi\otimes x)=\phi(x) \quad\text{when }|\phi|+|x|=0. \tag{5.2}$ There is no degree sign in either formula. The dual-basis identity makes their description intrinsic. The differential of the coevaluation cancels degree by degree: a coefficient from $d_K e_i$ has its opposite from $(-1)^{i+1}e_{i+1}\otimes d_{D(K)}e^*_{i+1}$, since the dual differential on that functional contributes $(-1)^i$. Evaluation is a chain map because the two coefficients on total degree minus one are $(-1)^{p+1}$ and $(-1)^p$.

Theorem 5.1. The canonical maps satisfy, globally in the derived categories, $(1_P\otimes\operatorname{ev}_P)(\eta_P\otimes1_P)=1_P,\qquad (\operatorname{ev}_P\otimes1_D)(1_D\otimes\eta_P)=1_D. \tag{5.3}$ Associators and unit constraints are understood.

Proof. The mate defining $\eta_P$ is $1_P$, so the first composite in (5.3) is $1_P$, by the inverse adjunction rule. For any bounded $A$, the mate of $1_A\otimes\eta_P$ is the tensor of $1_A$ with that same first composite, hence is $1_{A\otimes P}$. Uniqueness in the adjunction says its unit at $A$ is $1_A\otimes\eta_P$. Its second adjunction triangle, evaluated at $Q=\mathcal R$, is therefore the second equality in (5.3). These are identities of global adjunction maps; no local detection of equality in a derived category was required. $\square$

The inverse twist is essential: both $P\otimes D$ and $D\otimes P$ are neutral, so the coevaluation and evaluation have the indicated types. The degree signs in the symmetry have already been accounted for in $\chi$.

6. Signed biduality and its inverse coherence

Apply (3.1) to the evaluation followed by the symmetry: $P\otimes^L D\xrightarrow{\,c\,}D\otimes^L P \xrightarrow{\operatorname{ev}_P}\mathcal R.$ Its mate defines $\beta_P:P\to D_{L^{-1}}D_L(P)$. On a local model its actual formula is $\beta_K^i(x)(\phi)=(-1)^i\phi(x), \qquad \phi\in D(K)^{-i}. \tag{6.1}$ Indeed the symmetry multiplier is $(-1)^{i(-i)}=(-1)^i$. The outer dual differential, under the ordinary identification with $K^{**}$, is minus the original differential. Thus (6.1)'s consecutive signs make it a chain map. Its components are invertible because finite-free retracts are canonically bidual. The explicit chain and homotopy calculations are also proved in Polynomial rings, finite resolutions and derived duality, Sections 5–6; those formulas apply to finite projective models over any commutative ring.

Theorem 6.1. The maps $\beta$ give a natural isomorphism and contravariant equivalence $D_L:(\operatorname{Perf}^{\mathrm{unif}}_L)^{\mathrm{op}} \xrightarrow{\sim}\operatorname{Perf}^{\mathrm{unif}}_{L^{-1}}, \qquad D_L(\beta_P)\,\beta_{D_L(P)}=1_{D_L(P)}. \tag{6.2}$ In the coherence equation $D_L(\beta_P)$ means the dual arrow from the third dual to the first dual, with its canonically identified twist.

Proof. The mate construction makes $\beta$ natural for derived arrows: evaluation is natural by precomposition in its Hom argument, and the symmetry and adjunction are natural. Its local chain isomorphism proves global invertibility by the cone cohomology test.

For the global coherence use uniqueness of dual pairs, which we spell out. If $U,V$ are left duals of $K$, the map $U\xrightarrow{\,1_U\otimes\eta_V\,}U\otimes K\otimes V \xrightarrow{\,\operatorname{ev}_U\otimes1_V\,}V \tag{6.3}$ preserves evaluation and coevaluation. Substituting the two triangle identities proves those assertions: the inserted evaluation-coevaluation pair collapses to the relevant identity. The same formula with $U,V$ interchanged is inverse, by collapsing both inserted pairs. Any map preserving evaluation must equal (6.3), because inserting the $V$ coevaluation and using its triangle reconstructs the map from that evaluation equation.

The symmetric monoidal constraints make $P$ a left dual of $D$, with evaluation $\operatorname{ev}_P c_{P,D}$ and coevaluation $c_{P,D}\eta_P$. Their two triangles are (5.3) with the symmetries moved past the tensor factors; naturality and the hexagon make the two adjacent inverse symmetries cancel. The canonical second dual is another left dual of $D$. Formula (6.3), or equivalently its evaluation mate, identifies their comparison with $\beta_P$. Applying the same comparison to $D$, and dualizing the comparison for $P$, gives an automorphism of its original dual pair: the evaluation equations transpose under the same symmetry, and the coevaluation equations follow by the reconstruction just proved. Their composite therefore preserves the original pairing, so uniqueness forces it to be $1_D$. This proves the coherence in (6.2) globally. In local degree $j$ its multiplier is explicitly $(-1)^j(-1)^{-j}=1$.

Both functors preserve the uniform perfect condition and have composites identified with the identity by these natural bidual maps. They are consequently quasi-inverse contravariant equivalences. Their source variance and source-shift signs remain those of (2.2). No inference that all finite modules over an arbitrary ring are perfect has been made. $\square$

7. Exact pullback, descent and trace

Let $f:Y\to X$ be continuous, and use the unchanged coefficient sheaf $f^{-1}\mathcal R$ and the pulled-back twists. Exact inverse image, its compatibility with tensor and the coherent transported stack are proved in Relative pullback of locally module stacks, Sections 1 and 7 and the flat-tensor prerequisite.

Proposition 7.1. For uniformly perfect $P$ and bounded $Q$, the canonical mate comparison is an isomorphism $f^{-1}R\underline{\operatorname{Hom}}(P,Q) \xrightarrow{\sim} R\underline{\operatorname{Hom}}(f^{-1}P,f^{-1}Q). \tag{7.1}$ It preserves the maps $\Psi,\operatorname{ev},\eta,\beta$, and is compatible with composition of inverse images and refinement of the twisting cover.

Proof. Pull back evaluation (3.3), use the derived tensor comparison, and take its mate on $Y$. This defines (7.1) globally, without requiring inverse image to preserve injectives. On a local finite-projective model $K$, the canonical underived map $f^{-1}\underline{\operatorname{Hom}}(K^i,Q^j)\to \underline{\operatorname{Hom}}(f^{-1}K^i,f^{-1}Q^j)$ is invertible: for finite free modules it is a finite direct sum comparison, and for their retracts it retains the splitting maps. The differential (2.1) commutes with it. The finite models compute both derived values by Section 4, proving invertibility of (7.1) on an open covering and hence globally. Pulled-back finite models have the same interval, so $f^{-1}P$ remains uniformly perfect.

The defining mate equation preserves evaluation. It also preserves $\Psi$, since that is the mate of evaluation with the tensor symmetry. The identity endomorphism pulls back to the identity, so the bijection defining $\eta$, with its invertible comparison $\chi$, preserves $\eta$. The evaluation and symmetry mate similarly preserves $\beta$. For two inverse images the tensor comparisons compose, and transposing their composite evaluation gives exactly the direct composite mate; uniqueness of the adjunction comparison proves composition coherence, including three maps and the identity. All operations on line factors use their degree-zero contraction and associative multiplication. Consequently refinement pulls back exactly those maps; triple-overlap equations are the original line-cocycle equations, as checked in Section 2. These are comparisons of globally constructed maps, rather than assumed descent of arbitrary derived-category data. $\square$

For an endomorphism $u:P\to P$, define its trace as the neutral derived endomorphism of $\mathcal R$ $\mathcal R\xrightarrow{\eta_P}P\otimes^L D \xrightarrow{u\otimes1}P\otimes^L D \xrightarrow{c}D\otimes^L P \xrightarrow{\operatorname{ev}_P}\mathcal R. \tag{7.2}$ For a local chain representative $u:K\to K$, (5.2) and the symmetry give $\operatorname{tr}(u)=\sum_i(-1)^i\sum_r e^*_{i,r}(u^ie_{i,r})$. For finite free terms this is the alternating matrix trace. For a retract the dual-basis identity makes the same expression independent of the retract presentation. Formula (7.2) is a construction in the derived category, so is invariant under homotopies and all representative comparisons and is preserved by (7.1). An alternating trace can vanish even when the complex is nonzero.

8. Four graded exercises with complete solutions

Exercise 8.1 — A differential changes the bidual sign (introductory)

For a commutative ring $R$, let $K=(R\xrightarrow{r}R)$ in degrees $0,1$. Compute both duals, the bidual map, coevaluation and cohomology. Specialize to $R=\mathbb Z,r=2$.

Solution. The first dual is $R\xrightarrow{r}R$ in degrees $-1,0$, since (4.1) has sign $+1$ at degree minus one. The second dual is $R\xrightarrow{-r}R$ in degrees $0,1$. Thus $\beta^0=1,\beta^1=-1$, and its chain equation is $(-r)1=(-1)r$. Unsigned evaluation fails when $2r\ne0$.

With basis elements $e_0,e_1$ and their duals $\phi_0,\phi_1$, coevaluation sends $1$ to $e_0\otimes\phi_0+e_1\otimes\phi_1$. Its two nonzero boundary contributions are $r e_1\otimes\phi_0$ and $-r e_1\otimes\phi_0$. Evaluating the inserted pairs on $e_i$, or on $\phi_i$, leaves that same basis element; this checks both triangles. The identity trace is $1-1=0$. The cohomology of $K$ is $\operatorname{ann}(r)$ in degree zero and $R/rR$ in degree one; that of $D(K)$ has these groups in degrees minus one and zero. Over $\mathbb Z$ with $r=2$, only $\mathbb Z/2$ remains, respectively in degrees one and zero. Its dual is not the degree-zero ordinary Hom of that cohomology group into $\mathbb Z$, which is zero.

Exercise 8.2 — A two-variable perfect resolution (intermediate)

Let $R=k[x,y]$, and put $K^{-2}=R,K^{-1}=R^2,K^0=R$, with $d^{-2}=\begin{pmatrix}-y\\x\end{pmatrix},\qquad d^{-1}=(x,y). \tag{8.1}$ Prove that $K$ represents $k=R/(x,y)$, compute its dual and bidual, and compute the trace of the identity.

Solution. The differential product is zero. The first map is injective, since $R$ is a domain and $x\ne0$. If $xa+yb=0$, evaluation at $x=0$ gives $y b(0,y)=0$ in the domain $k[y]$, hence $b=xc$. Substitution and cancellation of $x$ give $a=-yc$. This proves exactness in degree minus one. The cokernel in degree zero is $R/(x,y)=k$, so the augmentation is a quasi-isomorphism.

The dual has $R,R^2,R$ in degrees $0,1,2$ with $d_D^0=\begin{pmatrix}-x\\-y\end{pmatrix},\qquad d_D^1=(-y,x). \tag{8.2}$ Its first kernel is zero. For $-ya+xb=0$, evaluation at $x=0$ shows $a=xc$; substitution gives $b=yc$. These vectors are exactly the image of $d_D^0$, replacing $c$ by $-c$. The final cokernel is $R/(x,y)=k$. Thus $D(K)$ has its only cohomology in degree two. Its bidual has differential minus each original differential, and $\beta$ has multipliers $+1,-1,+1$ in degrees $-2,-1,0$; the chain equations follow on both adjacent pairs. Its identity trace is $1-2+1=0$, although $K$ represents the nonzero module $k$. No inference from vanishing trace to vanishing cohomology is valid.

Exercise 8.3 — Biduality alone does not imply perfectness (advanced)

Put $A=k[\epsilon]/(\epsilon^2)$ and $S=A/(\epsilon)$. Prove that $A$ is injective over itself and $S$ is canonically derived-bidual, but not perfect. Compare the cohomology of $R\operatorname{Hom}_A(S,A)\otimes_A^L S$ and $R\operatorname{Hom}_A(S,S)$.

Solution. Let $\lambda:A\to k$ take the coefficient of $\epsilon$. The pairing $(a,b)\mapsto\lambda(ab)$ has matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ in the basis $1,\epsilon$, hence identifies $A$ with $\operatorname{Hom}_k(A,k)$ as an $A$-module. Evaluation at $1$ gives a natural bijection $\operatorname{Hom}_A(M,\operatorname{Hom}_k(A,k))\simeq\operatorname{Hom}_k(M,k)$, whose inverse sends $\alpha$ to $m\mapsto(a\mapsto\alpha(am))$. The right functor is exact, because any linear functional extends across a vector-space inclusion. This proves injectivity of $A$, as also established in Noetherian generators, filtered injectives and Ext, Exercise 8.4.

The injective target $A[0]$ computes the derived dual of $S$ without a higher target resolution. Its ordinary Hom is $\epsilon A$, identified with $S$ by $s\mapsto(1\mapsto s\epsilon)$. The same computation gives its second derived dual. Evaluation sends $s$ to the functional on $\epsilon A$ taking $t\epsilon$ to $st\epsilon$; under these identifications the bidual map is the identity of $S$.

There is an augmented projective resolution with an $A$ in every nonpositive degree and every negative differential multiplication by $\epsilon$. It is exact because $\ker(\epsilon)=\epsilon A=\operatorname{im}(\epsilon)$, and the degree-zero augmentation has that same kernel. Tensoring this resolution with $S$ makes every differential zero. Therefore $H^j\bigl(R\operatorname{Hom}_A(S,A)\otimes_A^L S\bigr)= \begin{cases}S&j\le0,\\0&j>0.\end{cases} \tag{8.3}$ Applying $\operatorname{Hom}_A(-,S)$ to the resolution also makes every differential zero, but reverses its degrees. The projective computation in the bounded Hom prerequisite gives $H^jR\operatorname{Hom}_A(S,S)= \begin{cases}S&j\ge0,\\0&j<0.\end{cases} \tag{8.4}$ A bounded finite projective model of $S$ would make the latter cohomology vanish in sufficiently high degrees. Thus $S$ is not perfect. Equations (8.3)–(8.4) also prevent the perfect tensor–Hom identity from extending to this bidual object. These computations use a bounded-above projective resolution and a bounded-below Hom value; they do not require a construction of derived Hom for arbitrary unbounded pairs.

Exercise 8.4 — Local rank one does not make global Hom exact (advanced)

Let $X$ have four points $a,b,u,v$, ordered by $a<u,a<v,b<u,b<v$ and no other strict relations. Give it the topology of upper sets, and use the constant field sheaf $k$. Construct a nonsplit extension of $\mathcal R$ by itself, although $\mathcal R[0]$ is uniformly perfect. Explain why the degree-one global derived Hom cannot be computed by the ordinary global Hom complex.

Solution. A sheaf on this space is the diagram of its sections on the least neighborhoods, with maps from a smaller point to a larger point. Conversely a diagram gives sections on an upper open as compatible families at its points; restriction forgets coordinates. Compatibility glues such families uniquely, and the least-neighborhood section at a point is its indicated module. This proves the diagram description, with stalkwise kernels and cokernels.

The constant sheaf has $k$ at every point and identity maps on the four arrows. Define $E$ to have $k^2$ at every point, with arrow maps $\begin{pmatrix}1&c_e\\0&1\end{pmatrix}, \qquad c_{a,u}=1,\quad c_{a,v}=c_{b,u}=c_{b,v}=0. \tag{8.5}$ There are no strict chains of length two, so these data satisfy every composition equation. Injection into the first coordinate and projection from the second coordinate give a pointwise exact, hence sheaf-exact, row $0\to\mathcal R\to E\to\mathcal R\to0$.

A section of this projection would take $1$ at point $z$ to $(t_z,1)$. Equivariance forces $t_u=t_a+1,\quad t_v=t_a,\quad t_u=t_b,\quad t_v=t_b. \tag{8.6}$ The last three equations give $t_u=t_a$, contradicting the first over any field. Thus the extension does not split. Short exact extensions and the Baer group, Proposition 1.2 proves that its boundary is a nonzero element of $\operatorname{Hom}_D(\mathcal R,\mathcal R[1])$.

Internally, $\underline{\operatorname{Hom}}(\mathcal R,-)$ is the identity functor, by evaluation at $1$, so $R\underline{\operatorname{Hom}}(\mathcal R,\mathcal R)=\mathcal R[0]$. The ordinary global Hom complex of the two degree-zero complexes has only a degree-zero term, and consequently has zero degree-one cohomology. It misses the extension just constructed. To compute global derived Hom one must use global Hom into an injective resolution, equivalently derive global sections of this internal Hom object. Local finite-projective terms justify exact internal Hom; they do not justify exactness of global sections.

Freely accessible references

The Stacks Project, Internal Hom in the derived category, for the ordinary sheaf internal-Hom framework and restriction to opens.
The Stacks Project, Duals, for the relation between perfect complexes and evaluation/coevaluation.
The Stacks Project, Lemma 15.76.15, for comparison with perfect duality over a ring.