Draft lesson.

Dualizable sheaves, finite covers and global projectivity

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

Local dual bases determine internal Hom, but global Hom also applies the global-sections functor. On a site with finite covers, this makes every dualizable module sheaf compact. It does not make every such sheaf projective. The tensor unit determines when all dualizable sheaves are compact projective, and explicit finite and infinite examples separate the two properties.

Use Tensor and Hom for twisted module stacks, Lemmas 1.1–1.2 and Theorem 6.1, for the complete internal Hom and local-dual-basis proofs, including both dual triangles and the converse. The neutral specialization is used throughout. Associated sheaves with general coefficients supplies exact sheafification for group coefficients; Finite-cover sheafification and locally nilpotent coefficients, §3, gives finite matching and directed colimits. Compare the group-valued Hom criterion in Finite projective bimodules and both adjoints of tensor, §1. No derived category occurs here.

1. Module sheaves and global sections

Fix a universe and a small site $\mathcal C$ with terminal object $X$. Covering families and their pullbacks are small, and the indicated cover intersections exist. Let $R$ be a sheaf of commutative unital rings. Write $\mathsf M=\mathsf{Mod}(R)$ for module sheaves and $[P,N]$ for internal Hom, whose value at $U$ consists of module maps on the entire slice over $U$.

The category $\mathsf M$ is cocomplete and abelian. Here are the constructions needed below. Module presheaf kernels and cokernels are computed over each $R(U)$; restrictions descend to the quotient groups and commute with the scalar restrictions. Their coimage–image maps are pointwise isomorphisms. A plus section is locally one of its matching-family entries; for double plus, refine twice. Thus every associated group-sheaf section locally has a presheaf representative. Define its multiplication by $r\in R(U)$ on representatives $p_i$ as $r|_{U_i}p_i$. Equality of representatives is witnessed on further covers, where scalar multiplication retains it; these values agree and glue uniquely. The associativity, addition and unit laws hold on representatives and hence globally. Extension of a map to a sheaf stays linear by the same local calculation. Thus module sheafification is left adjoint to inclusion.

Group sheafification is left exact and is an additive left adjoint, hence preserves kernels and cokernels and is exact. Consequently module kernels remain pointwise sheaf kernels, while cokernels are sheafifications of the presheaf cokernels. Applying exact sheafification to the pointwise coimage–image isomorphism proves the abelian assertion. Arbitrary limits are pointwise: limits commute with the matching limits, so these presheaf limits stay sheaves and retain their universal property. A small colimit is sheafification of the pointwise module-presheaf colimit; the scalar restrictions are induced by its universal cocone. This proves existence and its universal property. These are the group and module forms of the sheafification prerequisite.

Write $\Gamma(M)=M(X)$. Evaluation at the unit gives inverse natural group isomorphisms

\operatorname{Hom}_R(R,M)\simeq\Gamma(M), \qquad f\longmapsto f_X(1). \tag{1.1}

The inverse sends $m\in M(X)$ to the map whose component at $U$ is $r\mapsto r\,m|_U$. The unique arrow $U\to X$ defines the restriction; scalar compatibility gives naturality and linearity. Conversely naturality and $R(U)$-linearity force this formula for any $f$. Both composites are identities, including on the zero ring. In particular $\Gamma$ is left exact, since kernels are pointwise.

Compactness of $P$ means that its Hom functor preserves small filtered colimits. Projectivity means that every map from $P$ lifts through every epimorphism in $\mathsf M$. These are global categorical properties.

2. What the existing local duality theorem supplies

Put $P^*=[P,R]$. The sheaf-duality prerequisite proves that $P$ is dualizable exactly when it is locally a retract of a finite free module, on every slice. This condition includes an explicit local identity

p=\sum_{i=1}^n\alpha_i(p)p_i, \qquad p_i\in P(U),\quad \alpha_i:P|_U\to R|_U. \tag{2.1}

It need not be locally free on an arbitrary ringed site. We retain the proved canonical isomorphisms

P^*\otimes_R N\xrightarrow{\sim}[P,N], \qquad \alpha\otimes n\longmapsto(p\mapsto\alpha(p)n), \qquad P\xrightarrow{\sim}P^{**}. \tag{2.2}

For example, the local inverse of the first isomorphism sends $f$ to $\sum_i\alpha_i\otimes f(p_i)$. The dual-basis identity and its image under any functional prove both inverse equations. The canonical map's inverse is unique, so these local inverses glue and respect all restrictions. Coevaluation is the section corresponding to $1_P$, locally $\sum_i p_i\otimes\alpha_i$; the prerequisite verifies both triangles and their converse.

Global maps are exactly internal-Hom sections at $X$, since the slice over the terminal object is the original site. Thus (2.2) gives the natural global identity

\operatorname{Hom}_R(P,N)\simeq\Gamma(P^*\otimes_R N). \tag{2.3}

Tensor by $P^*$ preserves all colimits by its internal-Hom adjunction. For dualizable $P$, it preserves all limits as well: internal Hom from $P^*$ is tensor by $P^{**}\cong P$, and internal Hom from $P$ is tensor by $P^*$. These give tensor by $P$ and by $P^*$ as adjoints in both orders. The Hom bijections and triangles are precisely the proved tensor–Hom maps transported through (2.2). In particular both tensor functors are exact. Formula (2.3) still includes $\Gamma$, whose exactness has not been assumed.

3. Finite covers and filtered module colimits

For this section assume that the topology is specified by finite covering families, stable under pullback and composition, as in the finite-cover prerequisite. An empty cover is permitted where specified. The matching condition for a covering $\{U_i\to U\}_{i\in I}$ is

M(U)\simeq \operatorname{Eq}\left(\prod_iM(U_i) \rightrightarrows\prod_{i,j}M(U_i\times_UU_j)\right). \tag{3.1}

Both products are finite. The comparison is $R(U)$-linear through the scalar restrictions.

Theorem 3.1. Every small filtered colimit of $R$-module sheaves is computed pointwise. Every dualizable $R$-module sheaf is compact.

Proof of the colimit assertion. Let $(M_a)$ be a small filtered diagram and put $F(U)=\operatorname{colim}_aM_a(U)$, as an $R(U)$-module. The module colimit supplies the restrictions: each stage restriction is linear through $R(U)\to R(V)$, and its compatible cocones induce a map to $F(V)$. Identities, composition and scalar compatibility hold on every stage coprojection and hence on the colimit. This makes $F$ a module presheaf.

The underlying group of an $R(U)$-module filtered colimit is its group filtered colimit, by the same compatible-cocone construction of the action. Filtered colimits of groups commute with the finite matching limit in (3.1). Concretely, a finite tuple of colimit elements has representatives at one common stage. Its finitely many overlap equalities hold at a further common stage, so the tuple comes from that stage's matching group. If two stage tuples have the same colimit values, their finitely many coordinate equalities hold at a common later stage, proving injectivity of the comparison. For a general filtered category, its common cocones and coequalization of parallel arrows give these same finite representative and equality tests. For an empty cover the matching group is zero and its comparison stays invertible.

Therefore the matching group of $F$ is the filtered colimit of the stage matching groups, with their component maps. The matching comparison of $F$ is the colimit of the invertible comparisons for $M_a$, so is invertible. It is a sheaf. Its presheaf cocone has the colimit universal property among sheaves by fullness, including linearity of the induced map. Hence the sheaf colimit is the stated pointwise one.

Proof of compactness. Pointwise computation makes $\Gamma$ preserve filtered colimits. For dualizable $P$, tensor by $P^*$ preserves them by §2. Applying (2.3) yields inverse natural comparisons

\operatorname{colim}_a\operatorname{Hom}_R(P,M_a) \simeq\operatorname{colim}_a\Gamma(P^*\otimes M_a) \simeq\Gamma\!\left(P^*\otimes\operatorname{colim}_aM_a\right) \simeq\operatorname{Hom}_R\!\left(P,\operatorname{colim}_aM_a\right). \tag{3.2}

Every comparison respects stage maps, so their composite is the canonical compactness map. This proves compactness, without a claim about lifting through epimorphisms. $\square$

Finite covers here are a hypothesis on the site presentation, including its cover overlaps. Quasi-compactness of the terminal object alone is not substituted for it. The more general section-comparison hypotheses appear in Stacks, Lemma 7.17.7.

4. The tensor unit controls global compact projectivity

We first prove a categorical test. In any locally small cocomplete abelian category, an object $P$ is compact projective if and only if its group-valued Hom functor preserves all small colimits. Necessity of the two properties follows from preserving filtered colimits and cokernels. Conversely write a coproduct as the filtered colimit of its finite partial coproducts; this identity follows from their compatible-cocone property. Compactness and finite biproduct preservation give preservation of all coproducts by Hom. Projectivity makes Hom surjective on epimorphisms. Together with its usual kernel preservation, this makes Hom exact and cokernel-preserving. Every colimit is the cokernel of a difference map between coproducts, so all colimits are preserved. This proof uses categorical coproducts, not pointwise presheaf sums.

Theorem 4.1. On any ringed site with the structures in §§1–2, the following are equivalent: every dualizable module sheaf is compact projective; the tensor unit $R$ is compact projective; $\Gamma$ is exact and preserves small filtered colimits. Under the finite-cover hypothesis of §3, these reduce to exactness of $\Gamma$.

Proof. The unit is dualizable, with unit identifications as its evaluation and coevaluation, so the first condition implies the second. If $R$ is compact projective, (1.1) and the categorical test make $\Gamma$ preserve all colimits. For any dualizable $P$, (2.3) expresses its Hom functor as $\Gamma$ after tensor by $P^*$. Both functors preserve all colimits; the test makes $P$ compact projective. This proves the reverse implication.

By (1.1), compactness of $R$ is exactly preservation of filtered colimits by $\Gamma$. Projectivity of $R$ is exactly surjectivity of $\Gamma$ on epimorphisms. Since $\Gamma$ is left exact, that surjectivity is equivalent to exactness. This proves the three-way equivalence. Theorem 3.1 supplies the filtered-colimit condition under finite covers, proving the last assertion. $\square$

5. A finite site as an actual span category

Let $X=\{u,v,w\}$ have opens $\varnothing,W=\{w\},U=\{u,w\},V=\{v,w\},X$. Use its ordinary open covers and the constant field sheaf $R=k_X$. All covers have finite subcovers, and finite covers specify the same sheaves. Nonempty open values of $R$ are $k$, with identity restrictions.

An $R$-module sheaf is exactly a span of vector spaces

A\xrightarrow f C\xleftarrow g B, \tag{5.1}

with $A=M(U),B=M(V),C=M(W)$. Its remaining values are $M(\varnothing)=0$ and $M(X)=A\times_C B$, with the two projections and the forced restrictions to $W$. Conversely these formulas give a sheaf: a cover of $U$, $V$ or $W$ includes that whole open, so imposes only its identity condition; a cover of $X$ either includes $X$ or contains both $U,V$, whose matching is exactly the fibre product. Additional cover members restrict the same already determined sections. The sheaf property for every cover follows.

A map of spans is a commuting triple of linear maps. Its map on the fibre product is forced and exists, so these operations give inverse constructions on all module morphisms as well. Kernels and cokernels in spans are pointwise vector-space kernels and cokernels, with induced arrows; the pointwise coimage–image comparisons are natural isomorphisms. Thus this is an abelian equivalence, and epimorphisms are surjective on the three displayed values. Surjectivity of $M(X)\to N(X)$ is a different requirement.

In particular $\Gamma$ is the span fibre-product functor, and the unit corresponds to $k\xrightarrow1 k\xleftarrow1 k$. Theorem 3.1 makes it compact, while the explicit map in Exercise 3 shows it is not projective. This finite site already separates local duality and global projectivity without a derived argument.

6. Four graded exercises with complete solutions

Exercise 1 (grade 1). On the one-point space take $R=k\times k$, with $k\ne0$ a field, and $P=eR$ for $e=(1,0)$. Compute its duality maps and show that it is compact projective and dualizable but is not locally free on this ringed space.

Solution. Module sheaves on the point are ordinary $R$-modules. Evaluation at $e$ identifies $P^*$ with $eR$; the inverse sends $a\in eR$ to $er\mapsto ar$. Evaluation is multiplication $eR\otimes_R eR\to R$, and coevaluation sends $1$ to $e\otimes e$. The first triangle takes $p\in eR$ to $e\otimes e\otimes p$ and then to $ep=p$; the other takes $a$ to $a\otimes e\otimes e$ and then to $ae=a$. Thus both are identities.

The module is a finite retract of $R$, so is compact projective by the ordinary criterion. Its two factor components are $(k,0)$. A free module $R^n$ has components $(k^n,k^n)$; neither $n=0$ nor any positive $n$ gives $(k,0)$. A cover of the point contains the point, so local freeness would force this impossible free description there. Local finite projective retracts are the correct duality condition without local-ring hypotheses.

Exercise 2 (grade 2). On a point with ring $\mathbb Z$, let $E=\begin{pmatrix}1&1\\0&0\end{pmatrix}$ and $P=\operatorname{im}E\subset\mathbb Z^2$. Compute the dual as row functionals, the coevaluation and the inverse of $P^*\otimes N\to[P,N]$ for every module $N$. Verify invariance of the coevaluation under a change of finite free frame.

Solution. The image is the column axis $P=\{(b,0):b\in\mathbb Z\}$. A functional on $P$ extended by precomposition with $E$ is the row $(a,a)$, because rows $\lambda$ satisfying $\lambda E=\lambda$ have exactly that form. Evaluation sends $(a,a)$ paired with $(b,0)$ to $ab$. Thus the covariant column image and contravariant row image differ, although both have rank one.

Put $p=(1,0)$ and $\lambda=(1,1)$. Their pairing is 1; coevaluation is $1\mapsto p\otimes\lambda$. The two triangle maps multiply by that pairing and are identities. The comparison takes $\lambda_a\otimes n$ to $(b,0)\mapsto abn$; its inverse takes a map $f$ to $\lambda\otimes f(p)$. Evaluation at $p$ and scalar linearity verify both inverse maps for every $N$, naturally.

For a free frame $(p_i,\alpha_i)$, replace $p_j$ by $p'_j=\sum_i p_iG_{ij}$ with invertible matrix $G$, and replace its dual by $\alpha'_j=\sum_l(G^{-1})_{jl}\alpha_l$. Then $\sum_jp'_j\otimes\alpha'_j=\sum_{i,l}(\sum_jG_{ij}(G^{-1})_{jl})p_i\otimes\alpha_l=\sum_i p_i\otimes\alpha_i.$ Evaluation is unchanged by the same matrix inverse equation. This explains why compatible local frames do not affect the canonical coevaluation.

Exercise 3 (grade 3). In the three-point span model, classify all dualizable sheaves. Construct an epimorphism whose global sections are not surjective, and a compact projective sheaf that is not dualizable.

Solution. If a sheaf $A\xrightarrow f C\xleftarrow g B$ is dualizable, its restriction to $U$ is a finite-free retract because every cover of $U$ includes $U$. A finite free module on $U$ is the identity arrow $k^n\to k^n$ on $U,W$. Any idempotent endomorphism has the same matrix on those two values by naturality. Its image restriction is therefore an isomorphism. Thus $f$ is invertible and $A,C$ are finite-dimensional; the same argument on $V$ makes $g$ invertible and $B$ finite-dimensional. Conversely, when both arrows are invertible and $C$ is finite-dimensional, a basis of $C$ transported back by $f^{-1},g^{-1}$ identifies the whole span with $k^n\xrightarrow1 k^n\xleftarrow1 k^n$. It is $R^n$, hence dualizable. This proves the full classification.

Now take $F=(k\xrightarrow1 k\xleftarrow0 k)$ and $G=(k\to0\leftarrow k)$. The maps identity on the outer groups and zero on the middle group form a commuting triple $F\to G$, surjective on each value and hence an epimorphism. But $F(X)=\{(0,b):b\in k\}$, while $G(X)=k^2$, and the induced map is inclusion of the second coordinate. The global section $(1,0)$ does not lift. By (1.1), the corresponding map $R\to G$ does not lift, so the dualizable compact unit $R$ is not projective. Every nonzero dualizable sheaf is $R^n$ with $n>0$; it contains $R$ as a retract and therefore is not projective either.

Finally let $J=(k\xrightarrow1 k\leftarrow0)$. A map $J\to M$ is exactly its arbitrary map $k\to M(U)$: its middle component is forced by $f_M$, and its other outer component is zero. Hence $\operatorname{Hom}(J,M)\cong M(U)$ naturally. This evaluation preserves pointwise epimorphisms of spans and filtered colimits, making $J$ compact projective. It is not dualizable because its arrow $0\to k$ is not invertible. Neither global compact projectivity nor local duality implies the other in this example.

Exercise 4 (grade 4). Let $X=\mathbb N_{\ge1}$ be discrete and $R=k_X$. Put $P_n=k^n$ at the point $n$. Prove that $P$ is dualizable and globally projective but not compact. Show that the global coevaluation cannot be represented by one finite tensor sum in $P(X)\otimes_{R(X)}P^*(X)$. Determine the corresponding two properties of $R$.

Solution. A module sheaf on this discrete space is exactly a family $(M_n)$, with $M(U)=\prod_{n\in U}M_n$; sections restrict by forgetting coordinates. Conversely the singleton cover of any open gives exactly this description and all module maps are componentwise linear maps. This is an equivalence with $\prod_n\mathsf{Vect}_k$, where colimits are pointwise in the family coordinates. It does not say that colimits are pointwise on all opens.

On each singleton $P_n$ is finite free, so the sheaf duality theorem makes $P$ dualizable. Its dual has point values $(k^n)^*$, and the two duality maps are the usual finite-dimensional maps at each point. Given any map from $P$ through a target epimorphism, lift the images of a basis at each coordinate and assemble the chosen coordinate lifts. Epimorphisms of these families are coordinatewise surjective, so this proves global projectivity, using choice for the small collection of lifts.

Let $P^F$ be the family equal to $P_n$ on a finite subset $F\subset X$ and zero elsewhere. The directed colimit over finite subsets is $P$, coordinate by coordinate. Its identity cannot factor through any $P^F$, because at a point outside $F$ such a factor would be zero rather than the identity of $k^n$. Thus $P$ is not compact. The same argument with $R_n=k$ proves the unit is not compact. It is projective because global sections are the product of the coordinate groups and send every coordinatewise epimorphism to a surjection by choosing lifts; they are left exact as usual.

The sheaf coevaluation is the tuple of identity tensors in $P_n\otimes(P_n)^*$. Tensor sheaves on a discrete space have these coordinate values, so this is a genuine global section. If it came from a finite sum $\sum_{i=1}^r p_i\otimes\alpha_i$ in the sectionwise tensor at $X$, its value at $n$ would induce an endomorphism of $k^n$ with image in the span of the $r$ vectors $p_i(n)$. Its rank would be at most $r$. The identity has rank $n$, giving a contradiction for $n>r$. Locally on each singleton a finite tensor expression exists, but its length has no uniform global bound. Sheafification supplies the global coevaluation. The infinite singleton cover has no finite refinement, so this example is outside the hypothesis of Theorem 3.1.