Draft lesson.

Closed support and localization for locally module stacks

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

A twisted module may have no preferred global module sheaf representing it. Nevertheless, vanishing on an open set and having support on its closed complement are intrinsic conditions. The local support constructions must match through the Morita equivalences before their derived triangles become global. We make those comparisons explicit and distinguish a canonical pullback map from an invertible one.

Let $X$ be a topological space, $\mathcal R$ a sheaf of commutative rings, and $\mathfrak M$ an $\mathcal R$-linear stack locally equivalent to left module sheaves over unital, possibly noncommutative, algebras $\mathcal A_a$. Fix universes for all coverings, sums and Hom sets. For $f\colon Y\to X$, write $\mathfrak M_Y=f^\boxtimes\mathfrak M$ for the relative pullback stack and $f^{-1}$ for its exact functor on objects. Local coefficients are $f^{-1}\mathcal A_a$; we do not replace them by quotient rings. Put $Z\subset X$ closed, $U=X\setminus Z$, with inclusions $i\colon Z\to X$, $j\colon U\to X$.

Relative pullback of locally module stacks constructs these pullbacks, their direct-image adjunctions, extension by zero, and coherent Morita comparisons. Injective objects in locally module stacks proves the global abelian structure, enough injectives and exact inverse image. Injective replacements and right derived functors, Sections 1–3, and Derived morphisms through quasi-isomorphism roofs, Section 5, supply the bounded-below models and their Hom tests. Cone signs follow Chain homotopies, mapping cones and triangles, Sections 2–3. The quotient Hom argument is Orthogonal localization and the morphisms between pieces, Sections 1–2.

Throughout, $D^+$ means cohomology bounded below. All assertions concern these derived categories and retain their actual adjunction maps.

1. Closed pushforward and supported objects

Proposition 1.1. The functor $i_*\colon \mathfrak M_Z(Z)\to\mathfrak M(X)$ is exact and fully faithful. Its essential image consists precisely of objects whose local module stalks vanish outside $Z$. Its exact left adjoint $i^{-1}$ satisfies $i^{-1}i_*=1$, with the canonical counit identification.

Proof. In a module chart, direct image of $N$ on $Z$ has sections $N(V\cap Z)$ over $V$. At $z\in Z$, the intersections of neighborhoods in $X$ with $Z$ are cofinal among neighborhoods in $Z$. Its stalk is therefore $N_z$. Outside $Z$, choose a neighborhood disjoint from the closed set: the stalk is zero. These formulas use the inverse-image algebra coefficients, so they are formulas for the module stalks as well. They prove exactness and the counit identification stalkwise. The adjunction then proves full faithfulness.

If $F$ vanishes outside $Z$, its unit $F\to i_*i^{-1}F$ is an isomorphism at every chart stalk: it is the identity germ on $Z$, and both germs vanish off $Z$. Local stalks detect isomorphisms, giving the asserted essential image.

These local assertions are compatible on overlaps. A Morita equivalence commutes with restriction and inverse image by the coherent pullback construction. Its comparison with direct image is the unique comparison of right adjoints of those restriction functors. Both composites on a triple overlap have the same adjunction bijections, so they agree. The stalk comparisons and the unit and counit thus descend through the stack, proving the global assertions. $\square$

The condition is independent of a chart: an equivalence takes zero modules to zero modules. We refer to these objects as supported on $Z$, without imposing a closedness assertion on the support of an arbitrary object.

2. The intrinsic support right adjoint

Define an ambient object by the kernel $S_Z^0F=\ker\bigl(F\longrightarrow j_*j^{-1}F\bigr), \qquad i^!F=i^{-1}S_Z^0F. \tag{2.1}$ In a module chart, $S_Z^0F(V)$ is the module of sections of $F(V)$ whose restriction to $V\cap U$ is zero. A section has this property exactly when all its nonzero germs lie in $V\cap Z$: zero germs vanish on neighborhoods, and the sheaf axiom glues that vanishing. It is a module subsheaf and is zero off $Z$.

Proposition 2.1. There are natural identifications $i_*i^!F=S_Z^0F$ and a full adjunction $i_*\dashv i^!$. Both $i^!$ and $S_Z^0$ are additive and left exact. Their constructions, units and counits commute with the Morita overlap equivalences.

Proof. The first identification is Proposition 1.1 on the supported kernel. Any map $i_*N\to F$ has image zero off $Z$; its composite with $F\to j_*j^{-1}F$ is zero. This can also be checked by adjunction, since $j^{-1}i_*N=0$. The kernel property gives a unique factorization through $S_Z^0F$, and full faithfulness of $i_*$ identifies this with a unique map $N\to i^!F$. Conversely push that map forward and include the kernel in $F$. These operations are inverse and natural in both variables. Identity maps under this bijection give the unit and counit; naturality proves their triangle identities.

The supported-subsection description proves left exactness directly in each chart: a section of the middle object killed by a map lifts uniquely through a monomorphism, and the lift vanishes on $U$ because its image does. Additivity is immediate. Equivalently $i^!$, as an additive right adjoint, preserves kernels, and $i_*$ is exact.

For overlap compatibility, the Morita equivalences are exact and commute with $j^{-1},j_*$ and their units by uniqueness of the adjunction comparison. They therefore preserve the actual kernel (2.1) and its inclusion. Applying the coherent closed pullback gives the comparison for $i^!$. Alternatively it is the unique right-adjoint comparison of the already coherent $i_*$. All comparisons respect the displayed Hom bijection, so their triple-overlap and unit/counit equations agree. $\square$

The support object is defined by these morphisms. It is not obtained by choosing one module chart globally.

3. Injectives and the restriction epimorphism

Lemma 3.1. The functors $i_*,i^!,j_*,j^{-1}$ preserve injective objects. An injective $I\in\mathfrak M(X)$ gives a short exact sequence $0\longrightarrow i_*i^!I\longrightarrow I \xrightarrow{r_I} j_*j^{-1}I\longrightarrow0. \tag{3.1}$

Proof. An additive right adjoint of an exact left adjoint preserves injectives: transpose a map from a subobject, apply the exact left adjoint to that monomorphism, extend the transposed map into the injective object, and transpose back. Apply this to $i^{-1}\dashv i_*$, $i_*\dashv i^!$, $j^{-1}\dashv j_*$, and $j_!\dashv j^{-1}$. The left adjoints in question are exact by the stalk formulas and extension by zero.

The kernel in (3.1) is already (2.1). We prove the last map epic, rather than assume restriction is onto for every object. Restriction of $I$ to a chart is injective by the same open adjunction; transporting through its module equivalence preserves injectivity. For opens $V\subset W$ in that chart, the extension by zero $\mathcal A[V]$ of the regular module on $V$ represents sections on $V$. The natural map $\mathcal A[V]\to\mathcal A[W]$ is monic: its stalk is identity on $V$ and the inclusion from zero elsewhere. Extending maps into the injective chart module therefore makes its restriction map from $W$ to $V$ surjective.

Take $V=W\cap U$. Then $I(W)\to(j_*j^{-1}I)(W)$ is surjective in this chart. This proves epimorphy on all local stalks and hence in the global category. It also proves that all three objects in (3.1) are injective, by the first paragraph. Its termwise splitting need not be a splitting of complexes when applied to a complex. $\square$

4. Derive support and identify its image

For $F\in D^+(\mathfrak M(X))$, choose a bounded-below injective replacement $F\to I$. Define $Ri^!F=i^!I,\qquad S_ZF=i_*Ri^!F. \tag{4.1}$ Both $i^!I$ and $i_*i^!I$ are bounded-below injective complexes by Lemma 3.1.

Theorem 4.1. There is a natural derived adjunction $i_*\dashv Ri^!$, with invertible unit on $D^+(\mathfrak M_Z(Z))$. The exact functor $i_*$ gives an equivalence onto $D_Z^+(\mathfrak M(X))= \{F\mid H^nF\text{ vanishes on }U\text{ for every }n\}. \tag{4.2}$ The functor $S_Z$ is right adjoint to the inclusion of (4.2). Its counit $S_ZF\to F$ is universal among maps from objects of that subcategory, and is invertible for supported $F$.

Proof. For a complex representing a bounded-below derived object $N$ on $Z$, the degreewise adjunction respects the chain equations and the equations for homotopies. Since $I$ and $i^!I$ are K-injective, their Hom tests give natural bijections $\operatorname{Hom}_{D_X}(i_*N,F) =\operatorname{Hom}_{K_X}(i_*N,I) =\operatorname{Hom}_{K_Z}(N,i^!I) =\operatorname{Hom}_{D_Z}(N,Ri^!F). \tag{4.3}$ They are natural for all derived arrows: the K-injective test removes incoming quasi-isomorphism roofs uniquely. The resulting units and counits have the ordinary degreewise formulas on replacements and obey the triangle identities by this natural bijection.

Resolve $N$ on $Z$ by an injective complex $J$. Then $i_*J$ is an injective model on $X$, and $i^!i_*J=J$, with its ordinary unit identification. Thus the derived unit is invertible and $i_*$ is fully faithful. Its image has supported cohomology by exactness. Conversely for an object satisfying (4.2), the exact inverse image and exact closed pushforward give the map $F\to i_*i^{-1}F$. On cohomology it is the isomorphism of Proposition 1.1. Hence it is a quasi-isomorphism and supplies essential surjectivity. This proves the equivalence, not merely an identification of a kernel with an unspecified intrinsic derived category.

Transport (4.3) through that equivalence. It is exactly the right adjunction for inclusion of the supported subcategory, with the stated counit. A fully faithful left adjoint has invertible unit; its counit on its essential image is therefore invertible by the triangles. The category (4.2) is triangulated and retract closed, since exact restriction detects its zero cohomology and the long exact sequence and retract tests preserve this condition. $\square$

5. The natural localization triangle, with its sign

Theorem 5.1. Every $F\in D^+(\mathfrak M(X))$ has a natural distinguished triangle $S_ZF\longrightarrow F\longrightarrow Rj_*j^{-1}F \longrightarrow S_ZF[1]. \tag{5.1}$ The first arrow is the support counit, and the second is the derived open adjunction unit.

Proof. For the injective model $I$, Lemma 3.1 gives a termwise short exact sequence of complexes $0\longrightarrow A=i_*i^!I\longrightarrow I \xrightarrow r B=j_*j^{-1}I\longrightarrow0. \tag{5.2}$ Restriction preserves injectives, so $B$ computes $Rj_*j^{-1}F$. Define the fibre complex $\operatorname{Fib}(r)^n=I^n\oplus B^{n-1},\qquad d(x,y)=(d_Ix,rx-d_By). \tag{5.3}$ Its square is zero by the chain equation for $r$. It has maps $p(x,y)=x$ to $I$ and $c(y)=(0,y)$ from $B$ to $\operatorname{Fib}(r)[1]$. The chain isomorphism to $C_r[-1]$ sends $(x,y)$ to $(y,-x)$. The inverse rotation of the established cone triangle has first map $-p_r[-1]$, so this isomorphism gives exactly the distinguished triangle $\operatorname{Fib}(r)\xrightarrow p I\xrightarrow r B\xrightarrow c\operatorname{Fib}(r)[1]$.

The map $A\to\operatorname{Fib}(r)$, $a\mapsto(a,0)$, is a quasi-isomorphism. Here is the full local check. A fibre cycle $(x,y)$ satisfies $d_Ix=0$ and $rx=d_By$. Locally lift $y$ through the epimorphism $r$ to $z\in I^{n-1}$. Subtract the boundary $d(z,0)=(d_Iz,y)$. The remaining cycle $(x-d_Iz,0)$ comes from $A$. Conversely if $(a,0)=d(z,w)$, then $rz=d_Bw$. Locally lift $w$ to $v\in I^{n-2}$; now $z-d_Iv$ lies in $A$ and its differential is $a$. These arguments prove surjectivity and injectivity on every local cohomology stalk. The comparison respects the map to $I$, so the triangle becomes (5.1) with its specified first two arrows.

Naturality includes independence of the model. Maps between injective replacements are uniquely determined in the homotopy category. The functors and their restriction units take such a map to a strict square between the rows (5.2); the fibre map is its two diagonal components. A homotopy $h$ on $I$, with induced homotopy $k$ on $B$, acts on the fibre by $(x,y)\mapsto(hx,-ky)$. Its differential expression is the required difference of fibre maps, since $r'h=kr$. Thus homotopic representatives give the same triangle morphism. Replacement comparisons are homotopy equivalences and yield the unique compatible derived comparisons. This proves naturality on all derived arrows and preserves the actual connecting map. $\square$

Equation (5.3) fixes a useful coordinate convention: for a degree-zero map $M\to N$, the fibre has differential $+r$ from degree zero to degree one, and the displayed third arrow induces the positive quotient to its cokernel. Replacing that fibre differential by $-r$ requires the matching change to the coordinate formula for the third arrow.

6. The open complement is the actual Verdier quotient

Write $D_X=D^+(\mathfrak M(X))$, $D_U=D^+(\mathfrak M_U(U))$, and let $q\colon D_X\to D_X/D_Z^+$ be the Verdier quotient.

Theorem 6.1. Exact restriction and the derived direct image give an adjunction $j^{-1}\dashv Rj_*$ with invertible counit. There are exact equivalences $D_X/D_Z^+\simeq D_U\simeq(D_Z^+)^\perp\subset D_X, \tag{6.1}$ where the last embedding is $Rj_*$. In particular the canonical quotient Hom formula is $\operatorname{Hom}_{D_X/D_Z^+}(qF,qG[n]) \simeq\operatorname{Hom}_{D_X}(F,Rj_*j^{-1}G[n]) \simeq\operatorname{Hom}_{D_U}(j^{-1}F,j^{-1}G[n]). \tag{6.2}$

Proof. Exact restriction has the right adjoint $j_*$, which preserves injectives. Chain adjunction and the K-injective test give the derived adjunction exactly as in (4.3). On an injective replacement $J$ on $U$, its counit is the identity identification $j^{-1}j_*J=J$. Therefore $Rj_*$ is fully faithful.

For $T\in D_Z^+$, adjunction gives $\operatorname{Hom}(T,Rj_*N[n])=\operatorname{Hom}(j^{-1}T,N[n])=0$ for every shift. Thus every section object belongs to the right orthogonal. Triangle (5.1) is the orthogonal-decomposition triangle with supported first vertex and such a section as third vertex. The full roof and Hom proofs of the earlier orthogonal-localization theorem apply: the subcategory is thick by Theorem 4.1. They give its right-orthogonal equivalence with the Verdier quotient, and identify the quotient object of $G$ with that of its section. Derived adjunction then gives the second isomorphism in (6.2).

Every right-orthogonal object $F$ has zero support coreflection: its maps from all supported objects vanish, and the support adjunction tests $S_ZF$ against itself. The identity is zero, so $S_ZF=0$. Triangle (5.1) makes its unit $F\to Rj_*j^{-1}F$ invertible. Hence the image of the section is exactly the right orthogonal, completing (6.1). All identifications use the specified unit and counit and are natural on quotient denominators by the established roof argument. $\square$

7. Excision, successive supports and pullback

Proposition 7.1. The following comparisons are natural and coherent with further restrictions or inclusions.

For an open $v\colon V\hookrightarrow X$ containing $Z$, $Ri_X^!F\simeq Ri_V^!v^{-1}F \tag{7.1}$ as objects on $Z$. For closed inclusions $Z_0\xrightarrow{k}Z_1\xrightarrow{i_1}X$, with composite $i_0$, $Ri_0^!\simeq Rk^!Ri_1^!,\qquad S_{Z_0}S_{Z_1}\simeq S_{Z_0}. \tag{7.2}$ For any continuous $f\colon Y\to X$ and $Z'=f^{-1}Z$, there is a canonical map $f^{-1}S_ZF\longrightarrow S_{Z'}f^{-1}F. \tag{7.3}$ It commutes with the support counits and composes for successive pullbacks, but it need not be invertible.

Proof. Restrict (2.1) to $V$. The local supported-section description identifies its kernel with the support kernel on $V$, and their restrictions to $Z$ agree. Since $v^{-1}$ preserves injectives, applying this to an injective replacement proves (7.1). The comparison is literally restriction of sections with the same germs, so successive excisions give the same map.

For (7.2), the direct images compose as $(i_0)_*=(i_1)_*k_*$, with their coherent pullback identifications. Uniqueness of right adjoints gives $i_0^!=k^!i_1^!$ and its units and counits. Both right adjoints preserve injectives, so the degreewise equality on one injective replacement computes the derived composite as well. For three closed inclusions, both comparisons transpose the same composite Hom bijection, proving associativity. The corresponding composite counit is $\epsilon^{i_1}\circ(i_1)_*\epsilon^k Ri_1^!$, which is the counit of the composite inclusion.

More intrinsically, every supported object for $Z_0$ is supported for $Z_1$. Applying the two support adjunctions shows that $S_{Z_0}S_{Z_1}F$ and $S_{Z_0}F$ represent the same Hom functor on $D_{Z_0}^+$, with the same composite counit to $F$. This proves the second comparison and its compatibility with the first.

Exact inverse image takes an object supported on $Z$ to one supported on $Z'$: its cohomology stalks are the original stalks at $f(y)$ in the local inverse-image charts. Pull back the support counit to obtain $f^{-1}S_ZF\to f^{-1}F$. Its source is supported on $Z'$, so the right support adjunction factors it uniquely through $S_{Z'}f^{-1}F$. This is (7.3). The same uniqueness proves naturality, the counit square, and composition for two pullbacks using their coherent inverse-image identification. It proves no invertibility claim; Exercise 8.2 exhibits failure with unchanged coefficients. $\square$

8. Four graded exercises with complete solutions

Exercise 8.1 — Support commutes with an actual matrix Morita equivalence (intermediate)

Let $A=\operatorname{Mat}_m(k)$, $e=E_{11}$, and let $X=\{c,\eta\}$ have opens $\varnothing,\{\eta\},X$. A sheaf is an arrow $r\colon M\to N$ of left $A$-modules. For $Z=\{c\}$, compute its derived support and the open quotient Hom groups in matrix and $k$-module coordinates. Verify the comparison using the actual Morita maps.

Solution. Multiplication and matrix units give inverse maps $Ae\otimes_k eM\longrightarrow M,\quad ae\otimes u\longmapsto au, \qquad x\longmapsto\sum_{a=1}^m E_{a1}\otimes E_{1a}x. \tag{8.1}$ Their composite on $M$ is identity since $\sum_a E_{a1}E_{1a}=1$. On $Ae\otimes eM$, write the column $ae$ as $\sum_a a_{a1}E_{a1}$; the inverse composite becomes that same column tensor $u$. Thus these are the full Morita equivalence, natural on module maps. Applying $e$ is exact because it is the image of an idempotent acting on the underlying vector space, and the inverse is a finite column tensor.

Here $j_*N=(N\xrightarrow{1}N)$ is exact. The localization fibre on $c$ is the complex $[M\xrightarrow r N]$ in degrees $0,1$, and on $\eta$ it is contractible. Consequently $Ri^!F$ has $H^0=\ker r$, $H^1=\operatorname{coker}r$ and no other cohomology. Its differential is $+r$, and the connecting map takes $n$ to its positive cokernel class, by (5.3). Applying $e$ gives precisely $[eM\xrightarrow{er}eN]$; applying the maps (8.1) reconstructs the whole support complex and all its structural maps. This is an explicit instance of the overlap comparisons, including the boundary.

For arrows of bounded-below complexes $F=(M\to N)$, $G=(P\to Q)$, (6.2) gives quotient Hom groups $\operatorname{Hom}_{D^+(A)}(N,Q[n])$. Applying (8.1) identifies them with $\operatorname{Hom}_{D^+(k)}(eN,eQ[n])$: the exact inverse equivalences apply to complexes and their quasi-isomorphisms and hence to their derived morphisms. For degree-zero modules these are $\operatorname{Hom}_A(N,Q)$ in degree zero and zero in other degrees. Indeed every $k$-space is injective by extension of a basis, so its degree-zero complex computes all those groups. $\square$

Exercise 8.2 — Pullback to the closed point loses the support comparison (introductory)

On the same two-point space, use constant coefficient ring $k$ and the sheaf $F=(k\xrightarrow{1}k)$. Let $f\colon \{c\}\to X$ be the closed-point inclusion with unchanged coefficients. Compute both sides of (7.3). Explain why the failure also persists in matrix Morita coordinates.

Solution. The support fibre is $[k\xrightarrow1k]$, which is contractible: the identity from its degree-one term to degree zero is a contracting homotopy. Thus $S_ZF=0$. Its inverse image on the point is zero. But $f^{-1}F$ is the closed stalk $k$, and $Z'=\{c\}$ is the whole one-point space. Its support functor is identity, so the target of (7.3) is $k[0]$. The comparison is $0\to k[0]$, which is not an isomorphism. Exactness of inverse image has not made it commute with the support right adjoint.

In matrix coordinates take the column $Ae$ and its identity arrow. The same fibre is contractible while the pulled-back whole-support object is $Ae[0]\ne0$. Applying $e$ recovers the previous computation. This failure uses neither a changed coefficient ring nor a nonflat coefficient map. $\square$

Exercise 8.3 — A branching boundary and a failure of right exactness (intermediate)

Let $X=\{c,u,v\}$ have open subsets given by the upper sets of the order $c<u$, $c<v$. A sheaf of $k$-spaces is a pair of arrows $a\colon M\to N$, $b\colon M\to P$. For $Z=\{c\}$, compute derived support, its boundary, and a short exact sequence proving that ordinary support is not right exact.

Solution. The complement consists of the two discrete points. Its direct image has diagram $N\oplus P\to N,P$, with the two projections. Finite products are exact, so $j_*$ is exact here. The adjunction unit at $c$ is $r=(a,b)\colon M\to N\oplus P$. Equation (5.3) computes the closed support by $[M\xrightarrow{(a,b)}N\oplus P], \tag{8.2}$ in degrees $0,1$. The support cohomology is the kernel and cokernel of this actual map, and the boundary is its positive quotient. For the constant diagram $M=N=P=k$, $a=b=1$, the kernel is zero and the cokernel is $k$, with coordinate $(n,p)\mapsto p-n$. Thus the degree-one supported boundary is $\delta(n,p)=p-n$ in these coordinates.

Write a diagram as $(M;N,P)$. The objectwise exact sequence $0\to(0;k,k)\to(k;k,k)\to(k;0,0)\to0 \tag{8.3}$ has identity arrows in its middle object, and zero arrows in the other two. Its ordinary support kernels are $0,0,k$, respectively. The last induced map is not epic. The degree-one group in (8.2) retains precisely the information omitted by the ordinary kernel. All these computations commute with the matrix equivalence (8.1). $\square$

Exercise 8.4 — Successive closed supports and a contractible double-fibre summand (advanced)

For the upper topology of $0<1<2$, let a sheaf be $M_0\xrightarrow aM_1\xrightarrow bM_2$. Put $Z_1=\{0,1\}$, $Z_0=\{0\}$. Compute $Ri_0^!F$ directly and through $Rk^!Ri_1^!F$, displaying the comparison and homotopy. Then use $a\colon k\to k^2$, $a(t)=(t,0)$, $b\colon k^2\to k$, $b(x,y)=y$, to compute all support cohomology and the map between the two ambient support objects.

Solution. Direct images from the final point and from the final two-point chain are exact: on any upper neighborhood their section functor is evaluation at its initial point, or a finite product, or zero. The direct closed-point support is $C_a=[M_0\xrightarrow aM_1]$ in degrees $0,1$.

The support on $Z_1$ is the arrow of complexes $K_0=[M_0\xrightarrow{ba}M_2] \xrightarrow{(a,1)}K_1=[M_1\xrightarrow bM_2]. \tag{8.4}$ Its further closed-point support is the fibre of this arrow. Its three terms and differentials are $T^0=M_0,\quad T^1=M_2\oplus M_1,\quad T^2=M_2, \qquad d^0x=(bax,ax),\quad d^1(y,z)=y-bz. \tag{8.5}$ The composite is zero. Define $\alpha\colon C_a\to T$ by identity in degree zero and $z\mapsto(bz,z)$ in degree one; define $\beta\colon T\to C_a$ by identity in degree zero, $(y,z)\mapsto z$ in degree one and zero in degree two. They are chain maps and $\beta\alpha=1$. Let $h\colon T^2\to T^1$ be $y\mapsto(y,0)$, with its other components zero. Then $dh+hd$ is $(y,z)\mapsto(y-bz,0)$ in degree one and identity in degree two, and zero in degree zero. It equals $1-\alpha\beta$. This proves the full homotopy comparison, including the contractible summand, rather than only equality of cohomology counts. At the closed stalk the counit to $K_0$ is projection onto its fibre's first component; after $\alpha$ it is identity on $M_0$ and $b$ on $M_1$.

For the ambient diagram morphism, retain the contractible open-stalk complex $\operatorname{Fib}(1_{K_1})$. The restriction from $T=\operatorname{Fib}(K_0\to K_1)$ to it sends $(x,y)$ to $((a,1)x,y)$. Projection onto $K_0,K_1$ is now a strict map of diagram complexes. The map from this diagram to $(T\to0)$, identity at the closed stalk and zero at the open stalk, is a quasi-isomorphism. Inverting it gives the actual support counit roof. Thus the projected stalk formula comes with the required derived diagram map, without assuming its restriction square is strictly zero. These are the nested-support maps of (7.2).

For the specified matrices, $ba=0$. Hence $Ri_0^!F$ has zero degree-zero cohomology and degree-one cohomology $k^2/a(k)\simeq k$, identified by the second coordinate. In (8.4), $K_0$ has $k$ in degrees zero and one with zero differential. The complex $K_1$ has degree-zero cohomology $\ker b=k(1,0)$, and zero degree-one cohomology. The arrow on degree-zero cohomology is $a\colon k\to k(1,0)$, an isomorphism. The degree-one arrow is $k\to0$. Thus the ambient $S_{Z_1}F$ has degree-zero diagram $(k\xrightarrow{\sim}k\to0)$ and degree-one diagram $(k\to0\to0)$, and no other cohomology. The map $S_{Z_0}F\to S_{Z_1}F$ is zero on degree-zero cohomology and the identity on the displayed degree-one $k$, since its degree-one component is induced by $b$. The homotopy above verifies all cancellation and signs in the successive support computation. $\square$

9. Freely accessible further reading

The Stacks Project, Closed immersions and abelian sheaves, describes the ordinary supported-sheaf equivalence and right adjoint. Its Cohomology with support in a closed subset introduces the bounded-below derived support construction. These supply ordinary comparisons; the Morita-stack transport, derived quotient and explicit finite-space interfaces have their proofs above and in the linked prerequisites.

The existing programme lesson Sections with support and the localization triangle, Sections 1–3, treats ordinary module sheaves, including noncommutative coefficients and unbounded complexes. It also discusses changes of closed coefficients. Those constructions are complementary readings; this lesson keeps inverse-image coefficients and the coherently glued stack setting throughout.