AI review · 21 October 2026

Q1 · Which strata do functions detect?

IC coordinates

Function values become summand information only after a change of basis.

1. 1. Invert the IC matrix

   Fix a finite lower set \(L\) containing the support. With \(b_\nu^{(n)}=\operatorname{tr}_n I_\nu\), form

   $$B^{(n)}_{\mu\nu}=b_\nu^{(n)}(x_\mu),\qquad B^{(n)}_{\mu\nu}=0\ (\mu\not\le\nu),\qquad B^{(n)}_{\nu\nu}=(-1)^{\delta(\nu)}.$$

   For the value vector \(f_\lambda^{(n)}\), its IC coordinates are \(m_\lambda^{(n)}=(B^{(n)})^{-1}f_\lambda^{(n)}\).

   Proof

   1. The support of \(I_\nu\) contains only strata below \(\nu\). On its own open stratum its only stalk is \(k[\delta(\nu)]\), with Frobenius 1.
   2. Order the finite labels compatibly with the closure order. The matrix is triangular with nonzero diagonal, so descending substitution recovers the coordinates.
   3. The IC value matrix is required input: raw function values do not supply the lower-stratum IC stalk traces. There is no unrestricted infinite matrix inversion here.
   Sources

2. 2. Read the multiplicity traces

   Example · two strata. On \(\mathbb A^1=\{0\}\sqcup\mathbb G_m\), let \(I_0=k_{\{0\}}\), \(I_u=k_{\mathbb A^1}[1]\). With rows \((0,u)\) and columns \((I_0,I_u)\),

   $$B=\begin{pmatrix}1&-1\\0&-1\end{pmatrix}.$$

   Object IC of the lineIC of the line ⊕ IC of the point

   Orbit values
   

   →

   Apply \(B^{-1}\)

   →

   IC coordinates
   

   This is a finite-type illustration, not a claim that these objects are the original Satake convolutions.

   Put \(E_{\nu,i}(\lambda)=\operatorname{Hom}(I_\nu,{}^pH^i(M_\lambda))\), with induced Frobenius \(A_{\nu,i}\). Then $$m_{\nu\lambda}^{(n)}=\sum_i(-1)^i\operatorname{Tr}(A_{\nu,i}^{,n}),\qquad m_{\nu\lambda}^{(n)}\ne0\Longrightarrow\nu\in\Sigma_\lambda.$$
   Proof

   1. The geometric decomposition gives semisimple perverse cohomology. Evaluation identifies it with \(\bigoplus_\nu I_\nu\otimes E_{\nu,i}\).
   2. Using the chosen Weil structures, Frobenius acts on a map \(u\) by \(\Phi_i\circ\operatorname{Fr}^*(u)\circ\Phi_\nu^{-1}\). Evaluation respects this action.
   3. Additivity on perverse truncation triangles gives the alternating trace formula; no Frobenius-equivariant splitting of the whole derived object is needed.
   4. If every \(E_{\nu,i}\) were zero, the coordinate would be zero. Contraposition certifies the summand label.
   Sources

Trace cancellation

One field can hide multiplicities; purity lets extension fields reveal them.

1. 3. Check for cancellation

   Frobenius power

   1

   Eigenvalue 1
   multiplicity 1

   Eigenvalue −1
   multiplicity 1

   Trace
   

   A nonzero multiplicity space need not have nonzero first trace. For \(A=\operatorname{diag}(1,-1)\),

   $$\operatorname{Tr}(A^n)=1+(-1)^n.$$
   Proof

   1. These eigenvalues define a pure weight-zero Weil sheaf on a point. Its geometric multiplicity is two, but the first trace is zero.
   2. This refutes an inference from purity and decomposition alone. It is not an example of a canonical Satake convolution with cancelling eigenvalues.
   3. A sufficient fixed-field no-cancellation condition is common shift parity and positive real eigenvalues in one complex embedding. Even shifts alone do not exclude eigenvalues \(1,-1\) within the same degree.
   Sources

2. 4. Vary the residue field

   If \(M_\lambda\) is pure of weight \(w\) and \(I_\nu\) pure of weight \(w_\nu\), then $$\Sigma_\lambda=\bigcup_{n\ge1}\{\nu:m_{\nu\lambda}^{(n)}\ne0\}.$$
   Proof

   1. Perverse cohomology in degree \(i\) has weight \(w+i\). The eigenvalues of \(A_{\nu,i}\) have absolute value \(q^{(w+i-w_\nu)/2}\), so the same eigenvalue cannot occur in different degrees.
   2. Collect distinct eigenvalues \(\alpha_1,\ldots,\alpha_r\). Then \(m^{(n)}=\sum_j c_j\alpha_j^n\), where each \(c_j\) is a nonzero signed positive integer.
   3. The matrix \((\alpha_j^n)_{1\le n,j\le r}\) is invertible: its determinant is, up to sign, \(\prod_j\alpha_j\prod_{j<k}(\alpha_k-\alpha_j)\). Not all of the first \(r\) traces can vanish.
   4. Use the corresponding matrix \(B^{(n)}\) at each extension. A bound \(d_\nu=\sum_i\dim E_{\nu,i}\) gives a witness with \(n\le d_\nu\). Without a supplied bound, a finite prefix of zeros does not certify absence.
   5. Without single-weight purity, \(I\oplus I[1]\) has zero trace for every \(n\). All-extension traces recover a Grothendieck class, not every arbitrary complex.
   Sources

A symmetric example

The function and sheaf labels coincide in a split torus example, under the specified rationality conditions.

1. 5. Compute a torus convolution

   Example. Let \(G=\mathbb G_m^2\), let the involution swap the factors, and \(H=\operatorname{diag}\mathbb G_m\). Identify \(X=G/H\simeq\mathbb G_m\) by the quotient \(a/b\). Its arc orbits have valuation \(j\in\mathbb Z\).

   $$V_{(a,b)}*I_0=I_{a-b},\qquad T_{(a,b)}*\varphi_0=\mathbf1_{v_t(x)=a-b}.$$
   Proof

   1. The affine Grassmannian of the split torus is the discrete lattice \(\mathbb Z^2\). A Satake object is the point sheaf at \((t^a,t^b)\).
   2. Its action translates the unit component of \(LX\) to valuation \(a-b\), with no boundary and no extra shift in the relative-dimension-zero convention.
   3. The double coset has one right coset, so the normalized function convolution is \(\varphi_0(t^{b-a}x)\). This gives the same indicator. Taking \((a,b)=(j,0)\) reaches every label.
   Sources

2. 6. Separate labels from spanning

   Summand detection and module generation ask different questions:

   $$\text{a nonzero IC coordinate}\Rightarrow\text{an occurring summand},\qquad\text{function generation}\Longleftrightarrow\text{the IC-coordinate columns span}.$$
   Proof

   1. A single column \((1,1)^T\) has no zero row but spans only a line. Detecting every row is therefore insufficient to prove function generation.
   2. Comparing geometric labels with rational orbit labels needs the stated rationality and connected-stabilizer hypotheses. For the action \(g\cdot x=g^2x\) of \(\mathbb G_m\) on \(\mathbb G_m\), each geometric unit orbit splits into two rational square classes over an odd finite field.
   3. A non-split norm torus gives a different obstruction: geometric cocharacters can reach all valuations, while rational norm valuations are even. These examples lie outside the split/connected hypotheses used above.
   Sources

Proposed edit. Beside the quoted sentence, specify “the coefficient of \(\operatorname{tr}\mathrm{IC}_\mu\), not the value at \(x_\mu\)”. Link the matrix calculation and state any no-cancellation hypothesis explicitly. Regenerate the trace PDF after approval: its closing status remarks predate the current HTML.

Solve run: function-detected-strata.

Q2 · Can an unbounded heart generate?

Finite constructions

Finite t-bounds survive thick operations; truncation triangles give the converse.

1. 1. Keep finite bounds

   The t-bounded subcategory is thick and contains the heart.

   $$X,Y\in\mathcal C^{[a,b]}\implies\operatorname{cofib}(X\to Y)\in\mathcal C^{[a-1,b]}.$$
   Proof

   1. The two halves of a t-structure are closed under extensions and existing retracts, by their Hom-orthogonality descriptions.
   2. A finite list of bounded objects has common finite bounds. A shift changes those bounds by a finite amount, and the displayed cone bound follows from the triangle.
   3. Every finite construction from heart objects is therefore t-bounded. This uses actual t-bounds; it does not assume that vanishing cohomology detects zero in a degenerate t-structure.
   Sources

2. 2. Build the truncation tower

   Start
   \(0\)

   →

   Successive quotient
   \(H_t^q(M)[-q]\)

   →

   Last truncation
   \(M\)

   For \(M\in\mathcal C^{[a,b]}\), use finitely many triangles:

   $$\tau_{\le q-1}M\longrightarrow\tau_{\le q}M\longrightarrow H_t^q(M)[-q]\xrightarrow{+1},\quad a\le q\le b.$$
   Proof

   1. Begin with \(\tau_{\le a-1}M=0\).
   2. The third term is a shifted heart object. If the first term is in the thick span, the second is too.
   3. After \(b-a+1\) steps, the second term is \(\tau_{\le b}M=M\). The triangles need not split.
   Sources

Infinite amplitude

One copy of the coefficient field in every nonnegative degree gives the counterexample.

1. 3. Compute the counterexample

   Last degree of the finite truncation:

   3

   degree012345678…finite0\(M\)kkkkkkkkk→

   Example. In the standard derived category \(D(k)\) of all vector spaces over a field, let

   $$M^q=\begin{cases}k&q\ge0,\\0&q<0,\end{cases}\quad d_M=0,\qquad M=\bigoplus_{q\ge0}k[-q].$$

   Every \(H^q(M)\) is finite dimensional, but the amplitude is unbounded. Hence \(M\notin\operatorname{Thick}(\operatorname{Vect}_k)\).

   Proof

   1. All differentials are zero, so \(H^q(M)=M^q\).
   2. No integer is an upper bound for the nonzero degrees.
   3. Steps 1–2 identify the thick span of the whole heart with the bounded part, excluding \(M\).
   Sources

2. 4. Allow infinite coproducts

   Over a field, $$\operatorname{Thick}(k)=D^b(\operatorname{Vect}^{\mathrm{fd}}_k)\subsetneq\operatorname{Thick}(\operatorname{Vect}_k)=D^b(\operatorname{Vect}_k)\subsetneq\operatorname{Loc}(k)=D(k).$$
   Proof

   1. Every complex of vector spaces splits, up to quasi-isomorphism, as the zero-differential complex formed by its cohomology: choose complements of boundaries inside cycles.
   2. Thus \(N\simeq\bigoplus_q H^q(N)[-q]\), and choosing bases builds it from arbitrary coproducts of shifts of \(k\). Arbitrary sums of vector spaces are exact, so the termwise sums represent derived coproducts.
   3. An infinite-dimensional vector space in degree zero separates the first two spans; \(M\) separates the last two. This positive localizing statement is for \(D(k)\), not for every possible t-structure.
   Sources

Proposed edit. Retain boundedness in H3, and link this example after the truncation argument. Keep cohomological boundedness separate from finiteness of orbit supports.

Solve run: unbounded-heart-generation.

Q3 · Why does the lowest stalk survive?

Lowest IC sheaf

\(k_Z\hookrightarrow\mathcal H^{-d}V\), and inverse image preserves this ordinary-sheaf injection.

1. 1. Read the lowest sheaf

   $$V\in D^{\ge-d}(Z),\qquad\mathcal H^{-d}V=j_*k_U,\qquad k_Z\hookrightarrow j_*k_U.$$

   This is a monomorphism of ordinary sheaves; it is not a decomposition of the intersection complex.

   Proof

   1. Deligne’s construction starts with \(k_U[d]\), extends by derived direct image across boundary strata of dimension \(s\), and truncates in ordinary degrees at \(-s-1\).
   2. Since \(s\le d-1\), no truncation removes degree \(-d\). Right-derived direct image preserves the lower bound and its lowest sheaf is ordinary direct image. This gives \(\mathcal H^{-d}V=j_*k_U\).
   3. Every neighborhood of a point of \(Z\) meets dense \(U\). A nonzero constant germ therefore stays nonzero on restriction to \(U\), proving the sheaf injection.
   Sources

2. 2. Restrict the injection

   $$r^*V\in D^{\ge-d}(F_y),\qquad k_{F_y}\hookrightarrow\mathcal H^{-d}(r^*V)=r^*\mathcal H^{-d}V.$$
   Proof

   1. Here \(r^*\) is inverse image of sheaves of vector spaces. On stalks it preserves the vector space and is exact.
   2. It therefore commutes with ordinary cohomology sheaves and preserves the monomorphism from step 1.
   3. No perverse t-exactness of the closed inclusion is used.
   Sources

Lowest hypercohomology

Only one entry can contribute to total degree \(-d\).

1. 3. Isolate the corner term

   \(q\backslash p\)0123\(-d+1\)••••\(-d\)\(H^0\)•••\(-d-1\)0000

   For \(A=r^*V\), $$E_2^{p,q}=H^p(F_y,\mathcal H^q A)\Longrightarrow H^{p+q}(F_y,A),\qquad H^{-d}(F_y,A)\cong\Gamma(F_y,\mathcal H^{-d}A).$$
   Proof

   1. Every nonzero entry has \(p\ge0\) and \(q\ge-d\). On \(p+q=-d\), only \((0,-d)\) remains.
   2. An incoming differential would start at a negative column. An outgoing \(d_r\), \(r\ge2\), would end below row \(-d\). Both vanish.
   3. The total-degree filtration has one graded piece, so the edge map is an isomorphism. Finally \(\Gamma\) is left exact and preserves the injection of step 2.
   Sources

Proper fibre

The constant section \(1\) survives in the convolution stalk.

1. 4. Apply proper base change

   $$F_y\subseteq Z\text{ is closed and projective},\qquad (V*\mathrm{IC}_0)_y\simeq R\Gamma_c(F_y,r^*V)=R\Gamma(F_y,r^*V).$$

   Properness is essential: on the nonproper fibre \(\mathbb A^1\), \(H_c^0=0\) although \(H^0=k\).

   Proof

   1. The contracted product \(Z\widetilde\times L^+X\) identifies with the incidence space of pairs \((gL^+G,y)\) satisfying \(g^{-1}y\in L^+X\).
   2. Embed affine \(X\) equivariantly into a finite-dimensional representation \(W\). On bounded \(Z\), the lattices \(gW(\mathcal O)\) have uniform bounds \(t^bW(\mathcal O)\subseteq gW(\mathcal O)\subseteq t^{-a}W(\mathcal O)\). The incidence condition is closed and is visible after quotienting by \(t^bW(\mathcal O)\). The universal quotient has fixed rank on \(Z\): the determinant character pairs to zero with coroots, so its lattice index is constant.
   3. At that finite model the incidence projection is proper, because its other factor is projective \(Z\). Ordinary proper base change yields the stalk formula; properness identifies compactly supported with ordinary fibre cohomology.
   4. The coefficient on the arc factor is the unshifted constant sheaf, so there is no extra shift in this convention.
   Sources

2. 5. Keep the constant section

   Example. For \(f:\mathbb P^1\to\mathrm{pt}\), \(V=k[1]\),

   $$H^i(Rf_*V)=\begin{cases}k&i=-1,1,\\0&\text{otherwise.}\end{cases}$$

   The class in degree \(-1\) is exactly \(1\in H^0(\mathbb P^1,k)\). For a nodal irreducible curve, the lowest IC stalk at the node is \(k^2\); the constant class maps diagonally \(a\mapsto(a,a)\). Thus an injection is the right claim even when the lowest stalk has extra local branches.

   If \(F_y\ne\varnothing\), then \(1\in H^0(F_y,k)\) is nonzero. Steps 1–4 inject it into the stalk.

   Proof

   1. Evaluate the constant section at any point of \(F_y\). It has value \(1\), so cannot vanish.
   2. For an empty fibre the entire stalk is zero. Nonempty fibres are exactly the points of \(Z\cdot L^+X\), proving the support assertion.
   3. At \(y=t^\lambda x_0\), the coset \(t^\lambda L^+G\) belongs to the fibre. A top-orbit summand still requires the separate support bound and decomposition inputs.
   Sources

Proposed edit. Expand the previous review’s q1-s4 to show the full chain of maps, its ordinary-cohomology conventions and the proper-fibre hypothesis. Link the finite-dimensional example. This does not certify the separate global comparison inputs.

Solve run: lowest-degree-stalk.

Q4 · Can the orbit count be weakened?

Objectwise finiteness

A finite construction can involve only finitely many nonzero orbit restrictions.

1. 1. Count the active orbits

   For \(M\in\mathcal D(Y)\), define \(S(M)=\{\lambda:a_\lambda^*M\ne0\}\). Then $$S(\Delta_\lambda)=\{\lambda\},\quad S(\operatorname{cofib}(A\to B))\subseteq S(A)\cup S(B).$$
   Proof

   1. Extension by zero has zero restriction to every other orbit.
   2. Orbit restrictions are exact, so a cone can be nonzero only where at least one of its two inputs is nonzero. Shifts leave the active set unchanged.
   3. A retract has active set contained in that of the original object. Hence objects with finite active set form a thick subcategory containing every orbit standard.
   4. Thus any object nonzero on infinitely many orbits is excluded from the thick span, regardless of how the closure order is ranked.
   Sources

2. 2. Choose an open active orbit

   For finite nonempty \(S=S(M)\), put \(Z=\bigcup_{\lambda\in S}\overline{O_\lambda}\), and choose maximal \(\mu\in S\). Then \(O_\mu\subseteq Z\) is open and \(M\simeq r_*r^*M\).
   Proof

   1. Outside \(Z\), all orbit restrictions vanish. Joint conservativity and localization give \(M\simeq r_*r^*M\).
   2. No other active orbit closure contains \(O_\mu\), by maximality and the frontier condition.
   3. The complement is the closed set \((\overline{O_\mu}\setminus O_\mu)\cup\bigcup_{\lambda\in S\setminus\{\mu\}}\overline{O_\lambda}\). Each term can contain infinitely many orbits.
   4. The recollement assumption must cover these closed subspaces. H2 on the original page covers only closed finite unions, and does not by itself supply this extension.
   Sources

Finite induction

Induct on \(|S(M)|\), using the enlarged recollement condition.

1. 3. Remove one active orbit

   Active set
   \(S(M)\), size \(n\)

   →

   Open term
   one orbit

   +

   Closed term
   \(n-1\) active orbits

   On \(Z\), with \(j:O_\mu\hookrightarrow Z\) and closed complement \(i:D\hookrightarrow Z\), use $$j_!j^*r^*M\longrightarrow r^*M\longrightarrow i_*i^*r^*M\xrightarrow{+1}.$$
   Proof

   1. For no active orbits, joint conservativity gives zero.
   2. The open restriction is in the thick span of the constant on its orbit, by the orbitwise generation assumption. Exact extension carries it into the span of \(\Delta_\mu\).
   3. The closed restriction has exactly \(|S(M)|-1\) active orbits. Apply the induction inside \(\mathcal D(D)\), then extend to \(Y\).
   4. The triangle gives \(M\). There are finitely many orbit-removal steps, even if the ambient orbit set or some closures are infinite.
   Sources

2. 4. Test infinitely many orbits

   Example. Let \(Y=\bigsqcup_{n\ge0}\mathrm{pt}\), with trivial action and standard \(\delta_n\) at the \(n\)-th point. Then

   $$\operatorname{Thick}\{\delta_n\}=\bigoplus_{n\ge0}\operatorname{Perf}(k).$$

   The constant family \((k,k,\ldots)\), when admitted into the ambient category, lies outside this thick span.

   Proof

   1. Finite-support bounded finite-dimensional families are finite sums of point objects and are generated by the corresponding standards.
   2. The constant family has infinitely many active orbits, so step 1 excludes it. The closure order is an antichain, with every rank zero: even this well-founded order does not solve the finiteness problem.
   3. In the presentable category \(\prod_n D(k)\), arbitrary coproducts do build every family from the point generators. That proves localizing generation and uses a different closure operation.
   4. This example has infinitely many total orbits, all closed. It is not an example with infinite orbit closures; allowing such closures in the theorem remains conditional on the enlarged recollement.
   Sources

Infinite orbit closure

A restricted finite-activity category can be generated even when one orbit closure contains infinitely many other orbits.

1. 5. Allow an infinite closure

   Example. Let \(H=\mathbb G_a^2\) act on \(\mathbb P^2\) by

   $$ (a,b)[x:y:z]=[x+az:y+bz:z].$$

   The affine chart \(U=\{z\ne0\}\simeq\mathbb A^2\) is one orbit; every point of the boundary line \(L=\{z=0\}\simeq\mathbb P^1\) is fixed. Thus \(\overline U=\mathbb P^2\) contains infinitely many orbits.

   Restrict to the bounded equivariant constructible category \(\mathcal D_{\mathrm{fa}}\) of objects nonzero at only finitely many points of \(L\). Then

   $$\mathcal D_{\mathrm{fa}}=\operatorname{Thick}\bigl(\{j_!k_U\}\cup\{i_{p*}k:p\in L\}\bigr).$$
   Proof

   1. On \(z=1\), the action is translation on \(\mathbb A^2\), hence transitive and free. On \(z=0\), the displayed action fixes every point.
   2. The orbit category on \(U\) is that of a point. At a fixed boundary point the stabilizer is connected and contractible \(H(\mathbb C)\); the bounded equivariant heart consists of finite-dimensional constants. Finite truncation towers give orbitwise generation.
   3. All cuts in the argument are finite-type algebraic cuts. Their restrictions and extension by zero preserve finite activity. The usual triangle \(j_!j^*F\to F\to i_*i^*F\) therefore stays in the restricted category.
   4. The first term is generated by \(j_!k_U\). The boundary term is supported on finitely many points, and is built from their point standards. This proves the equality.
   5. The unrestricted constant sheaf \(k_{\mathbb P^2}\) is nonzero at every boundary point and is excluded from this category and from the displayed thick span. The example uses a unipotent group and is not a new reductive loop-space case.
   Sources

Placid levels

A finite-level piece can include infinitely many geometric orbits.

1. 6. Distinguish pieces from orbits

   For \(\mathbb A^n\), a finite jet level has shell pieces and a tail \(T_M=t^M\mathcal O^n\). The constant tail sheaf has infinitely many active geometric orbits.

   The theorem does not assert that arbitrary closed cuts preserve a placid constructible category.

   Proof

   1. The tail contains every sufficiently deep nonzero shell and the zero vector. Its constant sheaf restricts nontrivially to all of them.
   2. Step 1 prevents building this object from finitely many orbit standards in any common category with the stated exact restrictions.
   3. Finite-level dévissage instead uses the tail as one admissible piece. It can generate by standards of pieces without proving generation by geometric orbit standards.
   4. The prior review’s Q4 already treats the affine example. The new result here identifies the precise objectwise condition and the additional recollement check needed for an extension.
   Sources

Proposed edit. Offer this conditional finite-activity formulation beside H1, with the enlarged scope of recollement stated. Do not remove finiteness from the current theorem until the intended category satisfies these replacement assumptions. Link the existing affine Q4 rather than duplicating it.

Solve run: induction-without-finite-orbits.

Q5 · Is extension by zero exact?

Two exactness questions

Triangles and ordinary short exact sequences are preserved; a perverse heart adds a separate condition.

1. 1. Check ordinary extension

   For an ordinary sheaf \(A\) and \(j:U\hookrightarrow Z\) open, $$ (j_!A)_z=\begin{cases}A_z&z\in U,\\0&z\notin U.\end{cases}$$ Thus \(j_!\) is exact on ordinary sheaves.
   Proof

   1. Apply extension by zero to a short exact sequence on \(U\).
   2. At a point in \(U\), the stalk sequence is the original exact one. At a point of the complement, all three stalks are zero.
   3. Exactness of sheaves is checked on stalks, proving the claim. This differs from direct image \(j_*\), which is generally only left exact on ordinary sheaves.
   Sources

2. 2. Pass to stable exactness

   On complexes, \(j_!\) preserves shifts and cofiber sequences.

   $$A\longrightarrow B\longrightarrow C\xrightarrow{+1}\quad\mapsto\quad j_!A\longrightarrow j_!B\longrightarrow j_!C\xrightarrow{+1}.$$
   Proof

   1. Applying the exact sheaf functor termwise preserves quasi-isomorphisms and mapping cones, so it induces a triangulated functor.
   2. In stable-category language, \(j_!\) is a left adjoint between stable categories. It preserves zero and finite pushouts, hence cofibers.
   3. On the page’s small constructible subcategories, H2 explicitly supplies preservation and the adjunction. A finite-type theorem alone does not verify that hypothesis for an arbitrary loop-space cut.
   Sources

Thick generation

The challenged inference needs stable exactness alone.

1. 3. Transport finite constructions

   On \(U\)
   finite truncation triangles build \(j^*M\)

   → \(j_!\) →

   On \(Z\)
   their images are triangles built from \(\Delta_U\)

   For an exact \(F:\mathcal A\to\mathcal B\), $$F(\operatorname{Thick}_{\mathcal A}(S))\subseteq\operatorname{Thick}_{\mathcal B}(F(S)).$$ Hence $$j^*M\in\operatorname{Thick}(k_U)\Longrightarrow j_!j^*M\in\operatorname{Thick}(j_!k_U).$$
   Proof

   1. The inverse image under \(F\) of \(\operatorname{Thick}(F(S))\) contains \(S\).
   2. It is closed under shifts and cones by exactness. Any functor preserves a retract diagram and its splitting identity, so it is closed under retracts too.
   3. Minimality of thick closure proves the inclusion. No claim that \(F\) preserves a chosen heart occurs in this argument.
   Sources

A stronger claim

Perverse t-exactness can fail even when the derived functor is exact.

1. 4. Test the perverse heart

   Table 2. The meanings of exactness in this question.

   Category or structure	What is preserved?	Open \(j_!\)
   Ordinary sheaf heart	Short exact sequences	Yes
   Stable derived category	Cofiber sequences	Yes; used by the proof
   Middle-perverse t-structure	Both t-bounds	Not for every open immersion

   Example. Let \(Z=\mathbb C^2\), \(U=Z\setminus\{0\}\), and \(P=k_U[2]\), which is perverse on the smooth surface. For \(i:\{0\}\hookrightarrow Z\),

   $$H^{-1}(i^!j_!P)=k\ne0.$$

   Thus \(j_!P\) fails the perverse cosupport condition at \(0\).

   Proof

   1. Extension by zero gives \(j_!P\to k_Z[2]\to i_*k[2]\xrightarrow{+1}\).
   2. A real four-ball relative to its punctured ball has cohomology \(k\) only in degree 4. Consequently \(i^!k_Z[2]=k[-2]\), while \(i^!i_*k[2]=k[2]\).
   3. Apply \(i^!\). The long exact sequence for \(i^!j_!P\to k[-2]\to k[2]\) gives \(H^{-1}(i^!j_!P)\simeq k\).
   4. At the point stratum, middle-perverse cosupport requires no negative costalk cohomology. This proves failure of perverse t-exactness, fully compatible with steps 1–3.
   Sources

Proposed edit. At the challenged line, write: “Here exact means preserving cofiber sequences in the stable category; no perverse t-exactness is used. Hence (T2) applies.”

Solve run: open-extension-exactness.