Draft lesson.

Derived direct images and base change on finite posets

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

The value of a direct image at a point uses the inverse image of its smallest open neighbourhood. It need not use the literal point fibre. On finite posets, signed chain complexes compute those values and all maps between them. The same complexes make base change explicit: a monotone map pulls back cochains, and a collapsed chain contributes zero. Some squares give isomorphisms, while a closed-point square can fail even in degree zero.

Use Finite-poset sheaves and canonical homological resolutions, §§1–5, for the full sheaf/diagram comparison and the normalized injective resolution. Its Exercises 3–4 provide the cycle and second-cohomology computations used below. Injective objects in locally module stacks, §§6–7, proves the full bounded derived adjunction and coherent composition. Compare the explicit right Kan maps in Exactness through comma categories and small Kan values, §7. Relative pullback of locally module stacks, §1, identifies ordinary inverse image and its adjunction. Every vector-space value below may be infinite-dimensional.

1. Direct image and both inverse adjunction maps

Fix a field \(k\), universes and choice. Give finite posets their upper Alexandrov topology, and write \(\mathsf D_P=\mathsf{Fun}(P,\mathsf{Vect}_k)\). For a monotone map \(f:P\to Q\), put

P_q=\{p\in P:q\le f(p)\},\qquad (f^*N)(p)=N(f(p)),\qquad (f_*M)(q)=\Gamma(P_q,M|_{P_q}). \tag{1.1}

The subset \(P_q\) is upper. An arrow \(q\le r\) gives \(P_r\subset P_q\), and the direct-image transition forgets the omitted coordinates of a compatible tuple. These projections compose. A map of coefficient diagrams acts coordinatewise. Thus both expressions define functors on every object and morphism.

A map is continuous for these topologies exactly when it is monotone. Monotonicity makes inverse images of upper sets upper. Conversely, if \(p\le p'\), the inverse image of the upper neighbourhood of \(f(p)\) contains \(p\) and therefore \(p'\), forcing \(f(p)\le f(p')\). The ordinary sheaf direct image evaluated on that neighbourhood is exactly (1.1).

There are inverse natural linear bijections

\operatorname{Hom}_{\mathsf D_P}(f^*N,M) \simeq\operatorname{Hom}_{\mathsf D_Q}(N,f_*M). \tag{1.2}

For \(a:f^*N\to M\), send \(n\in N(q)\) to the tuple with coordinate \(a_pN_{q,f(p)}n\) at \(p\in P_q\). Naturality of \(a\) and the laws of \(N\) make this tuple compatible on every arrow of \(P_q\). Its restrictions give naturality in \(q\). Conversely, for \(b:N\to f_*M\), evaluate \(b_{f(p)}\) at coordinate \(p\) to define \(a_p\). For \(p\le p'\), naturality of \(b\) along \(f(p)\le f(p')\), followed by coordinate \(p'\), makes this family a diagram map. The tuple's compatibility also gives \(M_{pp'}a_p=a_{p'}N_{f(p),f(p')}\).

The second operation after the first returns \(a_p\), since \(N_{f(p),f(p)}=1\). Starting with \(b\), its naturality along \(q\le f(p)\) gives back every coordinate of \(b_q\), so the other composite is identity. Both constructions commute with coefficient maps. This proves the full adjunction. By the ordinary inverse-image adjunction in the prerequisite, its left adjoint \(f^*\) is the ordinary sheaf inverse image under the sheaf/diagram equivalence.

The unit sends \(n\in N(q)\) to \((N_{q,f(p)}n)_{p\in P_q}\); the counit at \(p\) extracts coordinate \(p\) of a section on \(P_{f(p)}\). Their first triangle is evaluation of that unit tuple at \(p\). For the other, start with a compatible tuple on \(P_q\): the unit restricts it to each \(P_{f(p)}\), and the counit picks its original \(p\)-coordinate. Thus both triangles hold on every component. Precomposition \(f^*\) is exact because exactness is pointwise. Consequently \(f_*\) preserves injectives: transpose an extension problem along a monomorphism, solve it after exact \(f^*\), and transpose back. It is left exact, but need not be exact.

2. One complex for every higher value and transition

For a diagram \(M\) in degree zero, define a complex of diagrams on \(Q\) by

K_f^n(M)(q)=\prod_{p_0<\cdots<p_n\text{ in }P_q}M(p_n) \quad(n\ge0). \tag{2.1}

The differential at a strict chain \(p_0<\cdots<p_{n+1}\) is

(ds)_{p_0\ldots p_{n+1}} =\sum_{i=0}^{n}(-1)^is_{p_0\ldots\widehat{p_i}\ldots p_{n+1}} +(-1)^{n+1}M_{p_np_{n+1}}s_{p_0\ldots p_n}. \tag{2.2}

For \(q\le r\), project to the chains lying in \(P_r\). Every face of such a chain also lies there, so projection commutes with (2.2). These projections compose and commute with maps on the terminal coefficients. The square-zero equation is the full signed coefficient calculation of the resolution prerequisite. Thus (2.1) is a specified complex and functor, not just a list of vector spaces.

Theorem 2.1. The complex \(K_f^\bullet(M)\) represents \(Rf_*M\). In particular

(R^nf_*M)(q)\simeq H^n(C^\bullet(P_q,M|_{P_q})), \tag{2.3}

and every transition is induced by the chain projection just described. This value is zero for \(n>\operatorname{height}(P_q)\), where height counts strict arrows and the empty poset has height minus one.

Proof. Take the actual injective resolution \(M\to J_P^\bullet(M)\) of the prerequisite. Its term for \(p_0<\cdots<p_n\) is \(I_{p_0}(M(p_n))\). Restriction to an upper subset \(S\) kills \(I_x(V)\) when \(x\notin S\): a point of \(S\) below \(x\) would force \(x\in S\). If \(x\in S\), that restriction is precisely \(I_x^S(V)\). Hence the restricted resolution is the canonical injective resolution on \(S\), term for term and with all its differential maps.

Apply this with \(S=P_q\). Global sections of each remaining \(I_{p_0}^{P_q}(M(p_n))\) identify with \(M(p_n)\), by the inverse maps in the prerequisite. Thus \(f_*J_P^n(M)(q)\) is exactly (2.1), and the differential is (2.2). The identifications commute with neighbourhood restriction and coefficient maps. Direct image applied to an injective resolution computes its right derived functor, proving the assertion as a complex on \(Q\). Each fibre-neighbourhood complex ends at its own height, proving the vanishing. Empty neighbourhoods give the zero complex. \(\square\)

The bound comes from inverse images of neighbourhoods in the source. The height of the target alone supplies no such bound.

3. Derived composition without a second arbitrary replacement

For \(f:P\to Q\) and \(t:Q\to T\), there are natural coherent isomorphisms

t_*f_*\simeq(tf)_*,\qquad Rt_*Rf_*\simeq R(tf)_* \tag{3.1}

on the appropriate bounded-below derived categories for the second assertion. Here are the ordinary comparison maps. A section of \(t_*f_*M\) at \(s\) is a compatible family of sections \(a_q\) on \(P_q\) for \(s\le t(q)\). Send it to the tuple whose coordinate at \(p\), with \(s\le t(f(p))\), is the \(p\)-coordinate of \(a_{f(p)}\). Conversely restrict a tuple on that inverse-image neighbourhood to every \(P_q\). The inequality \(s\le t(q)\le t(f(p))\) permits this restriction. Compatibility along \(q\le f(p)\) proves the two inverse identities. All coefficient and target transitions commute with these restrictions. For three maps, either composition picks the same innermost point coordinate, proving coherence.

For the derived comparison, use a common bounded-below injective complex \(J\) on \(P\). The complex \(f_*J\) is already termwise injective by §1. Applying \(t_*\) therefore computes its derived image without another replacement. Both sides of (3.1) are represented by the ordinary comparison \(t_*f_*J\simeq(tf)_*J\). The bounded Hom and resolution comparisons in the derived-adjunction prerequisite give its naturality on every derived arrow. The same three-map coordinate comparison on \(J\) gives coherence. This argument also retains the full adjunction \(f^*\dashv Rf_*\) from that prerequisite; it concerns bounded-below complexes.

4. Pull back normalized cochains, including collapsed chains

Let \(v:B\to A\) be any monotone map of finite posets and \(M\) a diagram on \(A\). Define

v^\sharp:C^\bullet(A,M)\longrightarrow C^\bullet(B,v^*M) \tag{4.1}

as follows. On an output strict chain \(b_0<\cdots<b_n\), use the coordinate of \(s\) at \(v(b_0)<\cdots<v(b_n)\) if those images are all distinct; otherwise use zero. Its value lies in \(M(v(b_n))\), the required terminal coefficient. In degree zero no collapse occurs.

This is a chain map. If the images of an output chain are strictly increasing, every face pulls back according to (2.2), and the terminal coefficient map is the same. If its image has exactly one adjacent equality, the two faces deleting those equal entries give the same coordinate with opposite signs. When the equality is at the terminal pair, the terminal coefficient map is identity, giving the same cancellation. All other faces remain collapsed and give zero. If there is more than one adjacent equality, deletion of one entry still leaves a repetition, so every face gives zero. These cases prove \(dv^\sharp=v^\sharp d\) for every output chain.

The rules commute with coefficient-diagram morphisms. They preserve identities and compose contravariantly: for \(w:D\to B\), the composite \(w^\sharp v^\sharp\) uses a coordinate exactly when both image chains are strict. That is equivalent to strictness of their composite image chain; a repetition at either stage forces a repetition after composition. Its value and last coefficient then agree with \((vw)^\sharp\). The rules also commute with restriction to upper subsets whenever the monotone maps and subsets are compatible.

There is an actual resolution comparison underlying (4.1). At \(b\), use the same rule on chains above \(b\) to define

v^*J_A^\bullet(M)\longrightarrow J_B^\bullet(v^*M). \tag{4.2}

The image chains start above \(v(b)\), so all required input coordinates exist. The calculation above proves the chain equation, and projecting to chains above a later point proves naturality in \(b\). In degree zero the augmentation maps commute, using \(M_{v(b),v(b_0)}\). Thus (4.2) extends the identity of \(v^*M\). Its source is a resolution because \(v^*\) is exact; its target is the explicit injective resolution. The bounded injective Hom test in the prerequisite identifies it with the canonical resolution comparison extending that identity. No injectivity of \(v^*J_A^n(M)\) is assumed. First pull a global section back to \(v^*J_A(M)\), then apply global sections of (4.2); this composite gives precisely (4.1).

5. The actual derived base-change map

Let \(g:Q'\to Q\) be monotone, and form the cartesian poset

P'=\{(p,r'):\ f(p)=g(r')\},\qquad g'(p,r')=p,\qquad f'(p,r')=r', \tag{5.1}

with the componentwise order. For each \(q'\), projection gives a monotone map

v_{q'}:P'_{q'}\longrightarrow P_{g(q')},\qquad(p,r')\longmapsto p. \tag{5.2}

It is defined because \(q'\le r'\) implies \(g(q')\le g(r')=f(p)\). The normalized maps (4.1) give a map of complexes on \(Q'\)

g^*K_f^\bullet(M)\longrightarrow K_{f'}^\bullet(g'^*M). \tag{5.3}

Neighbourhood restriction makes the maps \(v_{q'}\) compatible, so every transition square commutes. Naturality in \(M\) follows from §4. In degree zero, a compatible tuple \((m_p)\) is pulled back to \((m_p)_{(p,r')}\). This is the ordinary base-change map \(g^*f_*M\to f'_*g'^*M\): substituting the unit and counit formulas of §1 gives exactly these coordinates.

To identify (5.3) in the derived category, take \(J_P(M)\). Apply ordinary base change termwise, then the actual comparison (4.2) for \(g'\):

g^*f_*J_P(M)\longrightarrow f'_*g'^*J_P(M) \longrightarrow f'_*J_{P'}(g'^*M). \tag{5.4}

The second comparison extends identity on \(g'^*M\) and has an injective resolution as target, so it is the required derived comparison. The first is the ordinary adjunction comparison on that resolution. Hence (5.4) is the canonical map \(g^*Rf_*M\to Rf'_*g'^*M\). After the identifications of §2, its output coordinate is the input coordinate on the projected chain, or zero if that chain collapses. This is exactly (5.3), proving its canonical meaning as well as its cochain formula.

Theorem 5.1. Suppose \(g\) has unique upward lifts: for every \(q'\) and every \(q\ge g(q')\), there is exactly one \(r'\ge q'\) with \(g(r')=q\). Then (5.3) is an isomorphism of complexes for every \(f,M\), and derived base change is an isomorphism. In particular this holds when \(g\) is the inclusion of an upper open subset.

Proof. For \(p\in P_{g(q')}\), lift \(f(p)\) uniquely above \(q'\), obtaining \(r'_p\). This gives the inverse object map \(p\mapsto(p,r'_p)\) to (5.2). If \(p\le p_1\), lift \(f(p_1)\) above \(r'_p\). Its value is also a lift above \(q'\), so uniqueness makes it \(r'_{p_1}\); hence \(r'_p\le r'_{p_1}\). Thus both maps preserve order and are inverse poset isomorphisms. All strict chains and their coefficient transitions therefore correspond, so (5.3) is an actual complex isomorphism at every point. Its existing transition and coefficient naturalities make it an isomorphism of diagram complexes. For an upper inclusion, \(q\ge q'\) already belongs to the open subset and has itself as its unique lift. \(\square\)

The condition is sufficient. A square can give a derived isomorphism without it, as Exercise 2 shows. General closed-point base change does not follow from the cartesian property alone.

6. Four graded exercises with complete solutions

Exercise 1 (grade 3). Let \(L\) be the six-point cycle poset of the resolution lesson, and let \(P=L\cup\{z\}\) with \(z\) greater than every point. Map \(P\) to the chain \(Q=\{0<1\}\) by sending \(L\) to 0 and \(z\) to 1. For \(M=S_z\), compute every \(R^nf_*M\), including its transition map. Compare source and target height bounds.

Solution. At 0, \(P_0=P\); at 1, \(P_1=\{z\}\). The first complex is exactly \(k\to k^6\to k^6\) of the resolution lesson's Exercise 4: the first arrow is the constant tuple inclusion, and the second is the cycle differential of rank 5. Its cohomology is \(k\) in degree two only, with the explicit alternating cycle coordinate there. The second complex is \(k\) in degree zero only. The neighbourhood projection in degree zero is identity \(k\to k\); in higher degrees its target is zero. On degree-zero cohomology the source kernel is zero, so the induced map is \(0\to k\). In degree two it is \(k\to0\). Consequently

R^0f_*S_z=(0\to k),\qquad R^1f_*S_z=0,\qquad R^2f_*S_z=(k\to0),\qquad R^nf_*S_z=0\ (n>2). \tag{6.1}

This is the full direct-image diagram in each degree. The source neighbourhood at 0 has height 2 and reaches its bound. The target has height 1; that target height does not force \(R^2f_*\) to vanish.

Exercise 2 (grade 2). Take \(P=Q=\{*\}\), \(f=1\), and let \(g:\{0<1\}\to Q\) be constant. For an arbitrary vector space \(V\), compute the cochain base-change map. Show it is a quasi-isomorphism although unique upward lifting fails.

Solution. Here \(P'=\{0<1\}\), \(f'=1\), and \(g'\) is constant. At 0 the source complex is \(V\) in degree zero; the target is \(V^2\xrightarrow{(a,b)\mapsto b-a}V\) in degrees zero and one. The map in degree zero is \(v\mapsto(v,v)\); its degree-one map is zero, because the chain \(0<1\) projects to a repeated point. The differential kills that diagonal, so this is the specified chain map. Its target kernel is the diagonal and its target cokernel is zero: the differential is onto using \((0,v)\). The map therefore induces identity on the degree-zero copy of \(V\) and is a quasi-isomorphism.

At 1 both complexes are \(V\) in degree zero and the map is identity. The target's transition from 0 to 1 extracts the second coordinate, which sends the diagonal to its original vector. The source transition is identity, so the whole comparison is natural and is a quasi-isomorphism of diagrams. Above 0, both 0 and 1 lift the only point of \(Q\); uniqueness fails. Success of this square therefore does not make the sufficient lifting condition necessary.

Exercise 3 (grade 3). Adjoin a least point \(b\) below the six-point cycle \(L\), giving \(Q\), and let \(j:L\hookrightarrow Q\) be the upper open inclusion. Compute all \(R^nj_*k_L\). Compare \(j_*k_L\) with extension by zero and prove the latter's exact adjunction in this model.

Solution. The inverse neighbourhood at \(b\) is all of \(L\), so its cohomology is \(k\) in degrees zero and one and zero otherwise, by the complete cycle calculation in the resolution lesson. At \(x\in L\), the inverse neighbourhood is \(U_x\) inside \(L\), which has least point \(x\). Compatible tuples there are naturally their \(x\)-value, an exact evaluation functor; hence its constant diagram has \(k\) in degree zero and no positive groups. Restrictions of constant compatible tuples are identity. Thus

R^0j_*k_L=k_Q,\qquad R^1j_*k_L=S_b, \qquad R^nj_*k_L=0\quad(n>1). \tag{6.2}

Every outgoing transition from the degree-one value at \(b\) is zero because its target is zero.

For any diagram \(F\) on the upper subset \(L\), define \(j_!F\) to equal \(F\) on \(L\) and zero at \(b\), with zero maps from \(b\). There are no arrows from \(L\) to its complement, so these rules define a diagram. A map \(j_!F\to N\) is uniquely a map \(F\to N|_L\); the naturality squares starting outside \(L\) have both sides zero. This gives both inverse adjunction maps and \(j_!\dashv j^*\). It is exact since its values are the original exact rows or zero. The diagram \(j_!k_L\) has value zero at \(b\), whereas \(j_*k_L=k_Q\) has value \(k\) there. The two extensions therefore differ already in degree zero, and the nonzero \(R^1j_*\) supplies another distinction.

Exercise 4 (grade 4). Let \(f:L\to\{0<1\}\) be constant at 1, and pull back along the closed inclusion \(g:\{0\}\hookrightarrow\{0<1\}\). Use \(M=k_L\). Compute the canonical base-change maps in every degree and explain why the empty literal fibre does not compute the source value.

Solution. Both \(P_0\) and \(P_1\) are \(L\), so the complexes for the two target values are the same cycle complex and the transition is identity. Thus \(R^0f_*M\) and \(R^1f_*M\) are each \((k\xrightarrow1k)\), with all higher images zero. The cartesian poset \(P'\) is empty: the equations \(f(p)=1\) and \(g(0)=0\) cannot agree. The target of the base-change map is therefore the zero complex and every one of its higher groups is zero.

At the only point of \(Q'\), the source groups are \(k\) in degrees zero and one. Both canonical maps are \(k\to0\), and neither is an isomorphism. Higher-degree maps are \(0\to0\). Formula (5.3) gives exactly these maps by restriction to no chains. The singleton \(\{0\}\) is closed since its complement is upper, but its minimal open neighbourhood is the whole two-point chain. The inverse image of that neighbourhood is \(L\), although the inverse image of the literal point is empty. This proves the failure and identifies the mistaken fibre replacement in both degrees.