Draft lesson.

Slice sites, dependent products and base change

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

A map of sheaves has three associated functors between slice categories. One forgets part of the structure map, one pulls back, and one forms a dependent product. Their two adjunctions have different directions. A formula using compatible local sections constructs the dependent product without requiring fiber products in the original site, and makes its base-change maps explicit.

Let \(\mathcal E=\mathsf{Sh}(C,J)\) for a small sieve site, with set-valued sheaves in a fixed universe. All indexing sets and function sets below lie in that universe. Let \(p:A\to B\) be any sheaf map. Write \(r:X\to A\) and \(s:Y\to B\) for objects of the two slices. Pullbacks of sheaves exist and are computed sectionwise; this does not assume that \(C\) has fiber products.

We retain the complete element-site equivalence in Twisted modules, Section 7, the set fiber and slice-limit constructions in Fibres, slices and index adjunctions, Sections 3–4, the full local epimorphism criterion, and the adjunction and geometric-morphism conventions in Restriction and inverse images.

1. Element sites, pullback and dependent sum

The element site \(C/A\) has objects \((U,a)\) with \(a\in A(U)\). An arrow to \((U,a)\) is an arrow \(t:T\to U\) with the uniquely determined source label \(a|_t\). Thus sieves on \((U,a)\) are identified with sieves on \(U\); their covering, pullback and transitivity rules are exactly those in \(C\). This verifies all topology axioms and smallness. The retained equivalence sends \(X\to A\) to its fiber sheaf \((U,a)\mapsto X(U)_a\), with inverse the disjoint union over \(a\). Its sheaf gluing first glues the labels in \(A\), then the entries. These constructions and their inverse maps are already proved in the prerequisite.

The functor \(v:C/A\to C/B\) induced by \(p\) sends \((U,a)\) to \((U,p(a))\). Restriction of a fiber sheaf along \(v\) corresponds to

p^*Y=(A\times_B Y\longrightarrow A). \tag{1.1}

Indeed its fiber at \((U,a)\) is \(Y(U)_{p(a)}\), on every arrow as well as every object. Define the dependent sum by

\Sigma_pX=(X\xrightarrow{pr}B). \tag{1.2}

It acts on maps by their underlying maps; pullback acts by \((a,y)\mapsto(a,m(y))\). The adjunction \(\Sigma_p\dashv p^*\) has inverse Hom factors

h:X\to Y\quad\longleftrightarrow\quad [x\longmapsto(r(x),h(x))]:X\to A\times_B Y. \tag{1.3}

Here the left map is over \(B\), and the right is over \(A\); the inverse is projection to \(Y\). Both inverse equations follow on each section. Precomposition in \(X\) and postcomposition in \(Y\) commute with the same tuple formula, giving both naturalities. The unit is \(x\mapsto(r(x),x)\), and the counit is \((a,y)\mapsto y\). Substitution proves both triangle identities, including empty fibers.

2. Constructing the dependent-product sheaf

Define \((\Pi_pX)(U)\) as the set of pairs \((b,\sigma)\), where \(b\in B(U)\) and \(\sigma\) is the following entire compatible assignment. For each \(t:T\to U\) and \(a\in A(T)\) with \(p(a)=b|_t\), it specifies

\sigma_{t,a}\in X(T),\qquad r(\sigma_{t,a})=a, \quad \sigma_{t,a}|_w=\sigma_{tw,a|_w} \tag{2.1}

for every \(w:T'\to T\). These are small sets of data and equations. The map to \(B(U)\) is \((b,\sigma)\mapsto b\). Restriction along \(z:V\to U\) sends \(b\) to \(b|_z\) and uses the assignment \((t,a)\mapsto\sigma_{zt,a}\). The presheaf and structure-map laws follow from composition of arrows. A map \(m:X\to X'\) over \(A\) sends \(\sigma_{t,a}\) to \(m_T(\sigma_{t,a})\); its naturality preserves (2.1). This specifies the functor on all maps, with both functor laws.

Lemma 2.1. This presheaf is a sheaf.

Proof. Let compatible pairs be given over a covering sieve \(S\) on \(U\). Their \(B\)-labels glue uniquely to \(b\in B(U)\). Fix \(t:T\to U\) and \(a\in A(T)\) over \(b|_t\). For each \(w:T'\to T\) in the covering sieve \(t^*S\), the given pair over \(tw\) has an assignment. Its value at \((1_{T'},a|_w)\) is a section of \(X(T')\). These sections are compatible: restricting one uses (2.1), and the original matching pairs agree on their further restrictions. They glue uniquely to a section \(\sigma_{t,a}\in X(T)\). Their images under \(r\) are the restrictions of \(a\), so separation of \(A\) gives \(r(\sigma_{t,a})=a\).

For any further arrow \(z:T''\to T\), restriction of this glued section and the section constructed for \((tz,a|_z)\) agree on the covering sieve \(z^*t^*S\). Separation of \(X\) proves (2.1). On a member of \(S\), the constructed assignment recovers the given one by the same uniqueness. Conversely any global gluing has the specified \(b\), and each of its assigned sections is forced by this covering gluing in \(X\). Hence it is unique. Empty covers cause no exception: the sheaf axiom on the corresponding empty pullback cover supplies the unique required section. \(\square\)

Thus \(\Pi_pX\to B\) is an object of \(\mathcal E/B\), built without enlarging the site or assuming representability of \(A,B\).

3. The second adjunction and its triangles

Theorem 3.1. There is an adjunction \(p^*\dashv\Pi_p\).

Proof. Given \(h:A\times_B Y\to X\) over \(A\), send \(y\in Y(U)\) to the pair with \(b=s(y)\) and assignment

\sigma^y_{t,a}=h_T(a,y|_t). \tag{3.1}

Naturality of \(h\) proves (2.1). Restricting \(y\) restricts precisely this assignment, so it is a map \(Y\to\Pi_pX\) over \(B\). Conversely such a map \(k\) gives

h_U(a,y)=\sigma^y_{1_U,a},\qquad k_U(y)=(s(y),\sigma^y). \tag{3.2}

This has image over \(a\). To check naturality along \(t:T\to U\), restrict the right side using (2.1); naturality of \(k\) identifies the result with the value of (3.2) at \((a|_t,y|_t)\). Starting from \(h\), substitution with \(t=1\) returns \(h\). Starting from \(k\), the reconstructed value at \((t,a)\) is \(\sigma^{y|_t}_{1,a}=\sigma^y_{t,a}\), again by naturality of \(k\). This returns its entire assignment and its label. These are both inverse equations.

For a map \(Y'\to Y\), either naturality route replaces \(y\) by its image before applying \(h\). For a map \(X\to X'\), both routes apply it to the assigned value. Thus the two factors are natural in both variables. \(\square\)

The unit at \(Y\) sends \(y\) to the assignment \((t,a)\mapsto(a,y|_t)\) in \(\Pi_pp^*Y\). The counit is evaluation:

p^*\Pi_pX\longrightarrow X,\qquad (a,(p(a),\sigma))\longmapsto\sigma_{1,a}. \tag{3.3}

The triangle on \(p^*Y\) evaluates \((t,a)\mapsto(a,y|_t)\) at \((1,a)\), returning \((a,y)\). For the other triangle, apply \(\Pi_p\) of evaluation to the unit assignment of \((b,\sigma)\). Its value at \((t,a)\) is evaluation of the restricted assignment \(\sigma|_t\) at \((1,a)\), namely \(\sigma_{t,a}\). It returns the original pair at every entry. Both triangles are therefore checked on all sections.

4. The localization morphism and preservation properties

We have the full triple

\Sigma_p\ \dashv\ p^*\ \dashv\ \Pi_p. \tag{4.1}

Both slice categories are sheaf categories by Section 1. As a right adjoint to \(\Sigma_p\), pullback preserves all small limits; as a left adjoint to \(\Pi_p\), it preserves all small colimits. For either preservation assertion, the proved natural Hom factors convert a cone or cocone into one before applying the original universal factor, then transpose that unique factor back. The inverse Hom equation gives existence and uniqueness and keeps each structural component. This includes the empty diagrams.

Consequently \(p^*\dashv\Pi_p\) gives the canonical geometric morphism \(\mathcal E/A\to\mathcal E/B\). Its inverse image is pullback, and its direct image is dependent product. Restriction along the element-site functor \(v\) in Section 1 is that same pullback functor, but its left adjoint is \(\Sigma_p\). These are the two different adjunctions in (4.1).

In fact \(\Sigma_p\) preserves finite limits if and only if \(p\) is an isomorphism. Preservation of the terminal slice object would send \(1_A\) to a terminal object over \(B\). Its image is precisely \(p:A\to B\), so this forces \(p\) to be isomorphic over \(B\) to \(1_B\), equivalently forces \(p\) invertible. Conversely if \(p\) is invertible, composition with \(p\) and pullback along it are inverse equivalences, so preserve all existing limits. This explains why the left adjoint to element-site restriction need not itself be an inverse image.

5. Exponentials in every slice

For \(A,E\in\mathcal E\), take \(p:A\to1\) and set

E^A=\Pi_p(A\times E\to A). \tag{5.1}

A section over \(U\) is exactly a rule assigning \(e_{t,a}\in E(T)\) to every \(t:T\to U,a\in A(T)\), compatibly with restriction. The missing \(A\)-coordinate is the specified \(a\), so (2.1) gives precisely this description. The adjunction in Section 3 identifies maps \(Y\to E^A\) with maps \(Y\times A\to E\). Its forward and inverse factors are evaluation and \(y\mapsto[(t,a)\mapsto h(y|_t,a)]\). Substitution gives both inverse equations, and the naturalities in \(Y,E\) follow as in Section 3. Precomposing the \(a\)-argument proves contravariant naturality in \(A\).

More generally let \(A\to B,E\to B\) be objects of \(\mathcal E/B\). The exponential there is

\Pi_{A\to B}(A\times_B E\to A). \tag{5.2}

Indeed a map into this object transposes to a map \(A\times_B Y\to A\times_B E\) over \(A\). Such a map is exactly \((a,y)\mapsto(a,h(a,y))\) for a map \(A\times_B Y\to E\) over \(B\); projection and adding the \(a\)-coordinate give the inverse factors. These are the complete exponential universal maps, with evaluation and currying as above. Thus every slice is cartesian closed, including slices over nonrepresentable sheaves.

6. Both base-change comparisons

Let \(h:B'\to B\), let \(A'=B'\times_B A\), and write \(k:A'\to A\), \(p':A'\to B'\) for the projections. The dependent-sum comparison is

h^*\Sigma_pX\simeq\Sigma_{p'}k^*X. \tag{6.1}

At an object \(U\), its maps send \((b',x)\) to \(((b',r(x)),x)\) and project back to \((b',x)\). The common condition is \(h(b')=p(r(x))\); the displayed formulas are inverse and commute with restrictions and all maps in \(X\). The related Frobenius comparison \(\Sigma_p(X\times_Ap^*Y)\simeq\Sigma_pX\times_BY\) drops or inserts the repeated \(A\)-coordinate: both sides are exactly pairs \((x,y)\) with \(p(r(x))=s(y)\), including their map to \(B\).

The dependent-product comparison is

h^*\Pi_pX\simeq\Pi_{p'}k^*X. \tag{6.2}

A left-side section over \(U\) is a label \(b'\) and an assignment \(\sigma\) over \(h(b')\). A right-side assignment is indexed by \(t:T\to U\) and \(a'=(b'|_t,a)\) with \(p(a)=h(b')|_t\). Send its value to

\sigma'_{t,(b'|_t,a)}=((b'|_t,a),\sigma_{t,a}). \tag{6.3}

This belongs to \(k^*X\), with its required \(A'\)-label. Conversely project each value of \(\sigma'\) to its \(X\)-coordinate. These rules preserve every restriction equation (2.1) and are inverse on labels and values, proving (6.2). They also commute with maps in \(X\). The units and counits from Sections 1 and 3 use the same tuples or evaluation, so these comparisons respect them. For two successive base changes, either route gives the same tuple \((b'',a,x)\), or the same assignment indexed by \((b''|_t,a)\). Thus their composite comparisons agree, after the canonical pullback identifications; identity base change gives identity maps. This proves the base-change coherence as well as the isomorphisms themselves.

7. Composition of dependent products

For \(A\xrightarrow pB\xrightarrow qD\), dependent sums compose by composing structure maps. Pullbacks compose by the canonical rebracketing \((a,(p(a),z))\leftrightarrow(a,z)\). The right adjoints consequently have a comparison \(\Pi_{qp}\simeq\Pi_q\Pi_p\), whose inverse maps can also be written directly.

A section of \(\Pi_q\Pi_pX\) over \(U\) is a label \(d\in D(U)\) and assignments \(\tau_{t,b}\in(\Pi_pX)(T)\) for \(q(b)=d|_t\). Flatten it to the assignment for \(\Pi_{qp}X\) which at \((t,a)\) takes the inner value of \(\tau_{t,p(a)}\) at \((1_T,a)\). Conversely an assignment \(\sigma\) for \(\Pi_{qp}X\) gives \(\tau_{t,b}\) whose inner value at \((z:T'\to T,a)\), with \(p(a)=b|_z\), is \(\sigma_{tz,a}\). Equation (2.1) verifies both naturality requirements. Flattening after nesting returns each \(\sigma_{t,a}\). Nesting after flattening returns every inner value of \(\tau\), because restriction of \(\tau_{t,b}\) along \(z\) equals \(\tau_{tz,b|_z}\). These are the full inverse equations.

The comparison is natural in \(X\), and evaluating its units gives the iterated unit in Section 3. For three composable maps, both parenthesizations flatten to the same value at the same nested arguments, so their comparisons agree. Identity maps give the identity comparison. This provides composition coherence with the specified adjunctions, without asserting uniqueness of arbitrary natural isomorphisms.

8. Four exercises with full solutions

8.1 Empty fibers in a finite example

Exercise 1 (basic). In sets let \(A=\{a_0,a_1,a_2\}\), \(B=\{0,1,2\}\), with \(p(a_0)=p(a_1)=0\), \(p(a_2)=1\). Let \(X\to A\) have fibers of sizes \(2,3,0\). Compute dependent sums and products, their units and counits, and their base change along the inclusion \(\{0,2\}\hookrightarrow B\).

Solution. For any set map, the dependent-sum fiber at \(b\) is \(\coprod_{a\in p^{-1}b}X_a\), while the dependent-product fiber is \(\prod_{a\in p^{-1}b}X_a\). These follow directly from the compatible-assignment formula on the one-object identity site. The three sum fibers have sizes \(5,0,0\); the three product fibers have sizes \(6,0,1\). The last product is a singleton because there is one empty function. The product at \(1\) is empty because its one factor is empty.

For \(Y\to B\), the first unit sends \(x\) to \((r(x),x)\), and the first counit drops \(a\) from \((a,y)\). The second unit sends \(y\in Y_b\) to the constant family \((a,y)\) over \(p^{-1}b\), and its counit evaluates a family at its selected \(a\). In an empty fiber that family is the unique empty function; no evaluation input there exists. Currying and evaluating return every entry, checking the triangles including those fibers. Base change retains the fibers at \(0,2\), so its sum sizes are \(5,0\), and its product sizes are \(6,1\). The maps in (6.1) and (6.3) simply retain the identical entries with their selected base labels.

8.2 An open subterminal sheaf

Exercise 2 (intermediate). Let \(U\subseteq X\) be an open subspace, let \(A\) be its subterminal sheaf on \(X\), and let \(p:A\to1\). Identify the slice with sheaves on \(U\), and compute all three functors, including values on the empty open.

Solution. The subterminal has \(A(V)=1\) when \(V\subseteq U\), and \(A(V)=\varnothing\) otherwise. Hence an object over \(A\) has empty section set on every open not contained in \(U\); on opens contained in \(U\) it is exactly a sheaf \(G\) on \(U\). Conversely extending its section rule this way gives a sheaf on \(X\): a matching family can exist only when every member of its cover lies in \(U\), in which case their union also does and gluing is that of \(G\). This is the element-site equivalence in this case.

Pullback is ordinary restriction to \(U\). The dependent sum is extension by the empty set: its value on \(V\) is \(G(V)\) if \(V\subseteq U\) and is empty otherwise. In particular its value on the empty open is \(G(\varnothing)=1\), as the empty-cover axiom requires. The dependent product has value \(G(V\cap U)\). Indeed an assignment in (2.1) is required only on opens \(T\subseteq V\cap U\), and is determined by its value on the largest such open \(V\cap U\), with all other values its restrictions. This identifies \(\Pi_p\) with the ordinary direct image of the open inclusion, also giving a singleton on the empty open. Its counit is restriction/evaluation on opens in \(U\). The sum cannot preserve the terminal object unless \(U=X\), exactly as Section 4 proves.

8.3 A nonrepresentable slice and its dependent fibers

Exercise 3 (advanced). On the fork site with \(a\to c,b\to c\) and the covering sieve consisting of both leaves, let \(A\) have leaf sets \(\{0,1\},\{r\}\), and let \(B\) have leaf sets \(\{*\},\{r,s\}\). Map the first leaf set to \(*\), and include \(r\) into the second. An object \(X\to A\) has leaf fibers of sizes \(2,3\) over \(0,1\), and size \(4\) over \(r\). Compute \(\Sigma_pX\), \(\Pi_pX\), and base change to the subobject of \(B\) with leaf sets \(\{*\},\{s\}\).

Solution. The sheaf condition identifies a fork sheaf with its two leaf sets; its vertex is their product, with the two projections. The represented sheaves have leaf sizes \((1,0),(0,1),(1,1)\), so \(A\), of sizes \((2,1)\), is not representable. Its element site has the two vertex labels \((0,r),(1,r)\), the two first-leaf labels and the one second-leaf label. Each vertex is covered by its corresponding pair of leaves. A fiber sheaf there therefore has vertex fiber \(X_{a,i}\times X_{b,r}\) at \((i,r)\). This gives the slice equivalence on its actual sections and maps in this example.

Dependent sum at the first leaf has its one base fiber of size \(2+3=5\); at the second it has fiber sizes \(4,0\) over \(r,s\). Its vertex has \(20\) elements over \((*,r)\) and none over \((*,s)\). Dependent product has first-leaf fiber size \(2\cdot3=6\), and second-leaf fiber sizes \(4,1\). Its vertex fibers therefore have sizes \(24,6\), respectively, and its total vertex section set has size \(30\). The singleton over \(s\) is the empty product. All maps are componentwise functions on these fibers; their vertex maps are the corresponding products, so the calculation describes the functors on maps too.

After the specified base change, sum leaf sizes are \(5,0\), hence the vertex is empty. Product leaf sizes are \(6,1\), hence the vertex has size \(6\). These are also obtained by first pulling \(X\) back: its second-leaf base is empty, and its first-leaf fibers retain sizes \(2,3\). Formula (6.3) identifies each of the six product choices in the two calculations by keeping the same two entries. Thus both base-change comparisons hold even at the empty dependent fiber and for this nonrepresentable base.

8.4 A dependent product need not preserve epimorphisms

Exercise 4 (challenge). In sheaves of sets on \(\mathbb R\), put \(U_n=(-1/n,1/n)\), \(V=\mathbb R\setminus\{0\}\), for \(n\geq1\). Let \(E_n\to1\) be the sheaf of sections of the two-sheet open cover \(U_n\amalg V\to\mathbb R\). Prove that every \(E_n\to1\) is epic but \(\prod_nE_n\to1\) is not. Express this as failure of a dependent product to preserve an epimorphism.

Solution. A section of \(E_n\) over an open \(W\) is a partition of \(W\) into two open-and-closed pieces, one contained in \(U_n\) and the other in \(V\), carrying the two sheet labels. This follows by restricting a section to each label, and conversely gluing the two sections on that partition. The two opens cover \(\mathbb R\), so the map has a section locally at every point. The complete local lifting criterion therefore makes it epic.

There is no section of \(\prod_nE_n\) on any neighborhood of zero. If one existed, restrict it to an open interval \(I\) about zero inside that neighborhood. Each sheet-label function on \(I\) is locally constant and thus constant, and zero forces its label to be the \(U_n\)-sheet. To justify the interval assertion directly, a nonempty proper clopen subset of an interval would contain a point \(x\) and omit a point \(y\). Reflect if necessary so \(x<y\). The supremum of its intersection with \([x,y]\) belongs to that closed subset, is strictly below \(y\) by openness of the complement at \(y\), and can be increased while staying in the subset by openness. This contradicts the defining supremum. Completeness of the real numbers supplies the supremum. Thus the label is constant as claimed.

Consequently \(I\subseteq U_n\) for every \(n\), which would imply \(I\subseteq\{0\}\): any nonzero \(x\) is excluded by an integer \(n>1/|x|\). This is impossible for an open interval. The product map fails local lifting at zero and is not epic. Away from zero the common \(V\)-sheet supplies all the lifts. Any finite set of factors does lift near zero on the smallest corresponding \(U_n\); its product map is epic. The failure is therefore specific to the infinite family, not a failure of the finite lifting test.

Finally let \(A=\coprod_{n\geq1}1\), let \(p:A\to1\), and let \(Z\to A\) be the coproduct of the maps \(E_n\to1\) with their labels \(n\). This map is epic: any local section of \(A\) has a locally constant label, and on each label piece the covering \(U_n,V\) gives local lifts. In the slice \(\mathcal E/A\) it is a map to the terminal object. A section of \(\Pi_pZ\) is equivalent to one section of each \(E_n\): evaluate its assignment on the constant labels; conversely for a locally constant label glue the selected sections on its open label pieces. These two constructions are inverse and respect all restrictions. Thus \(\Pi_pZ\simeq\prod_nE_n\), whereas \(\Pi_p1_A=1\). Applying \(\Pi_p\) gives exactly the non-epic product map above, despite its preservation of all limits.