Universes and small categories

Written by GPT-6.1 Sol (OpenAI), October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Original text public domain (CC0).

A category can have small sets of maps between any two objects and still have too many objects for a small diagram to list them. This distinction matters when forming categories of presheaves, choosing generators and taking formal filtered colimits. A universe fixes what “small” means without putting an artificial finite bound on the constructions allowed.

We prove the closure properties behind this convention, then explain what they imply for categories and natural transformations. The prerequisites are elementary sets, functions and the definition of a category. Basic references are [Stacks, Set Theory], [Stacks, Categories] and [SGA 4, Univers]. The set-theoretic arguments needed below are given here. Existence of arbitrarily large Grothendieck universes is an additional axiom, which we state explicitly.

1. Smallness is invariant under relabelling

Fix a set \(\mathcal U\) satisfying the following conditions. It contains \(\varnothing\) and \(\mathbb N\), and it is transitive: if \(a\in\mathcal U\) and \(x\in a\), then \(x\in\mathcal U\). For \(a\in\mathcal U\), both \(\{a\}\) and its power set \(\mathcal P(a)\) belong to \(\mathcal U\). Finally, if \(J\in\mathcal U\) and \(a_j\in\mathcal U\) for every \(j\in J\), then

\[ \bigcup_{j\in J}a_j\in\mathcal U. \tag{1.1} \]

Such a set is a Grothendieck universe with the natural numbers included. The indexed-union condition does not require the family \((a_j)\), regarded as a function, to have already been encoded inside \(\mathcal U\).

A \(\mathcal U\)-set is a member of \(\mathcal U\). A set is \(\mathcal U\)-small if it admits a bijection with a \(\mathcal U\)-set. Thus being small concerns how many elements a set has; being a universe member also concerns the elements' set-theoretic encoding. All sets, functions and choices in this lesson are considered in an ambient setting with the axiom of choice. We sometimes choose a larger universe \(\mathcal V\) with \(\mathcal U\in\mathcal V\).

Our enlargement axiom is: for every set \(A\), there is a Grothendieck universe containing \(A\) as a member. In particular it permits the choice of \(\mathcal V\). This is an assumption, not a conclusion from the closure properties of one universe. The partial universes of [Stacks, Set Theory] are a different convention; their existence in ZFC does not supply this enlargement axiom.

Lemma 1.1. A Grothendieck universe \(\mathcal U\) is not \(\mathcal U\)-small. In particular, \(\mathcal U\notin\mathcal U\). The singleton \(\{\mathcal U\}\) is \(\mathcal U\)-small but is not a \(\mathcal U\)-set.

Proof. We first recall the needed form of Cantor's argument. There is no injection \(\mathcal P(B)\to B\). Indeed, an injection would give a surjection \(s:B\to\mathcal P(B)\): invert it on its image and send the remaining elements to \(\varnothing\). The subset \(D=\{b\in B:b\notin s(b)\}\) cannot equal \(s(b)\) for any \(b\), a contradiction.

Suppose there were a bijection \(\mathcal U\to a\) with \(a\in\mathcal U\). Every subset of \(a\) is an element of \(\mathcal P(a)\), and transitivity applied to \(\mathcal P(a)\in\mathcal U\) puts every such subset in \(\mathcal U\). Restricting the proposed bijection would therefore give an injection \(\mathcal P(a)\to a\), which is impossible. If \(\mathcal U\) were itself a member, its identity map would contradict this result.

The singleton of \(\varnothing\) is a member of \(\mathcal U\), so \(\{\mathcal U\}\) is bijective to a member. If \(\{\mathcal U\}\) were a member, transitivity would give \(\mathcal U\in\mathcal U\). \(\square\)

2. Building sets inside a universe

Proposition 2.1. For a Grothendieck universe \(\mathcal U\), the following constructions remain members of \(\mathcal U\):

Proof. If \(A\subseteq B\in\mathcal U\), then \(A\in\mathcal P(B)\in\mathcal U\), so transitivity gives \(A\in\mathcal U\). Every finite subset of \(\mathbb N\) is consequently a member. For \(a,b\in\mathcal U\), use the two-element index set \(\{0,1\}\) in (1.1) with values \(\{a\}\) and \(\{b\}\). This gives \(\{a,b\}\in\mathcal U\). Induction gives every finite set of members, including the empty set. If \(A\in\mathcal U\), its elements are members by transitivity, and (1.1) with the index set \(A\) gives \(\bigcup A\in\mathcal U\).

In particular \(A\cup B\in\mathcal U\) for \(A,B\in\mathcal U\). Encode an ordered pair by

\[ (a,b)=\{\{a\},\{a,b\}\}. \tag{2.1} \]

For \(a\in A\) and \(b\in B\), this pair belongs to \(\mathcal P(\mathcal P(A\cup B))\). The product \(A\times B\) is a subset of that universe member, hence is itself a member. A function \(A\to B\) is encoded by its graph. The set \(B^A\) of all such functions is a subset of \(\mathcal P(A\times B)\), so it too belongs to \(\mathcal U\). This includes empty domains and codomains; it uses no assertion that a function exists.

Let \(J\in\mathcal U\) and \(A_j\in\mathcal U\). Put \(B=\bigcup_j A_j\in\mathcal U\). The tagged disjoint union

\[ \coprod_{j\in J}A_j=\{(j,a):j\in J,\ a\in A_j\} \tag{2.2} \]

is a subset of \(J\times B\), and hence is a member. The product \(\prod_j A_j\) is the subset of \(B^J\) consisting of functions \(f\) with \(f(j)\in A_j\) for all \(j\). It is a member whether it is empty or nonempty. For \(J=\varnothing\), it is the singleton consisting of the empty function.

Finally, if \(\sim\) is an equivalence relation on \(A\in\mathcal U\), every equivalence class is a subset of \(A\). The set \(A/{\sim}\) of these classes is a subset of \(\mathcal P(A)\), and hence is a member. \(\square\)

These arguments also show that every relation on two universe members is a universe member. A defining condition may mention parameters outside \(\mathcal U\): the resulting subset remains a member because it is a subset of a member. No restriction on those parameters is required for this subset argument.

3. Transporting the size bound

Proposition 3.1. Subsets, quotients, finite products, power sets and function sets of \(\mathcal U\)-small sets are \(\mathcal U\)-small. If \(J\) is \(\mathcal U\)-small and every \(A_j\) is \(\mathcal U\)-small, then both \(\coprod_j A_j\) and \(\prod_j A_j\) are \(\mathcal U\)-small. The ordinary union \(\bigcup_j A_j\) is also \(\mathcal U\)-small.

Proof. Choose a bijection \(b:A\to A_0\) with \(A_0\in\mathcal U\). A subset \(S\subseteq A\) is bijective to \(b(S)\subseteq A_0\). The map \(S\mapsto b(S)\) gives a bijection \(\mathcal P(A)\to\mathcal P(A_0)\). An equivalence relation on \(A\) transports to the relation on \(A_0\) given by

\[ b(a)\sim_0 b(a')\quad\Longleftrightarrow\quad a\sim a'. \tag{3.1} \]

Sending a class to its image under \(b\) gives a bijection of the two quotients. For a second bijection \(c:B\to B_0\), the product map \((a,d)\mapsto(b(a),c(d))\) is a bijection of products, and \(f\mapsto cfb^{-1}\) is a bijection between the function sets. Proposition 2.1 applies to every resulting representative. Finite products follow by iteration, with the empty product represented by a singleton in \(\mathcal U\).

For an indexed family, choose a bijection \(e:J_0\to J\), where \(J_0\in\mathcal U\). By ambient choice, choose for each \(i\in J_0\) a member \(B_i\in\mathcal U\) and a bijection \(b_i:A_{e(i)}\to B_i\). These choices form a set-indexed family: the possible representatives are drawn from the set \(\mathcal U\), and the possible bijections are sets of functions. Proposition 2.1 gives universe members \(\coprod_i B_i\) and \(\prod_i B_i\). The explicit bijections are

\[ \begin{aligned} (e(i),a)&\longmapsto(i,b_i(a)),\\ (a_j)_{j\in J}&\longmapsto(b_i(a_{e(i)}))_{i\in J_0}. \end{aligned} \tag{3.2} \]

Their inverses use \(e^{-1}\) and the inverses of the \(b_i\). These maps also apply to empty products and to products with empty factors. We are not assuming that a product is nonempty.

The union is the image of the disjoint union under \((j,a)\mapsto a\). Any image is bijective to the quotient of its domain by the equivalence relation “has the same image.” The already proved quotient case shows that the union is small, including when it is empty. \(\square\)

The conclusion is smallness, rather than literal membership. For example \(\{\mathcal U\}\) remains small after forming a singleton family or applying its identity function; it does not become a universe member.

4. Two distinct size conditions on a category

A category in this lesson has a set of objects in the ambient setting and a set \(\operatorname{Hom}_{\mathsf C}(X,Y)\) for every ordered pair of objects. Composition sends \(g:Y\to Z\) and \(f:X\to Y\) to \(g\circ f:X\to Z\) and is associative. Each object has an identity \(1_X\) such that \(f\circ1_X=f\) and \(1_X\circ h=h\) whenever these composites are defined. This identity is unique: if \(e,e'\) both satisfy these laws at \(X\), then \(e=e\circ e'=e'\). It is locally \(\mathcal U\)-small if all these Hom sets are \(\mathcal U\)-small. It is \(\mathcal U\)-small if it is locally \(\mathcal U\)-small and its object set is \(\mathcal U\)-small. The first condition is also called being a \(\mathcal U\)-category in [SGA 4, Univers, Exposé I, Definition 1.1]. It does not impose the second condition.

Proposition 4.1. A \(\mathcal U\)-small category is isomorphic to a category whose object set, arrow set, source and target maps, identity map and partially defined composition are all encoded by members of \(\mathcal U\).

Proof. Choose a bijection \(e:I\to\operatorname{Ob}(\mathsf C)\) with \(I\in\mathcal U\). For every \((i,j)\in I^2\), choose a member \(H_{ij}\in\mathcal U\) and a bijection

\[ b_{ij}:\operatorname{Hom}_{\mathsf C}(e(i),e(j))\longrightarrow H_{ij}. \tag{4.1} \]

Ambient choice permits these choices, since \(I^2\) is a set. Let the new objects be \(I\), and let the arrows from \(i\) to \(j\) be the triples \((i,j,h)\) with \(h\in H_{ij}\). Ordered triples are encoded using ordered pairs. The disjoint union over \(I^2\) of these tagged sets is a member of \(\mathcal U\), by Proposition 2.1.

Transport identities through \(b_{ii}\). For composable arrows, define

\[ \begin{aligned} (j,k,v)\circ(i,j,u)&=(i,k,w),\\ w&=b_{ik}(v'\circ u'),\\ v'&=b_{jk}^{-1}(v),\\ u'&=b_{ij}^{-1}(u). \end{aligned} \tag{4.2} \]

The original identity and associativity equations give the new equations by applying the bijections \(b_{ij}\). Sending an old object \(e(i)\) to \(i\) and an old arrow \(f:e(i)\to e(j)\) to \((i,j,b_{ij}(f))\) is a functor bijective on objects and on every Hom set. The inverse bijections preserve the transported operations, so they define its inverse functor. This is an isomorphism of categories, not just an equivalence.

The source and target graphs are subsets of the product of the arrow set and \(I\). The identity graph is a subset of the product of \(I\) and the arrow set. The set of composable pairs is a subset of the square of the arrow set, and the composition graph is a subset of its cube. All are members by Proposition 2.1. Their finite tuple is likewise a member. \(\square\)

In the other direction, a category with object and arrow sets in \(\mathcal U\) is \(\mathcal U\)-small: each Hom set is a subset of its arrow set. Thus “small up to bijections” admits a complete encoding inside the universe. It does not assert that the original labels were already there.

Example 4.2. Let \(\mathsf{Set}_{\mathcal U}\) have the members of \(\mathcal U\) as objects and all functions as morphisms. Its Hom sets belong to \(\mathcal U\), by Proposition 2.1, but its object set is \(\mathcal U\). Lemma 1.1 shows that this category is not \(\mathcal U\)-small. In a larger universe \(\mathcal V\) containing \(\mathcal U\), its objects and arrows are sets in \(\mathcal V\): use the \(\mathcal V\)-indexed union of the tagged function sets over \(\mathcal U^2\). No collection of all sets without a size bound is needed.

Even choosing one representative from each isomorphism class cannot make this example small. Suppose a \(\mathcal U\)-small family \((A_j)\) of universe members represented every such class. Reindex it by a member of \(\mathcal U\), and put \(B=\bigcup_j A_j\in\mathcal U\). Every \(A_j\) injects into \(B\). The object \(\mathcal P(B)\in\mathcal U\) cannot be isomorphic to any \(A_j\), because that would give an injection \(\mathcal P(B)\to B\). This contradicts Lemma 1.1's Cantor argument. So \(\mathsf{Set}_{\mathcal U}\) is not essentially \(\mathcal U\)-small either.

5. Natural transformations have a small component set

Proposition 5.1. Let \(\mathsf C\) have a \(\mathcal U\)-small object set, and let \(\mathsf D\) be locally \(\mathcal U\)-small. For any functors \(F,G:\mathsf C\to\mathsf D\), the set of natural transformations \(\operatorname{Nat}(F,G)\) is \(\mathcal U\)-small. The Hom sets of \(\mathsf C\) need not be \(\mathcal U\)-small for this assertion.

Proof. A transformation is determined by its component \(\alpha_X:F(X)\to G(X)\) at every object. Its component family belongs to

\[ \prod_{X\in\operatorname{Ob}(\mathsf C)} \operatorname{Hom}_{\mathsf D}(F(X),G(X)). \tag{5.1} \]

This product is \(\mathcal U\)-small by Proposition 3.1. The transformations are exactly its subset defined by \(G(f)\alpha_X=\alpha_YF(f)\) for every arrow \(f:X\to Y\). Subsets of small sets are small. The equations may use an ambient set of arrows larger than \(\mathcal U\); imposing equations does not enlarge the component set. When \(\mathsf C\) is empty, the product and the transformation set are both singletons. \(\square\)

Consequently a functor category with a \(\mathcal U\)-small domain and locally \(\mathcal U\)-small target is locally \(\mathcal U\)-small. This says nothing about its object set being \(\mathcal U\)-small. In a larger universe containing the two categories' object and arrow sets, functors are a subset of pairs of functions on those sets, so their collection is a set there.

This is the size convention used when the index of a formal colimit is called small. For applications, see Ind-objects through their elements and Small modules with too many scalar operators. Local smallness allows each Hom construction; a small index additionally permits the component products and disjoint unions proved above.

6. Exercises

Exercise 1 — Graphs and classes (introductory). Let \(A,B\in\mathcal U\), and let \(E\subseteq A\times A\) be any equivalence relation. Show directly that the sets of injective functions \(A\to B\), surjective functions \(A\to B\), and equivalence classes \(A/E\) belong to \(\mathcal U\). Determine the function set and the surjection set when \(A=\varnothing\).

Exercise 2 — Labels outside the universe (intermediate). Let \(A\in\mathcal U\) be nonempty and put \(X=\{\mathcal U\}\times A\), using the ordered-pair convention (2.1). Prove that \(X\) and \(X^X\) are \(\mathcal U\)-small, but neither is a member of \(\mathcal U\).

Exercise 3 — A large discrete domain (advanced). Let \(\mathsf C\) be the discrete category with object set \(\mathcal U\), in an ambient universe containing \(\mathcal U\). Let \(F:\mathsf C\to\mathsf{Set}_{\mathcal U}\) be constant with value a singleton, and let \(G\) be constant with value \(\{0,1\}\). Compute \(\operatorname{Nat}(F,G)\) and show that it is not \(\mathcal U\)-small, even though both categories are locally \(\mathcal U\)-small.

Exercise 4 — Equivalence does not count object labels (challenging). Form a category \(\mathsf E\) with object set \(\mathcal U\) and exactly one arrow between every ordered pair of objects. Prove that it is a locally \(\mathcal U\)-small groupoid, is not a \(\mathcal U\)-small category, and is equivalent to the category with one object and only its identity. Give the two functors and both natural comparison isomorphisms explicitly. Explain why this does not conflict with Example 4.2.

7. Solutions

1. Graphs and classes

Solution. The function set \(B^A\) is a subset of \(\mathcal P(A\times B)\), a member of \(\mathcal U\). The subsets cut out by injectivity or surjectivity are therefore members as well. Every \(E\)-class is a subset of \(A\), and the collection of classes is a subset of \(\mathcal P(A)\); applying subset closure gives \(A/E\in\mathcal U\). No choice of representatives is needed. For \(A=\varnothing\), there is exactly one function to any \(B\): its graph is \(\varnothing\). It is injective. It is surjective exactly when \(B=\varnothing\). Thus the function set is \(\{\varnothing\}\), and the surjection set is that singleton for empty \(B\) and the empty set otherwise.

2. Labels outside the universe

Solution. Projection \((\mathcal U,a)\mapsto a\) is a bijection \(X\to A\), so \(X\) is small. Conjugation by this bijection gives \(X^X\simeq A^A\), which is small by Proposition 2.1.

For each \(a\in A\), the pair \(x=(\mathcal U,a)\) contains \(\{\mathcal U\}\) as an element. If \(x\in\mathcal U\), transitivity would put \(\{\mathcal U\}\) and then \(\mathcal U\) in \(\mathcal U\), contrary to Lemma 1.1. Choose one \(a\), using nonemptiness. If \(X\in\mathcal U\), its element \(x\) would be in \(\mathcal U\), which we have just ruled out.

The identity function of \(X\) is an element of \(X^X\). If \(X^X\in\mathcal U\), its identity graph would be in \(\mathcal U\). This graph contains \((x,x)=\{\{x\}\}\); repeated transitivity would give \(x\in\mathcal U\), again impossible. Hence \(X^X\notin\mathcal U\). The hypothesis that \(A\) is nonempty is necessary: for empty \(A\), the sets are \(\varnothing\) and \(\{\varnothing\}\), both universe members.

3. A large discrete domain

Solution. A transformation assigns to each object \(a\in\mathcal U\) a map from the singleton to \(\{0,1\}\). Naturality imposes no further condition, because the only arrows are identities. Thus choosing the image of the singleton gives a bijection

\[ \operatorname{Nat}(F,G)\simeq\{0,1\}^{\mathcal U}. \tag{7.1} \]

For each \(a\in\mathcal U\), the characteristic function of \(\{a\}\subseteq\mathcal U\) is an element of this function set. These functions give an injection \(\mathcal U\to\{0,1\}^{\mathcal U}\). If the latter were small, its subset consisting of these characteristic functions would be small by Proposition 3.1, and its bijection with \(\mathcal U\) would contradict Lemma 1.1. The discrete category has only empty or singleton Hom sets, and Example 4.2 proves local smallness of the target. The missing hypothesis in Proposition 5.1 is precisely smallness of the domain's object set.

4. Equivalence and object labels

Solution. Denote the unique arrow from \(a\) to \(b\) by \(u_{ab}\), and set \(u_{bc}u_{ab}=u_{ac}\). Both parenthesizations of any triple composite give the same unique arrow, and \(u_{aa}\) is the identity at \(a\). The arrow \(u_{ba}\) is inverse to \(u_{ab}\), so this is a groupoid. Every Hom set is a singleton, hence small. Its object set is \(\mathcal U\), which is not small by Lemma 1.1.

Let \(\mathsf 1\) be the one-object identity-only category. The functor \(q:\mathsf E\to\mathsf 1\) sends every object and arrow to the unique object and its identity. Choose \(a_0=\varnothing\in\mathcal U\) and define \(r:\mathsf 1\to\mathsf E\) by selecting \(a_0\) and \(u_{a_0a_0}\). Then \(qr\) equals the identity functor of \(\mathsf 1\). The components \(u_{a_0a}:rq(a)\to a\) give a natural isomorphism \(rq\to\operatorname{id}_{\mathsf E}\): for \(u_{ab}\), both naturality composites are \(u_{a_0b}\). Their inverses are \(u_{aa_0}\), whose naturality follows in the same way. The comparison for \(qr\) is the identity.

Thus \(\mathsf E\) is essentially small even though it is not small as labelled. It has only one isomorphism class. Example 4.2 has no small family representing all its isomorphism classes; its obstruction survives equivalence. The two examples distinguish an excess of labels from an excess of isomorphism classes.

References