Initial objects, universal colimits and pointed 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); referenced Mathlib proofs retain Apache-2.0.

A colimit describes how objects are assembled. Pullback restricts that assembly to a new base. Requiring these operations to commute is a strong geometric condition. Its smallest test is the empty diagram: an initial object must remain initial after every pullback. In a category with a zero object, that requirement already forces every object to be zero. Even restricting the test to binary coproducts gives the same obstruction.

We assume familiarity with initial and terminal objects, pullbacks and the universal properties of colimits. Dense probes and reconstruction from colimits, Section 3 introduces stability under base change; its Section 5 explains the fibrewise calculation for sets. Here we keep the index category arbitrary whenever its colimits exist. Basic references are [Mathlib], [Riehl] and these preceding lessons. The product consequence for a strict initial object uses the exact open proof identified in Section 2.

1. A colimit viewed over a base

Fix a category \(\mathsf C\) with pullbacks and an index category \(I\) whose colimits exist in \(\mathsf C\). Neither nonemptiness nor filteredness is assumed. For an object \(Z\), the slice \(\mathsf C/Z\) has objects \(a:X\to Z\); its arrows commute with the maps to \(Z\). An arrow \(u:Y\to Z\) defines the pullback functor

\[ \begin{gathered} u^*:\mathsf C/Z\longrightarrow\mathsf C/Y,\\ (X\to Z)\longmapsto(X\times_ZY\to Y). \end{gathered} \tag{1.1} \]

An \(I\)-diagram \((X_i\to Z)\) has a colimit in the slice computed by its underlying colimit in \(\mathsf C\). Indeed, let \(j_i:X_i\to L\) be that colimit. The compatible maps to \(Z\) give a unique \(a:L\to Z\). For any cocone to \(b:T\to Z\), the underlying unique map \(h:L\to T\) satisfies \(bhj_i=aj_i\) for all \(i\). Uniqueness in the colimit property gives \(bh=a\), so \(h\) is an arrow in the slice. This proves existence and uniqueness there as well. It also works for the empty diagram.

We say that \(I\)-colimits are stable under base change, or universal, if every functor (1.1) preserves them. Equivalently, for any diagram \(X:I\to\mathsf C\), any arrow \(a:L\to Z\) from its colimit and any \(u:Y\to Z\), the canonical comparison is invertible:

\[ \begin{gathered} \operatorname{colim}_{i\in I}(X_i\times_ZY)\\ \longrightarrow L\times_ZY,\\ L=\operatorname{colim}_{i\in I}X_i. \end{gathered} \tag{1.2} \]

The maps \(X_i\to Z\) in (1.2) are \(aj_i\). Their pullbacks give a cocone to \(L\times_ZY\), hence the comparison. Conversely, every diagram in the slice is of this form by the preceding calculation. Thus the displayed condition and preservation in slices are exactly equivalent. If all small colimits exist, requiring this for every small \(I\) defines universal small colimits. Requiring it just for finite \(I\) defines universal finite colimits; the empty category is included.

The issue is not whether a diagram happens to have a colimit. It is whether the specified colimit cocone still has its universal property over the new base.

2. Empty diagrams and strict initial objects

Let \(0\) be an initial object of \(\mathsf C\). It is strict initial if every arrow \(f:X\to0\) is invertible. Initiality alone asserts uniqueness of arrows from \(0\); it need not assert anything about arrows to \(0\).

In the slice \(\mathsf C/Z\), the unique arrow \(0\to Z\) is initial. To see this, the unique \(0\to T\) is automatically over \(Z\), because its composite to \(Z\) must be the unique arrow from \(0\) to \(Z\).

Theorem 2.1. In a category with pullbacks and an initial object \(0\), empty colimits are stable under every base change if and only if \(0\) is strict initial.

Proof. Suppose empty colimits are stable. Given \(f:X\to0\), pull back the initial object \(\operatorname{id}_0\) of \(\mathsf C/0\) along \(f\). The result is \(\operatorname{id}_X\), so it is initial in \(\mathsf C/X\). Put \(v:0\to X\). Initiality of \(\operatorname{id}_X\) gives an arrow from it to \(v\), namely \(r:X\to0\) satisfying

\[ vr=\operatorname{id}_X,\qquad rv=\operatorname{id}_0. \tag{2.1} \]

The second equality follows from initiality of \(0\). Also \(fv=\operatorname{id}_0\), so \(f=fvr=r\). Thus \(f\) is invertible, with inverse \(v\).

Conversely, suppose \(0\) is strict initial. For \(u:Y\to Z\), form \(P=0\times_ZY\). Its projection \(P\to0\) is invertible, so \(P\) is initial in \(\mathsf C\). Its map \(P\to Y\) is therefore initial in \(\mathsf C/Y\), by the slice argument above. This is precisely preservation of the empty colimit by \(u^*\). \(\square\)

Corollary 2.2. If \(\mathsf C\) has finite limits and finite colimits stable under base change, then every arrow to \(0\) is invertible and, for every \(X\),

\[ 0\times X\simeq0. \tag{2.2} \]

The strictness assertion is Theorem 2.1 applied to the empty diagram. For the product assertion we use [Mathlib], the exact declarations strictness transfer from a chosen initial object and isInitialMul. The first transfers strictness of a chosen initial object to all initial objects. The second gives the isomorphism (2.2), with the first projection as its forward map. Its hypotheses require only strict initiality and the existence of this binary product; finite colimits are used here solely to obtain strictness. These are complete open proofs by Bhavik Mehta and Mathlib contributors, released under Apache-2.0.

Sets illustrate the distinction. An arrow \(S\to\varnothing\) exists only for \(S=\varnothing\), so the initial set is strict. The singleton is terminal but is not initial: there is no function from it to the empty set. Thus strict initiality does not force an arbitrary category to collapse.

3. The obstruction in a pointed category

A category is pointed if it has a zero object \(0\), meaning an object that is both initial and terminal. There is then a distinguished zero arrow from \(X\) to \(Y\), obtained by composing \(X\to0\to Y\).

Theorem 3.1. Suppose \(\mathsf C\) is pointed and has pullbacks. The following conditions are equivalent:

  1. Empty colimits are stable under base change.
  2. Every object is isomorphic to \(0\).
  3. \(\mathsf C\) is equivalent to the category \(\mathsf{Pt}\) with one object and one arrow.

Proof. Condition 1 makes \(0\) strict by Theorem 2.1. Terminality supplies an arrow \(X\to0\) for every \(X\), and strictness makes every such arrow an isomorphism. This proves 2.

Under 2, every arrow \(X\to Y\) corresponds, using isomorphisms with \(0\), to an endomorphism of \(0\). There is exactly one such endomorphism. Hence every Hom collection in \(\mathsf C\) has exactly one element, and the functor from \(\mathsf{Pt}\) that selects \(0\) is fully faithful and essentially surjective. It is an equivalence, proving 3.

Under 3, the same full-faithfulness and essential-surjectivity statements show that every Hom collection has one element and every object is isomorphic to \(0\). In particular every arrow \(X\to0\) is invertible: the unique arrow \(0\to X\) is its inverse, since each endomorphism collection is a singleton. Theorem 2.1 now gives 1. \(\square\)

Corollary 3.2. An abelian category with finite colimits stable under base change is equivalent to \(\mathsf{Pt}\).

Proof. An abelian category has a zero object and finite limits and colimits. Universal finite colimits include universal empty colimits. Apply Theorem 3.1. \(\square\)

No smallness assumption on the collection of objects is needed for this conclusion. In a nontrivial abelian category, even the empty colimit fails the base-change test: pulling back \(0\to0\) along \(X\to0\) produces \(X\to X\), which is initial in the slice only if \(X\simeq0\).

4. Binary coproducts give a second test

The obstruction also has a nonempty form. Write \(j_1,j_2:X\to X\sqcup X\) for the coproduct arrows and \(h:X\sqcup X\to X\) for the fold map, defined by

\[ hj_1=\operatorname{id}_X=hj_2. \tag{4.1} \]

Theorem 4.1. Suppose \(\mathsf C\) is pointed and has pullbacks and binary coproducts. Binary coproducts are stable under every base change if and only if \(\mathsf C\simeq\mathsf{Pt}\).

Proof. In \(\mathsf C/0\), the coproduct of two copies of \(0\to0\) is again \(0\to0\): the unique maps from the two copies give a unique map to any object. Pulling back this cocone along \(X\to0\) gives two copies of \(\operatorname{id}_X\), with both cocone arrows equal to the identity. The comparison from their coproduct in \(\mathsf C/X\) is exactly the fold map (4.1). Stability makes \(h\) invertible. Cancelling \(h\) in (4.1) gives \(j_1=j_2\).

For any two arrows \(f,g:X\to Y\), the coproduct property gives \(a:X\sqcup X\to Y\) with \(aj_1=f\) and \(aj_2=g\). Equality of the coproduct arrows gives \(f=g\). There is at least one arrow between any two objects, namely the zero arrow. Thus every Hom collection is a singleton. In particular the arrows \(X\to0\) and \(0\to X\) are inverse, and the equivalence constructed in Theorem 3.1 applies.

Conversely, in a category equivalent to \(\mathsf{Pt}\), every Hom collection is a singleton. Any object, with the unique cocone arrows, is a colimit of any diagram. This remains true in every slice, whose Hom collections are again singletons. Therefore pulling back any binary-coproduct cocone leaves a colimit cocone. \(\square\)

This proof uses no additive structure. In an additive category, the failure can be measured by the kernel of the fold map; the second exercise computes it. The base-change condition concerns a pullback functor between slices, so the fact that direct sums have their familiar universal property in the original category does not imply this condition.

5. Four graded exercises with full solutions

Exercise 1 (introductory: adding a distinguished point). Let \(\mathsf C\) have a terminal object \(1\) and pullbacks. The category \(1/\mathsf C\) has objects \(a:1\to X\) and arrows that preserve these chosen points. Show that it has a zero object and pullbacks, and conclude that universal empty colimits in \(1/\mathsf C\) force it to be equivalent to \(\mathsf{Pt}\). For sets with a distinguished point, exhibit the failed empty-colimit test on a two-element pointed set.

Solution. The object \(\operatorname{id}_1:1\to1\) is initial: its only arrow to \(a:1\to X\) is \(a\). It is also terminal: the unique \(X\to1\) preserves the chosen point because its composite with \(a\) is the unique endomorphism of \(1\), hence the identity.

Given point-preserving arrows \((X,a)\to(Z,c)\leftarrow(Y,b)\), the pullback \(P=X\times_ZY\) in \(\mathsf C\) has a unique point \(d:1\to P\) whose two projections are \(a,b\). They are compatible because both composites to \(Z\) are \(c\). For any pointed cone, its unique underlying map to \(P\) preserves the point: compare its composite from \(1\) with \(d\) on both projections. This proves the pullback universal property in \(1/\mathsf C\). Theorem 3.1 gives the stated collapse.

For pointed sets, let \(X=\{*,x\}\) with \(x\ne*\). Pulling back the zero object \(1\to1\) along the unique pointed \(X\to1\) gives \(\operatorname{id}_X\). This is not initial in the slice over \(X\). Indeed, an arrow from it to the pointed inclusion \(1\to X\) would be a pointed map \(r:X\to1\) whose composite back to \(X\) is the identity. That composite is constant at \(*\), so it cannot fix \(x\). The empty-colimit test fails.

Exercise 2 (intermediate: the additive fold). In an additive category, prove that the kernel of \(h:X\oplus X\to X\), \(h=(1,1)\), is the map \(\kappa:X\to X\oplus X\) with components \((1,-1)\). Show that \(h\) is split surjective and is invertible exactly when \(X\) is a zero object. Interpret this calculation as the failed pullback comparison for the coproduct of two zero objects. The argument must work in characteristic two as well.

Solution. We have \(h\kappa=1_X-1_X=0\). If \(a:T\to X\oplus X\) has components \(u,v\) and \(ha=0\), then \(u+v=0\), so \(v=-u\) and \(a=\kappa u\). This factorization is unique because the first projection satisfies \(\pi_1\kappa=1_X\). Thus \(\kappa\) is a kernel, without any assumption that other kernels exist.

The first coproduct arrow \(j_1\) satisfies \(hj_1=1_X\), so \(h\) has a section. If \(h\) is invertible, \(h\kappa=0\) forces \(\kappa=0\), and hence \(1_X=\pi_1\kappa=0\). This makes \(X\) a zero object: every \(f:X\to Y\) equals \(f1_X=0\), and every \(g:Y\to X\) equals \(1_Xg=0\). Conversely, if \(X\) is a zero object, both \(X\oplus X\) and \(X\) are zero objects, and their unique connecting arrow is invertible.

As in Theorem 4.1, pulling back the coproduct of \(0\to0\) with itself along \(X\to0\) produces the identity cocone on two copies of \(X\to X\). Its comparison is \(h\). A nonzero \(X\) therefore gives an explicit failure, even for this nonempty finite diagram. When the Hom groups have characteristic two, \(-1_X=1_X\), but \(1_X-1_X=0\) and the component factorization remain valid. There is no division by two.

Exercise 3 (advanced: a strict initial object is not enough). Regard a bounded lattice \(L\) as a category, with one arrow \(A\to B\) when \(A\le B\). Prove that its initial object is strict and empty colimits are universal. Show that finite colimits are universal exactly when

\[ \begin{gathered} Y\wedge(A\vee B)\\ =(Y\wedge A)\vee(Y\wedge B) \end{gathered} \tag{5.1} \]

for all \(A,B,Y\in L\). Give a finite lattice in which empty colimits are universal but binary coproducts are not.

Solution. The least element \(0\) is initial. An arrow \(A\to0\) means \(A\le0\), hence \(A=0\), so initiality is strict. Pullbacks are meets; pulling back \(0\to Z\) along \(Y\le Z\) gives \(0\wedge Y=0\), proving empty stability directly.

In the slice over \(Z\), objects are the elements at most \(Z\). A finite diagram has as its colimit the join of its object values, with the empty join equal to \(0\). This follows because a cocone to \(T\) means that every object value is at most \(T\); the least such \(T\) is their join. The join remains at most \(Z\). Pullback along \(Y\le Z\) sends each value \(A\) to \(A\wedge Y\). Thus binary stability is the identity (5.1) for \(A,B,Y\le Z\). Taking \(Z\) to be the greatest element makes this necessary for all triples.

Conversely, (5.1), iterated through a finite list of object values, says that meet with \(Y\) preserves every nonempty finite join. It preserves the empty join because \(Y\wedge0=0\). The slice calculation therefore proves stability of all finite diagrams. This is the distributive-lattice condition in its meet-over-join form. It also implies the other distributive law: expand \((A\vee B)\wedge(A\vee C)\) first with respect to \(A\vee C\) and then with respect to \(A\vee B\). Absorption gives \(A\vee(A\wedge C)\vee(B\wedge C)=A\vee(B\wedge C)\).

For the counterexample take the five-element diamond \(\{0,a,b,c,1\}\), with three distinct incomparable atoms. Distinct atoms have meet \(0\) and join \(1\). Pull the coproduct \(a\vee b=1\) over the base \(1\) back along \(c\to1\). The pulled-back objects are \(a\wedge c=b\wedge c=0\), whose coproduct is \(0\); the pulled-back original colimit is \(1\wedge c=c\). Its comparison \(0\to c\) is not invertible. The strict initial object controls the empty test, but not this binary test.

Exercise 4 (advanced: a functor that preserves every colimit). Let \(\mathsf{Set}_*\) be pointed sets, and put \(L(S)=S\sqcup\{*\}\), with a new distinguished point. Prove that \(L:\mathsf{Set}\to\mathsf{Set}_*\) is left adjoint to forgetting the point and preserves every small colimit, including the empty one. Show that it preserves neither terminal objects nor binary products. Explain why its colimit preservation does not carry universal colimits from sets to pointed sets.

Solution. A pointed map \(S\sqcup\{*\}\to(X,x_0)\) is uniquely determined by its restriction \(S\to X\); its value on the new point must be \(x_0\). Restriction and this extension are inverse and commute with precomposition in \(S\) and postcomposition by pointed maps in \(X\). They give the natural adjunction bijection

\[ \begin{gathered} \operatorname{Hom}_{\mathsf{Set}_*}(L(S),X)\\ \simeq\operatorname{Hom}_{\mathsf{Set}}(S,U(X)). \end{gathered} \tag{5.2} \]

For a small diagram \(S_i\) with set colimit \(S\), a compatible family of pointed maps \(L(S_i)\to X\) corresponds to a compatible family of functions \(S_i\to U(X)\). The set colimit gives a unique function \(S\to U(X)\), whose unique pointed extension is \(L(S)\to X\). Thus the cocone \(L(S_i)\to L(S)\) has the colimit property in pointed sets. For the empty diagram, \(S=\varnothing\) and \(L(S)\) is the one-point set, the initial pointed set, so this case is included.

The terminal set is a singleton, and its image under \(L\) has two elements. The terminal pointed set has one element, so terminal objects are not preserved. Products in pointed sets are underlying products with the pair of distinguished points, as follows from the pullback calculation in Exercise 1. The comparison

\[ L(S\times T)\longrightarrow L(S)\times L(T) \tag{5.3} \]

sends \((s,t)\) to \((s,t)\) and the new point to \((*,*)\). When \(S,T\) are nonempty it misses every mixed pair \((s,*)\) and \((*,t)\), so it is not an isomorphism.

Sets have universal small colimits by the fibrewise proof in Dense probes, Section 5. Pointed sets fail even the empty test in Exercise 1. The functor \(L\) preserves colimits, but those facts alone give no preservation of the pullback comparisons (1.2): the product comparisons (5.3) already fail. Thus preserving colimits is insufficient to transport their stability under base change. The example tests exactly the extra pullback requirement.

6. References