nLab
skeleton

This entry is about the notion of skeleton in category theory. For the notion of (co)skeletal simplicial sets see at simplicial skeleton.

Definitions

A category is skeletal if objects that are isomorphic are necessarily equal. (So this is a notion irredeemably violating the principle of equivalence of category theory.)

Traditionally, a skeleton of a category $C$ is defined to be a skeletal subcategory of $C$ whose inclusion functor exhibits it as equivalent to $C$.

However, in the absence of the axiom of choice, it is more appropriate to define a skeleton of $C$ to be any skeletal category which is weakly equivalent to $C$.

Properties

Existence of skeletons

Theorem

If the axiom of choice holds, then every category $C$ has a skeleton (in the strongest sense).

Proof

Simply choose one object in each isomorphism class and one isomorphism to that object from each other object in that class.

In more detail, generate the full subcategory $\mathrm{sk}(C)$ containing just the chosen objects. Denote by $\mathrm{in}:\mathrm{sk}(C)\to C$ the inclusion. We exhibit a weak inverse of $\mathrm{in}$ as a functor $-\prime :x\mapsto x\prime$ constructed as follows. For every object $x$ one has chosen already the unique object $x\prime$ in $\mathrm{sk}(X)$ isomorphic to $x$, but one also needs to make a choice of isomorphism $i_x:x\to x\prime$ for every $x$. This enables to conjugate between $C(x,y)$ and $C(x\prime ,y\prime )$ by

$$(x\stackrel{f}{\to }y)\mapsto (x\prime \stackrel{i_x^{-1}}{\to }x\stackrel{f}{\to }y\stackrel{i_y}{\to }y\prime ).$$(x\stackrel{f}\to y)\mapsto (x'\stackrel{i_x^{-1}}\to x\stackrel{f}\to y\stackrel{i_y}\to y').

The rule for morphisms $-\prime :f\mapsto f\prime :=i_y\circ f\circ i_x^{-1}$ is clearly functorial. Let us show that $-\prime$ is a weak inverse of $\mathrm{in}$. In one direction, given $y\in \mathrm{sk}(C)$ we compute $({\mathrm{in}}_y)\prime =y$ (strict equality); in another direction, given $x\in C$, notice that $i_{{\mathrm{in}}_{x\prime }}^{-1}:{\mathrm{in}}_{x\prime }\cong x$ for $x\in C$ is an isomorphism. It suffices to show that these isomorphisms for all $x\in C$ together form a natural isomorphism $i_{\mathrm{in}}^{-1}:{\mathrm{in}}_{-\prime }\to {\mathrm{id}}_C$; the naturality diagram is commutative precisely because of the conjugation formula for the functor $-\prime$ for morphisms. This completes the proof that $-\prime$ is indeed a weak inverse of $\mathrm{in}$.

In fact, the statement that every (possibly small) category has a skeleton is equivalent to the axiom of choice if “subcategory” and “equivalence” have their naive (‘strong’) meanings. For given a surjection $p:A\to B$ in $\mathrm{Set}$, make $A$ into a category with a unique isomorphism $a\cong a\prime$ iff $p(a)=p(a\prime )$; then a skeleton of $A$ supplies a splitting of $p$.

Even with the more general notion of weak or ana-equivalence of categories, some amount of choice is required to show that every category has a skeleton. It would be interesting to know the precise strength of the statement “every category is weakly equivalent to a skeletal one.” One thing we can say without any choice is:

Theorem

Any thin category (i.e. any preordered set) has a skeleton (in the sense of weak equivalence).

Proof

In this case, we can take the objects of the skeleton of $C$ to be the isomorphism classes of $C$. If $C$ is thin, then we can define a partial ordering on its set of isomorphism classes, making them into a skeleton of $C$.

Notice that the axiom of choice fails in general when one considers internal categories. Hence not every internal category has a skeleton. A necessary condition for an internal category $X_1\rightrightarrows X_0$ to have a skeleton is the existence quotient $X_0/X_1$ - the object of orbits under the action of the core of $X$. If the quotient map $X_0\to X_0/X_1$ has a section, then one could consider $X$ to have a skeleton, but this condition isn’t sufficient for the induced inclusion functor to be a weak equivalence of internal categories when this makes sense (i.e. if the category is internal to a site).

David Roberts: The claim above about the necessity of the existence of the quotient needs to be checked.

Equivalents of choice

Define a coskeleton of a category $C$ to be a skeletal category $S$ with a surjective equivalence $C\to S$. In Categories, Allegories it is shown that the following are equivalent.

1. Any two ana-equivalent categories are strongly equivalent.

   I removed ‘non-ana’, since I don't think that ‘strongly equivalent’ would ever be used in an ‘ana-’ sense. —Toby

   Addendum: Actually, I don't know why I asked whether you meant weakly or strongly here, since obviously one can prove that two ana-equivalent categories are weakly equivalent! It seems that the discussion above used the terms ‘equivalence’ and ‘ana-equivalence’ where equivalence of categories uses ‘strong equivalence’ and ‘weak equivalence’ or ‘ana-equivalence’; so I just changed it. And then I added another entry, which maybe you should remove if Freyd & Scedrov don't actually address it. On the other hand, if they really talk about weak equivalence instead of ana-equivalence (although if they define it in elementary terms, it's hard to tell the difference), maybe there's no need to say ‘ana-’ at all on this page.

2. Any two weakly equivalent categories are strongly equivalent.

3. Every small category has a skeleton.

4. Every small category has a coskeleton.

5. Any two skeletons of a given small category are isomorphic.

6. Any two coskeletons of a given small category are isomorphic.

For convenience we add:

1. Given an inhabited family $\{S_i{\}}_I$ of equinumerous sets there exists $0\in I$ and a family of isomorphisms of the permutation groups $\{\mathrm{Aut}(S_0)\to \mathrm{Aut}(S_i){\}}_I$.
2. Given a family $\{S_i{\}}_I$ of inhabited equinumerous sets, there exists a family $(x_i{)}_I$ such that $x_i\in S_i$ for all $i\in I$.

Uniqueness of constructions

It is well-known that objects defined by universal properties in a category, such as limits and colimits, are not unique on the nose, but only unique up to unique canonical isomorphism. It can be tempting to suppose that in a skeletal category, where any two isomorphic objects are equal, such objects will in fact be unique on the nose. However, under the most appropriate definition of “unique,” this is not true (in general), because of the presence of automorphisms.

More explicitly, consider the notion cartesian product in a category. Although we colloquially speak of “a product” of objects $A$ and $B$ as being the object $A\times B$, strictly speaking a product consists of the object $A\times B$ together with the projections $A\times B\to A$ and $A\times B\to B$ which exhibit its universal property. Thus, even if the category in question is skeletal, so that there can be only one object $A\times B$ that is a product of $A$ and $B$, in general this object can still “be the product of $A$ and $B$” in many different ways.

For example, given any projections $A\times B\to A$ and $A\times B\to B$ that exhibit $A\times B$ as a product of $A$ and $B$, we can compose them both with any automorphism of $A\times B$ to get a new, different, pair of projections that also exhibit $A\times B$ as a product of $A$ and $B$. In fact, the universal property of a product implies that any two pairs of projections are related by an automorphism of $A\times B$, so this example is generic. Thus, even in a skeletal category, we cannot speak of “the” product of $A$ and $B$, except in the same generalized sense that makes sense in any category. A formal way to say this is that the “category of products of $A$ and $B$,” while still equivalent to the trivial category, as it is in any category with products, will not be isomorphic to the trivial category even when the ambient category is skeletal.

(It is true in a few cases, though, that skeletality implies uniqueness on the nose. For instance, a terminal object can have no nonidentity automorphisms, so in a skeletal category, a terminal object (if one exists) really is unique on the nose.)