Category theory

Equality and Equivalence

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



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 CC is defined to be a skeletal subcategory of CC whose inclusion functor exhibits it as equivalent to CC.

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


Existence of skeletons


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


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 sk(C)sk(C) containing just the chosen objects. Denote by in:sk(C)Cin\colon sk(C)\to C the inclusion. We exhibit a weak inverse of inin as a functor :xx-':x\mapsto x' constructed as follows. For every object xx one has chosen already the unique object xx' in sk(X)sk(X) isomorphic to xx, but one also needs to make a choice of isomorphism i x:xxi_x\colon x\to x' for every xx. This enables to conjugate between C(x,y)C(x,y) and C(x,y)C(x',y') by

(xfy)(xi x 1xfyi yy). (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 :ff:=i yfi x 1-': f\mapsto f' := i_y\circ f\circ i_{x}^{-1} is clearly functorial. Let us show that -' is a weak inverse of inin. In one direction, given ysk(C)y\in sk(C) we compute (in y)=y(in_{y})' = y (strict equality); in another direction, given xCx\in C, notice that i in x 1:in xxi^{-1}_{in_{x'}}:in_{x'}\cong x for xCx\in C is an isomorphism. It suffices to show that these isomorphisms for all xCx\in C together form a natural isomorphism i in 1:in id Ci^{-1}_{in}:in_{-'}\to id_C; the naturality diagram is commutative precisely because of the conjugation formula for the functor -' for morphisms. This completes the proof that -' is indeed a weak inverse of inin.

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:ABp:A\to B in SetSet, make AA into a category with a unique isomorphism aaa\cong a' iff p(a)=p(a)p(a)=p(a'); then a skeleton of AA supplies a splitting of pp.

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:


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


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

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 1X 0X_1 \rightrightarrows X_0 to have a skeleton is the existence quotient X 0/X 1X_0/X_1 - the object of orbits under the action of the core of XX. If the quotient map X 0X 0/X 1X_0 \to X_0/X_1 has a section, then one could consider XX 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 CC to be a skeletal category SS with a surjective equivalence CSC\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\{S_i\}_I of equinumerous sets there exists 0I0 \in I and a family of isomorphisms of the permutation groups {Aut(S 0)Aut(S i)} I\{Aut(S_0) \to Aut(S_i)\}_I.
  2. Given a family {S i} I\{S_i\}_I of inhabited equinumerous sets, there exists a family (x i) I(x_i)_I such that x iS ix_i \in S_i for all iIi \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 AA and BB as being the object A×BA\times B, strictly speaking a product consists of the object A×BA\times B together with the projections A×BAA\times B\to A and A×BBA\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×BA\times B that is a product of AA and BB, in general this object can still “be the product of AA and BB” in many different ways.

For example, given any projections A×BAA\times B\to A and A×BBA\times B\to B that exhibit A×BA\times B as a product of AA and BB, we can compose them both with any automorphism of A×BA\times B to get a new, different, pair of projections that also exhibit A×BA\times B as a product of AA and BB. In fact, the universal property of a product implies that any two pairs of projections are related by an automorphism of A×BA\times B, so this example is generic. Thus, even in a skeletal category, we cannot speak of “the” product of AA and BB, 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 AA and BB,” 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.)

Last revised on November 28, 2012 at 14:07:26. See the history of this page for a list of all contributions to it.