skeleton

Given a category $\mathcal{C}$, a skeleton is a subcategory that is equivalent to $\mathcal{C}$ but has no “redundant” objects. You can think of a skeleton as the most “economical” model for the category $\mathcal{C}$.

Definition

A skeleton of a category $\mathcal{C}$ is a subcategory $\mathcal{S}$ for which each object in $\mathcal{C}$ is isomorphic to exactly one object in $\mathcal{S}$ and the canonical inclusion functor

$$i:\mathcal{S}⟶\mathcal{C}$$

is fully faithful.

Note, by this lemma $\mathcal{S}$ and $\mathcal{C}$ are equivalent.

Existence of skeleton

Every category has a skeleton. This is because we can define equivalence classes on $\text{ob}(\mathcal{C})$ using isomorphism of objects (which defines an equivalence relation by this).

Then for each equivalence class choose a representative (using the Axiom of Choice and a few other intense details).

Using these representatives, take all the morphism from $\mathcal{C}$ between those objects to ensure that $i:\mathcal{S}\to \mathcal{C}$ is fully faithful looking at morphisms between objects in $\mathcal{S}$.

Examples

• ${\text{Set}}_0^{\text{fin}}$ (the finite sets of natural numbers $\{1,\dots ,n\}$ is a skeleton of ${\text{Set}}^{\text{fin}}$ (finite sets).

• ${\text{vect}}_{k,0}^{\text{fin}}$ is a skeleton of ${\text{vect}}_k^{\text{fin}}$

Equivalence using skeletons

Let ${\mathcal{C}}_1,{\mathcal{C}}_2$ be categories with skeletons ${\mathcal{S}}_1,{\mathcal{S}}_2$. Then ${\mathcal{C}}_1$ and ${\mathcal{C}}_2$ are equivalent if and only if ${\mathcal{S}}_1$ and ${\mathcal{S}}_2$ are isomorphic (i.e. equivalent with only identity natural isomorphisms).

Remark

This means that if you are only using skeletal categories (those that no two items are isomorphic to each other) then you can use a more naive notion of equivalence without quasi-inverses. However, most (interesting) categories aren’t skeletal.

Proof (sketch)

Let $F:{\mathcal{C}}_1\to {\mathcal{C}}_2$ be an equivalence of categories. We aim to build an isomorphism (call it $\tilde{F}$) between ${\mathcal{S}}_1$ and ${\mathcal{S}}_2$.

For each object $z\in {\mathcal{C}}_2$ pick an isomorphism $g_z:z\to s\in {\mathcal{S}}_2$. The object $s\in S_2$ is unique since ${\mathcal{S}}_2$ is a skeleton. Now for $x\in {\mathcal{S}}_1$, let $\tilde{F}(x)$ be the unique object in ${\mathcal{S}}_2$ that is isomorphic to $F(x)$. Next, we need to define what the (proposed) functor does to morphisms. We do this using the maps we know we have coming from the equivalence $F$ and the isomorphism Given a morphism $x\stackrel{f}{\to }y$ in ${\mathcal{S}}_1$, define $\tilde{F}(f):\tilde{F}(x)\to \tilde{F}(y)$ to be the composition

$$\tilde{F}(x)\stackrel{g_{F(x)}^{-1}}{\to }F(x)\stackrel{F(f)}{\to }F(y)\stackrel{g_{F(y)}}{\to }\tilde{F}(y)$$

Since we built everything using a functor and isomorphisms $\tilde{F}$ is a functor.

We can now build an inverse functor $\tilde{G}:{\mathcal{S}}_2\to {\mathcal{S}}_1$ that is strict (has no need for a quasi-inverse the objects will be identical, not simply isomorphic). Given $z\in {\mathcal{S}}_2$, let $\tilde{G}(z)\in {\mathcal{S}}_1$ be the unique object of ${\mathcal{S}}_1$ such that

$$\tilde{F}(\tilde{G}(z))\simeq z.$$

The fact that this exists uses that $F$ is essentially surjective.

Given a morphism $h:z\to w$ in ${\mathcal{S}}_2$, define $\tilde{G}(h):\tilde{G}(z)\to \tilde{G}(w)$ so that

$$\tilde{F}(\tilde{G}(h))=h.$$

This uses that $F$ is fully faithful.

Therefore,

$$\tilde{F}\circ \tilde{G}={\text{id}}_{{\mathcal{S}}_2}\text{and}\tilde{G}\circ \tilde{F}={\text{id}}_{{\mathcal{S}}_1}$$

For the other direction, assume that ${\mathcal{S}}_1$ and ${\mathcal{S}}_2$ are isomorphic. We can define an equivalence

$${\mathcal{C}}_1⟶{\mathcal{C}}_2$$

using the composition

$${\mathcal{C}}_1\stackrel{f}{\to }{\mathcal{S}}_1\stackrel{g}{\to }{\mathcal{S}}_2\stackrel{h}{\to }{\mathcal{C}}_2$$

where

• $f$ is any quasi-inverse of the inclusion ${\mathcal{S}}_1\to {\mathcal{C}}_1$,

• $g$ is the assumed isomorphism between skeletons

• $h$ is the canonical inclusion (which we know is an equivalence).