Cayley's Theorem (Category Theory)

Theorem
Let $\mathbf C$ be a small category.

Then there exists a category $\mathbf D$, subject to:


 * $(1): \quad $ The objects of $\mathbf D$ are sets.
 * $(2): \quad $ The morphisms of $\mathbf D$ are mappings.
 * $(3): \quad \mathbf C \cong \mathbf D$, i.e. $\mathbf C$ and $\mathbf D$ are isomorphic.

Proof
Denote with $\mathbf{Set}$ the category of sets.

Define a functor $H: \mathbf C \to \mathbf{Set}$ by:


 * $H C := \left\{{f \in \operatorname{mor} \mathbf C: \operatorname{cod} f = C}\right\}$
 * $H f: H A \to H B, g \mapsto f \circ g$

for $f: A \to B$ a morphism of $\mathbf C$.

It is immediate by the definition of identity morphism that:


 * $H \left({\operatorname{id}_A}\right) = \operatorname{id}_{H A}$

For $f: A \to B$ and $g: B \to C$, observe:

Thus by Equality of Mappings, $H \left({g \circ f}\right) = H g \circ H f$.

It follows that $H$ is a functor.

It is clear that $H$ is injective on objects.

Suppose now that $H$ were not faithful.

Then there would be morphisms $g, h: A \to B$ of $\mathbf C$ such that $H g = H h$.

Since $\operatorname{id}_A \in H A$, this means in particular that:


 * $g \circ \operatorname{id}_A = h \circ \operatorname{id}_A$

by Equality of Mappings.

But the definition of identity morphism then reduces this to $g = h$.

Hence, $H$ is faithful.

By Functor is Isomorphism onto Image iff Faithful and Injective on Objects, it follows that $\mathbf C \cong H \mathbf C$.

Here $H \mathbf C$ is the image of $H$.

Also see

 * Cayley's Theorem (Group Theory)

Although Cayley did not prove this result, it is very similar in both statement and proof to Cayley's Theorem in group theory.

In fact, as Steve Awodey states it in :


 *  [ Cayley's ] theorem may be generalized to show that any category that is not "too big" can be represented as a [...] category of sets and functions.

The contributor Lord_Farin subsequently was as audacious as to name the general result after Cayley as well.