Cayley's Theorem (Category Theory)

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

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

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.

That is, $\mathbf C$ is isomorphic to a subcategory of $\mathbf{Set}$.

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


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

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

It is immediate by the definition of identity morphism that:


 * $\map H {\operatorname{id}_A} = \operatorname{id}_{H A}$

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

Thus by Equality of Mappings:
 * $\map H {g \circ f} = H g \circ H f$

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 Embedding iff Faithful and Injective on Objects, it follows that $H$ is an embedding.

Thus $\mathbf C$ is isomorphic to a subcategory of $\mathbf{Set}$.

Also see

 * Cayley's Representation Theorem