Let $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$.
-
1.
The representable presheaf associated to $A$ is the presheaf
\[ h_{A}\colon \mathcal{C}^{\mathsf{op}}\to \mathsf{Sets} \]where
-
•
Action on Objects. For each $X\in \operatorname {\mathrm{Obj}}(\mathcal{C})$, we have
\[ h_{A}(X) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,A). \] -
•
Action on Morphisms. For each $X,Y\in \operatorname {\mathrm{Obj}}(\mathcal{C})$, the action on morphisms
\[ h_{A|X,Y}\colon \operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,Y)\to \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}(h_{A}(Y),h_{A}(X)) \]of $h_{A}$ at $(X,Y)$ is given by sending a morphism
\[ f\colon X\to Y \]of $\mathcal{C}$ to the map of sets
\[ h_{A}(f) \colon \underbrace{h_{A}(Y)}_{\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Hom}}_{\mathcal{C}}(Y,A)} \to \underbrace{h_{A}(X)}_{\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,A)} \]defined by
\[ h_{A}(f) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f^{*}, \]where $f^{*}$ is the precomposition by $f$ morphism of Chapter 11: Categories, Item 1 of Definition 11.1.4.1.1.
-
•
-
2.
A representing object for a presheaf $\mathcal{F}\colon \mathcal{C}^{\mathsf{op}}\to \mathsf{Sets}$ on $\mathcal{C}$ is an object $A$ of $\mathcal{C}$ such that we have $\mathcal{F}\cong h_{A}$.
-
3.
A presheaf $\mathcal{F}\colon \mathcal{C}^{\mathsf{op}}\to \mathsf{Sets}$ on $\mathcal{C}$ is representable if $\mathcal{F}$ admits a representing object.
12.1.2 Representable Presheaves
Let $\mathcal{C}$ be a category.
The representable presheaf on the delooping $\mathsf{B}{A}$ of a monoid $A$ associated to the unique object $\bullet $ of $\mathsf{B}{A}$ is the left regular representation of $A$ of 
, .
Let $\mathcal{F}\colon \mathcal{C}^{\mathsf{op}}\to \mathsf{Sets}$ be a presheaf. If there exist $A,B\in \operatorname {\mathrm{Obj}}(\mathcal{C})$ such that we have natural isomorphisms
\begin{align*} h_{A} & \cong \mathcal{F},\\ h_{B} & \cong \mathcal{F}, \end{align*}
then $A\cong B$.
Proof of Proposition 12.1.2.1.3.
By composing the isomorphisms $h_{A}\cong \mathcal{F}\cong h_{B}$, we get a natural isomorphism $h_{A}\cong h_{B}$. By Item 2 of Proposition 12.1.4.1.3, we have $A\cong B$.
