Let $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$.
-
1.
The corepresentable copresheaf associated to $A$ is the copresheaf
\[ h^{A}\colon \mathcal{C}\to \mathsf{Sets} \]where
-
•
Action on Objects. For each $X\in \operatorname {\mathrm{Obj}}(\mathcal{C})$, we have
\[ h^{A}(X) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,X). \] -
•
Action on Morphisms. For each $X,Y\in \operatorname {\mathrm{Obj}}(\mathcal{C})$, the action on morphisms
\[ h^{A}_{X,Y}\colon \operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,Y)\to \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}(h^{A}(X),h^{A}(Y)) \]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
\[ h^{A}(f) \colon \underbrace{h^{A}(X)}_{\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,X)} \to \underbrace{h^{A}(Y)}_{\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,Y)} \]defined by
\[ h^{A}(f) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f_{*}, \]where $f_{*}$ is the postcomposition by $f$ morphism of Chapter 11: Categories, Item 2 of Definition 11.1.4.1.1.
-
•
-
2.
A corepresenting object for a copresheaf $F\colon \mathcal{C}\to \mathsf{Sets}$ on $\mathcal{C}$ is an object $A$ of $\mathcal{C}$ such that we have $F\cong h^{A}$.
-
3.
A copresheaf $F\colon \mathcal{C}^{\mathsf{op}}\to \mathsf{Sets}$ on $\mathcal{C}$ is corepresentable if $F$ admits a corepresenting object.
12.2.2 Corepresentable Copresheaves
Let $\mathcal{C}$ be a category.
The corepresentable copresheaf on the delooping $\mathsf{B}{A}$ of a monoid $A$ associated to the unique object $\bullet $ of $\mathsf{B}{A}$ is the right regular representation of $A$ of 
, .
Let $F\colon \mathcal{C}\to \mathsf{Sets}$ be a copresheaf. If there exist $A,B\in \operatorname {\mathrm{Obj}}(\mathcal{C})$ such that we have natural isomorphisms
\begin{align*} h^{A} & \cong F,\\ h^{B} & \cong F, \end{align*}
then $A\cong B$.
Proof of Proposition 12.2.2.1.3.
By composing the isomorphisms $h^{A}\cong F\cong h^{B}$, we get a natural isomorphism $h^{A}\cong h^{B}$. By Item 2 of Proposition 12.2.4.1.2, we have $A\cong B$.
