Definition 4.40.1. Let \mathcal{C} be a category. A category fibred in groupoids p : \mathcal{S} \to \mathcal{C} is called representable if there exist an object X of \mathcal{C} and an equivalence j : \mathcal{S} \to \mathcal{C}/X (in the 2-category of groupoids over \mathcal{C}).
4.40 Representable categories fibred in groupoids
Here is our definition of a representable category fibred in groupoids. As promised this is invariant under equivalences.
The usual abuse of notation is to say that X represents \mathcal{S} and not mention the equivalence j. We spell out what this entails.
Lemma 4.40.2. Let \mathcal{C} be a category. Let p : \mathcal{S} \to \mathcal{C} be a category fibred in groupoids.
\mathcal{S} is representable if and only if the following conditions are satisfied:
\mathcal{S} is fibred in setoids, and
the presheaf U \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)/\cong is representable.
If \mathcal{S} is representable the pair (X, j), where j is the equivalence j : \mathcal{S} \to \mathcal{C}/X, is uniquely determined up to isomorphism.
Proof. The first assertion follows immediately from Lemma 4.39.5. For the second, suppose that j' : \mathcal{S} \to \mathcal{C}/X' is a second such pair. Choose a 1-morphism t' : \mathcal{C}/X' \to \mathcal{S} such that j' \circ t' \cong \text{id}_{\mathcal{C}/X'} and t' \circ j' \cong \text{id}_\mathcal {S}. Then j \circ t' : \mathcal{C}/X' \to \mathcal{C}/X is an equivalence. Hence it is an isomorphism, see Lemma 4.38.6. Hence by the Yoneda Lemma 4.3.5 (via Example 4.38.7 for example) it is given by an isomorphism X' \to X. \square
Lemma 4.40.3. Let \mathcal{C} be a category. Let \mathcal{X}, \mathcal{Y} be categories fibred in groupoids over \mathcal{C}. Assume that \mathcal{X}, \mathcal{Y} are representable by objects X, Y of \mathcal{C}. Then
More precisely, given \phi : X \to Y there exists a 1-morphism f : \mathcal{X} \to \mathcal{Y} which induces \phi on isomorphism classes of objects and which is unique up to unique 2-isomorphism.
Proof. By Example 4.38.7 we have \mathcal{C}/X = \mathcal{S}_{h_ X} and \mathcal{C}/Y = \mathcal{S}_{h_ Y}. By Lemma 4.39.6 we have
By the Yoneda Lemma 4.3.5 we have \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ X, h_ Y) = \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(X, Y). \square
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