Lemma 8.2.1. This actually does give a presheaf.

## 8.2 Presheaves of morphisms associated to fibred categories

Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category, see Categories, Section 4.32. Suppose that $x, y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ are objects in the fibre category over $U$. We are going to define a functor

In other words this will be a presheaf on $\mathcal{C}/U$, see Sites, Definition 7.2.2. Make a choice of pullbacks as in Categories, Definition 4.32.6. Then, for $f : V \to U$ we set

Let $f' : V' \to U$ be a second object of $\mathcal{C}/U$. We also have to define the restriction map corresponding to a morphism $g : V'/U \to V/U$ in $\mathcal{C}/U$, in other words $g : V' \to V$ and $f' = f \circ g$. This will be a map

This map will basically be $g^\ast $, except that this transforms an element $\phi $ of the left hand side into an element $g^\ast \phi $ of $\mathop{Mor}\nolimits _{\mathcal{S}_{V'}}(g^\ast f^\ast x, g^\ast f^\ast y)$. At this point we use the transformation $\alpha _{g, f}$ of Categories, Lemma 4.32.7. In a formula, the restriction map is described by

Of course, nobody thinks of this restriction map in this way. We will only do this once in order to verify the following lemma.

**Proof.**
Let $g : V'/U \to V/U$ be as above and similarly $g' : V''/U \to V'/U$ be morphisms in $\mathcal{C}/U$. So $f' = f \circ g$ and $f'' = f' \circ g' = f \circ g \circ g'$. Let $\phi \in \mathop{Mor}\nolimits _{\mathcal{S}_ V}(f^\ast x, f^\ast y)$. Then we have

which is what we want, namely $\phi |_{V''} = (\phi |_{V'})|_{V''}$. The first equality holds because $\alpha _{g', g}$ is a transformation of functors, and hence

commutes. The second equality holds because of property (d) of a pseudo functor since $f' = f \circ g$ (see Categories, Definition 4.28.5). The last equality follows from the fact that $(g')^*$ is a functor. $\square$

From now on we often omit mentioning the transformations $\alpha _{g, f}$ and we simply identify the functors $g^* \circ f^*$ and $(f \circ g)^*$. In particular, given $g : V'/U \to V/U$ the restriction mappings for the presheaf $\mathit{Mor}(x, y)$ will sometimes be denoted $\phi \mapsto g^*\phi $. We formalize the construction in a definition.

Definition 8.2.2. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category, see Categories, Section 4.32. Given an object $U$ of $\mathcal{C}$ and objects $x$, $y$ of the fibre category, the *presheaf of morphisms from $x$ to $y$* is the presheaf

described above. It is denoted $\mathit{Mor}(x, y)$. The subpresheaf $\mathit{Isom}(x, y)$ whose values over $V$ is the set of isomorphisms $f^*x \to f^*y$ in the fibre category $\mathcal{S}_ V$ is called the *presheaf of isomorphisms from $x$ to $y$*.

If $\mathcal{S}$ is fibred in groupoids then of course $\mathit{Isom}(x, y) = \mathit{Mor}(x, y)$, and it is customary to use the $\mathit{Isom}$ notation.

Lemma 8.2.3. Let $F : \mathcal{S}_1 \to \mathcal{S}_2$ be a $1$-morphism of fibred categories over the category $\mathcal{C}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x, y\in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{S}_1)_ U)$. Then $F$ defines a canonical morphism of presheaves

on $\mathcal{C}/U$.

**Proof.**
By Categories, Definition 4.32.9 the functor $F$ maps strongly cartesian morphisms to strongly cartesian morphisms. Hence if $f : V \to U$ is a morphism in $\mathcal{C}$, then there are canonical isomorphisms $\alpha _ V : f^*F(x) \to F(f^*x)$, $\beta _ V : f^*F(y) \to F(f^*y)$ such that $f^*F(x) \to F(f^*x) \to F(x)$ is the canonical morphism $f^*F(x) \to F(x)$, and similarly for $\beta _ V$. Thus we may define

by $\phi \mapsto \beta _ V^{-1} \circ F(\phi ) \circ \alpha _ V$. We omit the verification that this is compatible with the restriction mappings. $\square$

Remark 8.2.4. Suppose that $p : \mathcal{S} \to \mathcal{C}$ is fibred in groupoids. In this case we can prove Lemma 8.2.1 using Categories, Lemma 4.35.4 which says that $\mathcal{S} \to \mathcal{C}$ is equivalent to the category associated to a contravariant functor $F : \mathcal{C} \to \textit{Groupoids}$. In the case of the fibred category associated to $F$ we have $g^* \circ f^* = (f \circ g)^*$ on the nose and there is no need to use the maps $\alpha _{g, f}$. In this case the lemma is (even more) trivial. Of course then one uses that the $\mathit{Mor}(x, y)$ presheaf is unchanged when passing to an equivalent fibred category which follows from Lemma 8.2.3.

Lemma 8.2.5. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category, see Categories, Section 4.32. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and let $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. Denote $x, y : \mathcal{C}/U \to \mathcal{S}$ also the corresponding $1$-morphisms, see Categories, Lemma 4.40.1. Then

the $2$-fibre product $\mathcal{S} \times _{\mathcal{S} \times \mathcal{S}, (x, y)} \mathcal{C}/U$ is fibred in setoids over $\mathcal{C}/U$, and

$\mathit{Isom}(x, y)$ is the presheaf of sets corresponding to this category fibred in setoids, see Categories, Lemma 4.38.6.

**Proof.**
Omitted. Hint: Objects of the $2$-fibre product are $(a : V \to U, z, a : V \to U, (\alpha , \beta ))$ where $\alpha : z \to a^*x$ and $\beta : z \to a^*y$ are isomorphisms in $\mathcal{S}_ V$. Thus the relationship with $\mathit{Isom}(x, y)$ comes by assigning to such an object the isomorphism $\beta \circ \alpha ^{-1}$.
$\square$

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