## 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 4.33. 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

$\mathit{Mor}(x, y) : (\mathcal{C}/U)^{opp} \longrightarrow \textit{Sets}.$

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 4.33.6. Then, for $f : V \to U$ we set

$\mathit{Mor}(x, y)(f : V \to U) = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ V}(f^\ast x, f^\ast y).$

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

$\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ V}(f^\ast x, f^\ast y) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{V'}}({f'}^\ast x, {f'}^\ast y), \quad \phi \longmapsto \phi |_{V'}$

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 $\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{V'}}(g^\ast f^\ast x, g^\ast f^\ast y)$. At this point we use the transformation $\alpha _{g, f}$ of Categories, Lemma 4.33.7. In a formula, the restriction map is described by

$\phi |_{V'} = (\alpha _{g, f})_ y^{-1} \circ g^\ast \phi \circ (\alpha _{g, f})_ x.$

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 $\phi \in \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ V}(f^\ast x, f^\ast y)$. Then we have

\begin{eqnarray*} & & (\alpha _{g \circ g', f})_ y^{-1} \circ (g \circ g')^\ast \phi \circ (\alpha _{g \circ g', f})_ x \\ & = & (\alpha _{g \circ g', f})_ y^{-1} \circ (\alpha _{g', g})_{f^*y}^{-1} \circ (g')^*g^\ast \phi \circ (\alpha _{g', g})_{f^*x} \circ (\alpha _{g \circ g', f})_ x \\ & = & (\alpha _{g', f'})_ y^{-1} \circ (g')^*(\alpha _{g, f})_ y^{-1} \circ (g')^* g^\ast \phi \circ (g')^*(\alpha _{g, f})_ x \circ (\alpha _{g', f'})_ x \\ & = & (\alpha _{g', f'})_ y^{-1} \circ (g')^*\Big( (\alpha _{g, f})_ y^{-1} \circ g^\ast \phi \circ (\alpha _{g, f})_ x \Big) \circ (\alpha _{g', f'})_ x \end{eqnarray*}

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

$\xymatrix{ (g \circ g')^*f^*x \ar[rr]_{(g \circ g')^\ast \phi } \ar[d]_{(\alpha _{g', g})_{f^*x}} & & (g \circ g')^*f^*y \ar[d]^{(\alpha _{g', g})_{f^*y}} \\ (g')^*g^*f^*x \ar[rr]^{(g')^*g^\ast \phi } & & (g')^*g^*f^*y }$

commutes. The second equality holds because of property (d) of a pseudo functor since $f' = f \circ g$ (see Categories, Definition 4.29.5). 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 4.33. 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

$(f : V \to U) \longmapsto \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ V}(f^*x, f^*y)$

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

$\mathit{Mor}_{\mathcal{S}_1}(x, y) \longrightarrow \mathit{Mor}_{\mathcal{S}_2}(F(x), F(y))$

on $\mathcal{C}/U$.

Proof. By Categories, Definition 4.33.9 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

$\xymatrix{ \mathit{Mor}_{\mathcal{S}_1}(x, y)(f : V \to U) \ar@{=}[r] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{1, V}}(f^\ast x, f^\ast y) \ar[d] \\ \mathit{Mor}_{\mathcal{S}_2}(F(x), F(y))(f : V \to U) \ar@{=}[r] & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_{2, V}}(f^\ast F(x), f^\ast F(y)) }$

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 4.36.4 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 4.33. 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 4.41.2. Then

1. 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

2. $\mathit{Isom}(x, y)$ is the presheaf of sets corresponding to this category fibred in setoids, see Categories, Lemma 4.39.6.

Proof. Omitted. Hint: Objects of the $2$-fibre product are $(a : V \to U, z, (\alpha , \beta ))$ 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 $\beta \circ \alpha ^{-1}$. $\square$

## Comments (2)

Comment #3996 by Praphulla Koushik on

This is about $\phi|_{V'}$. Let $x,y\in \mathcal{S}(U)$. We have $g:V'\rightarrow V$ such that $f\circ g=f'$. We have $g_x:f'^*x\rightarrow f^*x$ that projects to $g:V'\rightarrow V$.

Similarly, we have $g_y:f'^*y\rightarrow f^*y$. Given this set up, we want to associate a map $f'^*x\rightarrow f'^*y$ for each $\phi:f^*x\rightarrow f^*y$. One obvious choice is to look for a map $g^*\phi:f'^*x\rightarrow f'^*y$ such that $\phi\circ g_x=g_y\circ g^*\phi$. Recall that the same thing we do in defining the functor $h^*:\mathcal{S}(W)\rightarrow \mathcal{S}(W')$ given an arrow $h:W'\rightarrow W$ in $\mathcal{C}$.

I think it is easier to think in this way. Comments are welcome.

I am not able to render diagrams here, so not writing anything. If some one can let me know how to render using tikzcd or otherwise, I can add diagrams. Sorry for spamming, posting same comment three times.

Comment #4114 by on

@#3994, 3995, 3996. I think what you are saying is exactly what is being said in the introduction of this section. So I do not see the need to change anything.

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