## 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

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

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

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

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 $\mathop{Mor}\nolimits _{\mathcal{S}_{V'}}(g^\ast f^\ast x, g^\ast f^\ast y)$. 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 $\phi \in \mathop{Mor}\nolimits _{\mathcal{S}_ V}(f^\ast x, f^\ast y)$. 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

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

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

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

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 4.41.1. 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 $(a : V \to U, z, a : V \to U, (\alpha , \beta ))$ 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 $\beta \circ \alpha ^{-1}$. $\square$

## Comments (4)

Comment #3994 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 $g_x:f'^*x\rightarrow f^*x$ from following diagram

$

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 $f'^*x\rightarrow f'^*y$ giving following commutative diagram

$

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

Comment #3995 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 $g_x:f'^*x\rightarrow f^*x$ from following diagram

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 (the same think we do in defining the functor $h^*:\mathcal{S}(W)\rightarrow \mathcal{S}(W')$ given an arrow $h:W'\righatrrow W$ in $\mathcal{C}$) is to look for a map $f'^*x\rightarrow f'^*y$ giving following commutative diagram

I think it is easier to think in this way.

I do not know why tikzcd is not rendering. So, I am just writing the code.

Comments are welcome.

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.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02Z9. Beware of the difference between the letter 'O' and the digit '0'.