The Stacks project

Proof by naive method. In this proof method we proceed in stages:

First, given $x$ lying over $U$ and any object $y$ of $\mathcal{S}$, we say that two morphisms $a, b : x \to y$ of $\mathcal{S}$ lying over the same arrow of $\mathcal{C}$ are locally equal if there exists a covering $\{ f_ i : U_ i \to U\} $ of $\mathcal{C}$ such that the compositions

are equal. This gives an equivalence relation $\sim $ on arrows of $\mathcal{S}$. If $b \sim b'$ then $a \circ b \circ c \sim a \circ b' \circ c$ (verification omitted). Hence we can quotient out by this equivalence relation to obtain a new category $\mathcal{S}^1$ over $\mathcal{C}$ together with a morphism $G^1 : \mathcal{S} \to \mathcal{S}^1$.

One checks that $G^1$ preserves strongly cartesian morphisms and that $\mathcal{S}^1$ is a fibred category over $\mathcal{C}$. Checks omitted. Thus we reduce to the case where locally equal morphisms are equal.

Next, we add morphisms as follows. Given $x$ lying over $U$ and any object $y$ of lying over $V$ a locally defined morphism from $x$ to $y$ is given by

1. a morphism $f : U \to V$,

2. a covering $\{ f_ i : U_ i \to U\} $ of $U$, and

3. morphisms $a_ i : f_ i^*x \to y$ with $p(a_ i) = f \circ f_ i$

with the property that the compositions

are equal. Note that a usual morphism $a : x \to y$ gives a locally defined morphism $(p(a) : U \to V, \{ \text{id}_ U\} , a)$. We say two locally defined morphisms $(f, \{ f_ i : U_ i \to U\} , a_ i)$ and $(g, \{ g_ j : U'_ j \to U\} , b_ j)$ are equal if $f = g$ and the compositions

are equal (this is the right condition since we are in the situation where locally equal morphisms are equal). To compose locally defined morphisms $(f, \{ f_ i : U_ i \to U\} , a_ i)$ from $x$ to $y$ and $(g, \{ g_ j : V_ j \to V\} , b_ j)$ from $y$ to $z$ lying over $W$, just take $g \circ f : U \to W$, the covering $\{ U_ i \times _ V V_ j \to U\} $, and as maps the compositions

We omit the verification that this is a locally defined morphism.

One checks that $\mathcal{S}^2$ with the same objects as $\mathcal{S}$ and with locally defined morphisms as morphisms is a category over $\mathcal{C}$, that there is a functor $G^2 : \mathcal{S} \to \mathcal{S}^2$ over $\mathcal{C}$, that this functor preserves strongly cartesian objects, and that $\mathcal{S}^2$ is a fibred category over $\mathcal{C}$. Checks omitted. This reduces one to the case where the morphism presheaves of $\mathcal{S}$ are all sheaves, by checking that the effect of using locally defined morphisms is to take the sheafification of the (separated) morphisms presheaves.

Finally, in the case where the morphism presheaves are all sheaves we have to add objects in order to make sure descent conditions are effective in the end result. The simplest way to do this is to consider the category $\mathcal{S}'$ whose objects are pairs $(\mathcal{U}, \xi )$ where $\mathcal{U} = \{ U_ i \to U\} $ is a covering of $\mathcal{C}$ and $\xi = (X_ i, \varphi _{ii'})$ is a descent datum relative $\mathcal{U}$. Suppose given two such data $(\mathcal{U}, \xi ) = (\{ f_ i : U_ i \to U\} , x_ i, \varphi _{ii'})$ and $(\mathcal{V}, \eta ) = (\{ g_ j : V_ j \to V\} , y_ j, \psi _{jj'})$. We define

as the set of $(f, a_{ij})$, where $f : U \to V$ and

are morphisms of $\mathcal{S}$ lying over $U_ i \times _ V V_ j \to V_ j$. These have to satisfy the following condition: for any $i, i' \in I$ and $j, j' \in J$ set $W = (U_ i \times _ U U_{i'}) \times _ V (V_ j \times _ V V_{j'})$. Then

commutes. At this point you have to verify the following things:

1. there is a well defined composition on morphisms as above,

2. this turns $\mathcal{S}'$ into a category over $\mathcal{C}$,

3. there is a functor $G : \mathcal{S} \to \mathcal{S}'$ over $\mathcal{C}$,

4. for $x, y$ objects of $\mathcal{S}$ we have $\mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(x, y) = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}'}(G(x), G(y))$,

5. any object of $\mathcal{S}'$ locally comes from an object of $\mathcal{S}$, i.e., part (2) of the lemma holds,

6. $G$ preserves strongly cartesian morphisms,

7. $\mathcal{S}'$ is a fibred category over $\mathcal{C}$, and

8. $\mathcal{S}'$ is a stack over $\mathcal{C}$.

This is all not hard but there is a lot of it. Details omitted. $\square$

Less naive proof. Here is a less naive proof. By Categories, Lemma 4.36.4 there exists an equivalence of fibred categories $\mathcal{S} \to \mathcal{S}'$ where $\mathcal{S}'$ is a split fibred category, i.e., one in which the pullback functors compose on the nose. Obviously the lemma for $\mathcal{S}'$ implies the lemma for $\mathcal{S}$. Hence we may think of $\mathcal{S}$ as a presheaf in categories.

Consider the $2$-category $\textit{Cat}$ temporarily as a category by forgetting about $2$-morphisms. Let us think of a category as a quintuple $(\text{Ob}, \text{Arrows}, s, t, \circ )$ as in Categories, Section 4.2. Consider the forgetful functor

Then $forget$ is faithful, $\textit{Cat}$ has limits and $forget$ commutes with them, $\textit{Cat}$ has directed colimits and $forget$ commutes with them, and $forget$ reflects isomorphisms. We can sheafify presheaves with values in $\textit{Cat}$, and by an argument similar to the one in the first part of Sites, Section 7.44 the result commutes with $forget$. Applying this to $\mathcal{S}$ we obtain a sheafification $\mathcal{S}^\# $ which has a sheaf of objects and a sheaf of morphisms both of which are the sheafifications of the corresponding presheaves for $\mathcal{S}$. In this case it is quite easy to see that the map $\mathcal{S} \to \mathcal{S}^\# $ has the properties (1) and (2) of the lemma.

However, the category $\mathcal{S}^\# $ may not yet be a stack since, although the presheaf of objects is a sheaf, the descent condition may not yet be satisfied. To remedy this we have to add more objects. But the argument above does reduce us to the case where $\mathcal{S} = \mathcal{S}_ F$ for some sheaf(!) $F : \mathcal{C}^{opp} \to \textit{Cat}$ of categories. In this case consider the functor $F' : \mathcal{C}^{opp} \to \textit{Cat}$ defined by

1. The set $\mathop{\mathrm{Ob}}\nolimits (F'(U))$ is the set of pairs $(\mathcal{U}, \xi )$ where $\mathcal{U} = \{ U_ i \to U\} $ is a covering of $U$ and $\xi = (x_ i, \varphi _{ii'})$ is a descent datum relative to $\mathcal{U}$.

2. A morphism in $F'(U)$ from $(\mathcal{U}, \xi )$ to $(\mathcal{V}, \eta )$ is an element of

   \[ \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Mor}}\nolimits _{DD(\mathcal{W})}(a^*\xi , b^*\eta ) \]

   where the colimit is over all common refinements $a : \mathcal{W} \to \mathcal{U}$, $b : \mathcal{W} \to \mathcal{V}$. This colimit is filtered (verification omitted). Hence composition of morphisms in $F(U)$ is defined by finding a common refinement and composing in $DD(\mathcal{W})$.

3. Given $h : V \to U$ and an object $(\mathcal{U}, \xi )$ of $F'(U)$ we set $F'(h)(\mathcal{U}, \xi )$ equal to $(V \times _ U \mathcal{U}, \text{pr}_1^*\xi )$. More precisely, if $\mathcal{U} = \{ U_ i \to U\} $ and $\xi = (x_ i, \varphi _{ii'})$, then $V \times _ U \mathcal{U} = \{ V \times _ U U_ i \to V\} $ which comes with a canonical morphism $\text{pr}_1 : V \times _ U \mathcal{U} \to \mathcal{U}$ and $\text{pr}_1^*\xi $ is the pullback of $\xi $ with respect to this morphism (see Definition 8.3.4).

4. Given $h : V \to U$, objects $(\mathcal{U}, \xi )$ and $(\mathcal{V}, \eta )$ and a morphism between them, represented by $a : \mathcal{W} \to \mathcal{U}$, $b : \mathcal{W} \to \mathcal{V}$, and $\alpha : a^*\xi \to b^*\eta $, then $F'(h)(\alpha )$ is represented by $a' : V \times _ U\mathcal{W} \to V \times _ U\mathcal{U}$, $b' : V \times _ U\mathcal{W} \to V \times _ U\mathcal{V}$, and the pullback $\alpha '$ of the morphism $\alpha $ via the map $V \times _ U \mathcal{W} \to \mathcal{W}$. This works since pullbacks in $\mathcal{S}_ F$ commute on the nose.

There is a map $F \to F'$ given by associating to an object $x$ of $F(U)$ the object $(\{ U \to U\} , (x, triv))$ of $F'(U)$. At this point you have to check that the corresponding functor $\mathcal{S}_ F \to \mathcal{S}_{F'}$ has properties (1) and (2) of the lemma, and finally that $\mathcal{S}_{F'}$ is a stack. Details omitted. $\square$

Proof. Omitted. Hint: Suppose that $x' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}'_ U)$. By the result of Lemma 8.8.1 there exists a covering $\{ U_ i \to U\} _{i \in I}$ such that $x'|_{U_ i} = G(x_ i)$ for some $x_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{U_ i})$. Moreover, there exist coverings $\{ U_{ijk} \to U_ i \times _ U U_ j\} $ and isomorphisms $\alpha _{ijk} : x_ i|_{U_{ijk}} \to x_ j|_{U_{ijk}}$ with $G(\alpha _{ijk}) = \text{id}_{x'|_{U_{ijk}}}$. Set $y_ i = F(x_ i)$. Then you can check that

agree on overlaps and therefore (as $\mathcal{X}$ is a stack) define a morphism $\beta _{ij} : y_ i|_{U_ i \times _ U U_ j} \to y_ j|_{U_ i \times _ U U_ j}$. Next, you check that the $\beta _{ij}$ define a descent datum. Since $\mathcal{X}$ is a stack these descent data are effective and we find an object $y$ of $\mathcal{X}_ U$ agreeing with $G(x_ i)$ over $U_ i$. The hint is to set $H(x') = y$. $\square$

Proof. Omitted. $\square$

Proof. Let us denote $\mathcal{X}', \mathcal{Y}', \mathcal{Z}'$ the stackifications and $\mathcal{W}$ the stackification of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$. By construction of $2$-fibre products there is a canonical $1$-morphism $\mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{Z}'$. As the second $2$-fibre product is a stack (see Lemma 8.4.6) this $1$-morphism induces a $1$-morphism $h : \mathcal{W} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{Z}'$ by the universal property of stackification, see Lemma 8.8.2. Now $h$ is a morphism of stacks, and we may check that it is an equivalence using Lemmas 8.4.7 and 8.4.8.

Thus we first prove that $h$ induces isomorphisms of $\mathit{Mor}$-sheaves. Let $\xi , \xi '$ be objects of $\mathcal{W}$ over $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. We want to show that

is an isomorphism. To do this we may work locally on $U$ (see Sites, Section 7.26). Hence by construction of $\mathcal{W}$ (see Lemma 8.8.1) we may assume that $\xi , \xi '$ actually come from objects $(x, z, \alpha )$ and $(x', z', \alpha ')$ of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ over $U$. By the same lemma once more we see that in this case $\mathit{Mor}(\xi , \xi ')$ is the sheafification of

and that $\mathit{Mor}(h(\xi ), h(\xi '))$ is equal to the fibre product

where $i : \mathcal{X} \to \mathcal{X}'$, $j : \mathcal{Y} \to \mathcal{Y}'$, and $k : \mathcal{Z} \to \mathcal{Z}'$ are the canonical functors. Thus the first displayed map of this paragraph is an isomorphism as sheafification is exact (and hence the sheafification of a fibre product of presheaves is the fibre product of the sheafifications).

Finally, we have to check that any object of $\mathcal{X}' \times _{\mathcal{Y}'} \mathcal{Z}'$ over $U$ is locally on $U$ in the essential image of $h$. Write such an object as a triple $(x', z', \alpha )$. Then $x'$ locally comes from an object of $\mathcal{X}$, $z'$ locally comes from an object of $\mathcal{Z}$, and having made suitable replacements for $x'$, $z'$ the morphism $\alpha $ of $\mathcal{Y}'_ U$ locally comes from a morphism of $\mathcal{Y}$. In other words, we have shown that any object of $\mathcal{X}' \times _{\mathcal{Y}'} \mathcal{Z}'$ over $U$ is locally on $U$ in the essential image of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{X}' \times _{\mathcal{Y}'} \mathcal{Z}'$, hence a fortiori it is locally in the essential image of $h$. $\square$

Proof. This follows from the fact that stackification is compatible with $2$-fibre products by Lemma 8.8.4 and the fact that there is a formula for the inertia in terms of $2$-fibre products of categories over $\mathcal{C}$, see Categories, Lemma 4.34.1. $\square$