3.2. 群胚纤维范畴

定义 3.2.0.1. A caregory fibered in sets over $\mathcal{C}$ is a fibered category $\mathcal{F}\to \mathcal{C}$ such that for all $U\in \mathcal{C}$, $\mathcal{F}(U)$ is a set, i.e. only maps are identity.

We note in category theory, a set means a category where the only maps are identity (this is definition).

Note for such $\mathcal{F}$, we have a well-defined pullback map. Indeed, we get diagramThen we get $y^{\prime }\to y$ lying over ${\mathrm{Id}}_U$, i.e. $y^{\prime }\to y$ is morphism in $\mathcal{F}(U)$. But $\mathcal{F}(U)$ is a set, so it must be identity by definition, i.e. $y^{\prime }=y$.

So, given $U\stackrel{f}{\to }V$, we get $f^{\ast }:\mathcal{F}(V)\to \mathcal{F}(U)$ compatible with composition, i.e. $\mathcal{F}$ naturally yields a presheaf $F_{\mathcal{F}}$ where $F_{\mathcal{F}}(U):=\mathcal{F}(U)$.

The next task is to show categories fibered in sets are the same as presheaves.

引理 3.2.0.2. If $\mathcal{F}\to \mathcal{C}$ is a category fibered in sets and $\mathcal{G}\to \mathcal{C}$ is any fibered category, then ${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{G},\mathcal{F})$ is a set.

We note again that in higher category theory, sets are, by definition, categories with only identity morphism.

证明. For $f,g:\mathcal{G}\to \mathcal{F}$ morphisms of fibered category, and $\alpha :f\to g$ morphism, we want to show $\alpha =\mathrm{Id}$. For all $x\in \mathcal{G}(X)$, ${\alpha }_x:f(x)\to g(x)$ is in $\mathcal{F}(X)$, so ${\alpha }_x=\mathrm{Id}$. Also, given $\varphi :y\to x$, we get diagramand hence $f(\varphi )=g(\varphi )$. Hence $f=g$ and $\alpha$ is ${\mathrm{Id}}_f$.

$\square$

推论 3.2.0.3. Categories fibered in sets over $\mathcal{C}$ is a (locally small) category, i.e. ${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{G},\mathcal{F})$ is a set.

例 3.2.0.4. Given a presheaf $F:{\mathcal{C}}^{opp}\to \mathsf{Sets}$, let $\mathcal{F}:={\mathcal{F}}_F$ be the following category: objects are $(U,u)$ with $U\in \mathcal{C}$, $u\in F(U)$, morphisms $(U^{\prime },u^{\prime })\to (U,u)$ are $g:U^{\prime }\to U$ in $\mathcal{C}$ such that $g^{\ast }:F(U)\to F(U^{\prime })$ so $u\mapsto u^{\prime }$.

This is a fibered category because: we just let $p:\mathcal{F}\to \mathcal{C}$ be $(U,x)\mapsto U$, and we need to show pullback exists. Suppose we havewe claim this is a pullback, i.e. all maps are Cartesian.

To check this, we have the following diagramIn this diagram, we get $x\mapsto U,g^{\ast }x\mapsto U^{\prime }$ and suppose we are given $h^{\ast }x$,We are trying to show there is unique map $h^{\ast }x\to g^{\ast }x$, i.e. we want

Why this map exists? By definition, $h^{\ast }x\to x$ means $(gf{)}^{\ast }x=f^{\ast }{g}^{\ast }x=f^{\ast }(g^{\ast }x)$ as we are working with presheaf. Hence we indeed get the desired arrow $h^{\ast }x\to g^{\ast }x$.

We should also check $\mathcal{F}(U)$ is a set, so that $\mathcal{F}\to \mathcal{C}$ is fibered in sets.

The objects of $\mathcal{F}(U)$ are $(U,x)$ with $x\in F(U)$. The morphisms are $(U,y)\to (U,x)$ lying over identity ${\mathrm{Id}}_U$, i.e.but $y={\mathrm{Id}}_U^{\ast }(x)$ and hence $y=x$. Thus morphisms in $\mathcal{F}(U)$ are identity, i.e. $\mathcal{F}(U)$ is a set.

命题 3.2.0.5. There is an equivalence of categories between$$\left\{\begin{array}{c}\text{presheaves on }\\ \mathcal{C}\end{array}\right\}\leftrightarrow \left\{\begin{array}{c}\text{categories fibered}\\ \text{ in sets over }\mathcal{C}\end{array}\right\}$$given by$$\begin{array}{c}F\mapsto {\mathcal{F}}_F\\ F_{\mathcal{F}}\text{leftmapsto}\mathcal{F}\end{array}$$

We defined the maps already, so one should check these are quasi-inverse maps

Throughout the course, we will identify $\mathcal{F}$ with $F_{\mathcal{F}}$.

定义 3.2.0.6. A category is a groupoid if all morphisms are isomorphisms.

例 3.2.0.7. If $G$ is a group, then the category with one object and morphisms being $G$ is a groupoids (i.e. if object is $\bullet$, then we get an arrow $\bullet \stackrel{g}{\to }\bullet$ for each $g\in G$). Thus groups are examples of groupoids.

定义 3.2.0.8. A category fibered in groupoids is a fibered category $\mathcal{F}\to \mathcal{C}$ such that all $\mathcal{F}(U)$ are groupoids.

命题 3.2.0.9. If $\mathcal{F},{\mathcal{F}}^{\prime }$ are categories fibered in groupoids over $\mathcal{C}$, then the category ${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{F},{\mathcal{F}}^{\prime })$ is a groupoid.

证明. Let $f,g:\mathcal{F}\to {\mathcal{F}}^{\prime }$ with $\xi :f\to g$. We need to show $\xi$ is isomorphism, i.e. forall $x\in \mathcal{F}$, we need to show ${\xi }_x:f(x)\stackrel{\sim }{\to }g(x)$. Let $X=p_{\mathcal{F}}(x)$, then since $\xi$ is base-preserving natural trans, then ${\xi }_x$ lies over ${\mathrm{Id}}_X$, i.e. ${\xi }_x\in {\mathcal{F}}^{\prime }(X)=\text{groupoid}$, i.e. ${\xi }_x$ is isomorphism.

$\square$

We note in the proof, we only need ${\mathcal{F}}^{\prime }$ being fibered in groupoids. Also, by the same argument, one can show that if $\mathcal{F}\to \mathcal{C}$ is category fibered in groupoids, then all arrows are Cartesian.

Next we consider a nice example of categories fibered in groupoids.

定义 3.2.0.10. A groupoid in $\mathcal{C}$ is $(X_0,X_1,s,t,ϵ,i,m)$ such that $X_0,X\in \mathcal{C}$ andand we get Cartesian squareWe note in the above, we used the notation $X_1{\times }_{s,X_0,t}{X}_1$, which is just the fibered product, but we put emphasis on the two maps $s$ and $t$ that defines the fibered product.

So, here is the intuition: Here:

1.	$s$: source
2.	$t$: target
3.	$ϵ$: identity
4.	$i$: inverse
5.	$m$: multiplication/composition

Here is how you supposed to think of this set of data.

$X_0$ is supposed to be like objects in a category, $X_1$ is supposed to be arrows in the category. Then what’s going on with $s$ and $t$ inis that, $s$ takes an arrow and sends it to the source, while $t$ takes an a rrow and sends it to the target.

$ϵ:X_0\to X_1$ takes objects to the identity arrow.

$i:X_1\to X_1$ takes an arrow to its inverse (this supposed to exists because its a groupoid).

$m:X_1{\times }_{s,X_0,t}{X}_1\to X_1$ is like an arrow $(\alpha ,\beta )$ such that the source of $\alpha$ is equal the target of $\beta$. In other word, $(\alpha ,\beta )$ under $m$ is supposed to be “$\alpha \circ \beta$”.

The above is the intuition, and let’s give the actual axioms about groupoids in $\mathcal{C}$.

Axioms:

1.	$s\circ ϵ=\mathrm{Id}=t\circ ϵ$
2.	$t\circ i=s$ and $s\circ i=t$.
3.	$s\circ m=s\circ p_2$
4.	$t\circ m=t\circ p_1$
5.	(Associativity): the following two maps (we dropped the $s,X_0,t$ in all the fibered products here)are equal.
6.	(Identity):This says $\alpha \circ \mathrm{Id}=\alpha$ and $\mathrm{Id}\circ \alpha =\alpha$.
7.	(Inverse): we get diagramsThis says that ${\alpha }^{-1}\circ \alpha =\mathrm{Id}$ and $\alpha \circ {\alpha }^{-1}=\mathrm{Id}$.

Last time we end up listing the axioms of groupoids in $\mathcal{C}$. The definition seems complicated, but the idea is not bad. Basically, $X_0$ should be objects, $X_1$ be morphisms, $s$ takes morpihsms to its source, and $t$ to the target, $ϵ$ sends objects to identity map, $i$ to inverse arrow, and $m$ is composition of arrows.

We remark that, groupoids are sort of generalization of groups, where groups in category consists of only one object, and groupoids consists of more than one objects.

Today we are going to show we can get from groupoids in $\mathcal{C}$ to categories fibered in groupoids over $\mathcal{C}$.

Given $U\in \mathcal{C}$, let $\{X_0(U)/X_1(U)\}$ be a category defined as follows: objects are $X_0(U)={\mathrm{Hom}}_{\mathcal{C}}(U,X_0)$, and morphisms for $u\to u^{\prime }$ is an element $\alpha \in X_1(U)$ such that $s(\alpha )=u$ and $t(\alpha )=u^{\prime }$ (here $u,u^{\prime }:U\to X_0$, $\alpha :U\to X_1$ and hence $s(\alpha )=s\circ \alpha$ is an arrow $U\to X_0$, i.e. it make sense to ask $s(\alpha )=u$ and so on). This is a category, where composition of arrows are given by apply $m$, i.e. say we have $\eta :u^{\prime \prime }\to u^{\prime }$ and $\xi :u^{\prime }\to u$, where $u^{\prime \prime }\to u$ is given by $m(\xi ,\eta )$.

Next, we define a fibered category $\mathcal{F}=\{X_0/X_1\}$ over $\mathcal{C}$ as follows: objects are $(U,u)$, where $U\in \mathcal{C}$ and $u\in \{X_0(U)/X_1(U)\}$ (recall objects of this category is just $X_0(U)$). To get the morphisms, note given $f:V\to U$, we get the arrow (which is a functor)$$\{X_0(U)/X_1(U)\}\stackrel{f^{\ast }}{\to }\{X_0(V)/X_1(V)\}$$is well-defined (i.e. $f:U\to V$ induces $f^{\ast }:X_0(U)\to X_0(V)$ and $f^{\ast }:X_1(U)\to X_1(V)$ and hence $f^{\ast }:\{X_0(U)/X_1(U)\}\to \{X_0(V)/X_1(V)\}$). Then, morphisms $(V,v)\to (U,u)$ will be given by pairs $f:V\to U$ and $\alpha :v\stackrel{\sim }{\to }{f}^{\ast }u$ an isomorphism in $\{X_0(V)/X_1(V)\}$.

Then the projection $p:\mathcal{F}\to \mathcal{C}$ is going to be $p(U,u)=U$.

It remains to check $p:\mathcal{F}\to \mathcal{C}$ is a category fibered in groupoids.

The fiber $\mathcal{F}(U)$ is the category defined by: objects are, by definition, just $X_0(U)$ (as it is the same as the objects of $\{X_0(U)/X_1(U)\}$). The morphism for $(U,v)\to (U,u)$ is given by $U\stackrel{f}{\to }U$ and $\alpha :v\stackrel{\sim }{\to }{f}^{\ast }u$. However, since $f$ must live over the identity, $f=\mathrm{Id}$ and hence $f^{\ast }u=u$. In other word, morphisms are just $X_1(U)$, i.e. $\mathcal{F}(U)=\{X_0(U)/X_1(U)\}$ is a groupoid, as desired.

Aside, if $\mathcal{F}\stackrel{p}{\to }\mathcal{C}$ is category fibered in groupoids, for $X\in \mathcal{C}$ we can define $p/X:\mathcal{F}/X\to \mathcal{C}/X$, which is a category fibered in groupoids where $\mathcal{F}/X$ behaves like objects of $\mathcal{F}$ over $X$. This notion is hardly been used, so if we need it in the future we will define it, but for now its just aside.

The next notion is rather important.

定义 3.2.0.11. Given $\mathcal{F}\stackrel{p}{\to }\mathcal{C}$ a category fibered in groupoids, $x,x^{\prime }\in \mathcal{F}(X)$. We define a preasheaf$$\mathrm{Isom}(x,x^{\prime }):(\mathcal{C}/X{)}^{opp}\to \mathsf{Sets}$$as follows.

Let $f:Y\to X$ in $\mathcal{C}$,$$\mathrm{Isom}(x,x^{\prime })(f):={\mathrm{Isom}}_{\mathcal{F}(Y)}(f^{\ast }x,f^{\ast }{x}^{\prime })={\mathrm{Hom}}_{\mathcal{F}(Y)}(f^{\ast }x,f^{\ast }{x}^{\prime })$$because $\mathcal{F}(Y)$ is a groupoid (hence all arrows are isomorphisms). This depends on choices of pullbacks but we just fix one for all $f$.

Why is $\mathrm{Isom}(x,x^{\prime })$ a presheaf?

Say we have have our arrows $Y\stackrel{f}{\to }X$, and $Z\stackrel{g}{\to }Y$. Then we getwhere $(gf{)}^{\ast }$ and $g^{\ast }{f}^{\ast }$ are two choices of pullback. However, note all pullbacks are isomorphic, we see that we get In particular, this means that $\beta$ is canonical isomorphism and hence we get (canonical) arrow$$\begin{array}{c}\mathrm{Isom}(f^{\ast }x,f^{\ast }{x}^{\prime })\to \mathrm{Isom}((gf{)}^{\ast }x,(gf{)}^{\ast }{x}^{\prime })\\ \alpha \mapsto {\gamma }^{-1}\alpha \beta \end{array}$$which concludes $\mathrm{Isom}(x,x^{\prime })$ is a presheaf (as it is compatible with composition of arrows).

Next, we define fibered products of groupoids. So, unlike normal fibered products in $1$-category, now we are working with $2$-categories, hence we also need to consider arrows between arrows.

We will start with a diagram of groupoidsWe are going to define ${\mathcal{G}}_1{\times }_{\mathcal{G}}{\mathcal{G}}_2$ so that we get the following diagram:where $\Sigma$ is arrow between arrows.

Next, we will define some arbitrary category ${\mathcal{G}}_1{\times }_{\mathcal{G}}{\mathcal{G}}_2$, then we talk about universal properties that will convince us this is what fibered products should be for groupoids.

定义 3.2.0.12. For groupoids ${\mathcal{G}}_1,{\mathcal{G}}_2,\mathcal{G}$ with $f:{\mathcal{G}}_1\to \mathcal{G}$ and $g:{\mathcal{G}}_2\to \mathcal{G}$, let ${\mathcal{G}}_1{\times }_{\mathcal{G}}{\mathcal{G}}_2$ be the following category.

The objects are $(x,y,\sigma )$ where $x\in {\mathcal{G}}_1,y\in {\mathcal{G}}_2$ and $f(x)\underset{\sigma }{\overset{\sim }{\to }}g(y)$.

The morphisms for $(x^{\prime },y^{\prime },{\sigma }^{\prime })$ to $(x,y,\sigma )$ will be a pair $(a:x^{\prime }\to x,b:y^{\prime }\to y)$ so that we get diagram

Next, we need to define $p_1$ and $p_2$. They are given by $p_1(x,y,\sigma )=x$ and $p_2(x,y,\sigma )=y$, and $\Sigma (x,y,\sigma )=\sigma$.

To add a few words on $\Sigma$, we note $\Sigma :f\circ p_1\to g\circ p_2$, hence by definition this means we want that, inside ${\mathcal{G}}_1\times {\mathcal{G}}_2$, for any $(x^{\prime },y^{\prime },{\sigma }^{\prime })\to (x,y,\sigma )$, we get the following commutative squareBut if you expand the definition of $p_i,\Sigma$, this becomes exactly the square in definition of morphisms in ${\mathcal{G}}_1\times {\mathcal{G}}_2$. Hence $\Sigma$ is indeed natural trans as desired.

The universal property will be that, for all diagramwhere $\mathcal{H}$ is a groupoid with $\mathcal{H}\stackrel{\alpha }{\to }{\mathcal{G}}_1$, $\mathcal{H}\stackrel{\beta }{\to }{\mathcal{G}}_2$ and isomorphism (which is natural transformation) $\gamma :f\circ \alpha \to g\circ \beta$, we get unique $(h,{\lambda }_1,{\lambda }_2)$ with diagramwhere the two arrows ${\lambda }_1,{\lambda }_2$ are natural trans between arrows $\alpha$ ($\beta$, respectively) and the composing arrows of $h$ and $p_1$ ($h$ and $p_2$, respectively), so that the diagramcommutes.

Well, why is this object exists? To answer this, we want to construct the unique $h:\mathcal{H}\to {\mathcal{G}}_1{\times }_{\mathcal{G}}{\mathcal{G}}_2$. This will be the most “obvious” thing to do, which is $z\mapsto (\alpha (z),\beta (z),\gamma (z))$. We left the details of how $h$ acts on morphisms, as it should be natural.

Now we want to make sure we get the natural transformations ${\lambda }_1,{\lambda }_2$. That is, we want a natural trans ${\lambda }_1:\alpha \to p_1\circ h$. This is the same as, for arbitrary $z$ we want to get ${\lambda }_1(z):\alpha (z)\to p_1(h(z))=\alpha (z)$. Well, there is only one natural thing to do, which is take ${\lambda }_1(z)$ to be identity between $\alpha (z)$ and $\alpha (z)$. We do the same for ${\lambda }_2$.

Next we need to check commutativity of the following diagramwhere $\gamma$ is given by definition:Now for any $z$, we getbut then in particular $\Sigma (h(z))=\Sigma (\alpha (z),\beta (z),\gamma (z))=\gamma (z)$ by definition. Hence it is indeed commutative.

This concludes the definition of fibered products of groupoids, and we are heading to define fibered products of categories fibered in groupoids.

For this, let ${\mathcal{F}}_i\to \mathcal{C}$ be categories fibered in groupoids. Then we want to have a diagramso that for all $\mathcal{H}\to {\mathcal{F}}_1$ and $\mathcal{H}\to {\mathcal{F}}_2$ we get unique arrow $\mathcal{H}\to {\mathcal{F}}_1{\times }_{\mathcal{F}}{\mathcal{F}}_2$ with additional arrows between the arrows.

We want $\mathcal{G}={\mathcal{F}}_1{\times }_{\mathcal{F}}{\mathcal{F}}_2$ to have the property that$${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{H},\mathcal{F})={\mathrm{Hom}}_{\mathcal{C}}(\mathcal{H},{\mathcal{F}}_1){\times }_{{\mathrm{Hom}}_{\mathcal{C}}(\mathcal{H},\mathcal{F})}{\mathrm{Hom}}_{\mathcal{C}}(\mathcal{H},{\mathcal{F}}_2)$$However, on the RHS, they are just fibered products of groupoids, and by $2$-Yoneda lemma, this determines $\mathcal{G}$ if the RHS exists.

命题 3.2.0.13 (Olsson, Prop 3.4.13). The fibered product $\mathcal{G}$ exists.