3.1. 基础定义和米田引理

The next topic is fibered category. At this point, we talked about descents. Then, descents plus fibered category gives categorical stacks, then plus geometry then we get algebraic stacks. In particular, categorical stacks are special kind of fibered categories which satisfy descent.

Frequently, given a diagram$$ZXY$$we say take the fibered product and we get(where we use $\square$ to indicate Cartesian diagram) However, $X{\times }_{Y}Z$ is only defined up to (canonical) isomorphism.

Thus, any two objects $W_1,W_2$ that claim to be the fibered product are isomorphic up to unique isomorphism. Usually it is good enough to make a choice between $W_1$ and $W_2$ and the choice does not make a difference.

However, since stacks are about automorphisms, we need to keep track of the choices we made, which is bad. Thus, we can speak of whenis Cartesian, rather than saying $W$ is “the” fibered product.

定义 3.1.0.1. Let $\mathcal{C}$ be a category, then a category over $\mathcal{C}$ is a category $\mathcal{F}$ and a functor $p:\mathcal{F}\to \mathcal{C}$.

A morphism $\varphi :U\to V$ in $\mathcal{F}$ is Cartesian if for the following diagram and for all $\psi :W\to V$ and factorization, there exists unique $\lambda :W\to U$ withsuch that $\varphi \lambda =\psi$ and $p(\lambda )=h$.

The point of this is that, with this, the diagramlooks like a pullback.

定义 3.1.0.2. Let $\mathcal{F}$ be category over $\mathcal{C}$ and $\varphi :U\to V$ be Cartesian, then we say $U$ is a pullback of $\mathbf{V}$ along $\mathbf{p}(\mathbf{\varphi })$.

If $U^{\prime }\stackrel{{\varphi }^{\prime }}{\to }V$ and $U\stackrel{\varphi }{\to }V$ are both pullbacks along the samething $p(\varphi )=p({\varphi }^{\prime })$, then we haveand hence $p(\lambda )=\mathrm{Id}$ which implies $\lambda$ is an isomorphism (with ${\varphi }^{\prime }\circ \lambda =\varphi$).

注 3.1.0.3. Now given $p:\mathcal{F}\to \mathcal{C}$, for any $U\in \mathcal{C}$, let $\mathcal{F}(U)$ be a category defined as follows: the objects are $X\in \mathcal{F}$ such that $p(X)=U$ and morphisms are $X\stackrel{\varphi }{\to }Y$ such that $p(\varphi )={\mathrm{Id}}_U$.

The idea above is that we are supposed to think of $\mathcal{F}$ as a map from $\mathcal{C}$ to categories, where we input $U$ and output a “category” $\mathcal{F}(U)$.

定义 3.1.0.4. Let $p:\mathcal{F}\to \mathcal{C}$ be a category over $\mathcal{C}$, then we say $\mathcal{F}$ is a fibered category if “pullbacks exist”, i.e. given diagramthen there exists Cartesian arrow $\varphi :u\to v$ such that $p(\varphi )=h$, i.e.is a pullback.

例 3.1.0.5. Let $\mathcal{C}=\mathsf{Sch}$ be the category of schemes, and let $M_g$ be the category of genus $g$ curves. In other words, objects of $M_g$ are $C\stackrel{\pi }{\to }S$ where $\pi$ is smooth and geometric fibers of $\pi$ are genus $g$ curves(recall geometric fiber means $X\times \mathrm{Spec}(\overline{\kappa (x)})$). The morphisms are diagrams

Then, our $p:M_g\to \mathsf{Sch}$ is going to be $(C\stackrel{\pi }{\to }S)\mapsto S$. Since pullback exists because they exists in $\mathsf{Sch}$, we see $M_g$ is a fibered category. Indeed, just take the fibered product in $\mathsf{Sch}$, saythen we getwhere since $\pi$ is smooth, then ${\pi }^{\prime }$ is smooth, and ${\pi }^{\prime }$ has the same geometric fibers as $\pi$.

Thus, we just constructed the moduli space of genus $g$ curves.

例 3.1.0.6. If $\mathcal{C}$ is any category, $X\in \mathcal{C}$, then consider $\mathcal{C}/X$ as the category with objects $Y\to X$ and morphismsThen $p:\mathcal{C}/X\to \mathcal{C}$ with $Y\mapsto Y$ gives $\mathcal{C}/X$ the structure of a fibered category. This is because every arrow in $\mathcal{C}/X$ is Cartesian. Indeed, take an arrow $\varphi :Y^{\prime }\to Y$ in $\mathcal{C}/X$, and let $\psi :Y^{\prime \prime }\to Y$ be any arrow in $\mathcal{C}/X$ with a factorization $Y^{\prime \prime }\stackrel{h}{\to }{Y}^{\prime }\stackrel{\varphi }{\to }Y$, then we see we indeed have dashed arrow in the following diagramThat is, we just take the dashed arrow be $h$, which is indeed unique and it exists.

例 3.1.0.7. Let’s consider the category $\mathsf{QCoh}$ over $\mathsf{Sch}$. Here objects of $\mathsf{QCoh}$ are pairs $(S,\mathcal{F})$ where $\mathcal{F}$ is quasi-coherent sheaf on $S$, and morphisms between $(S^{\prime },{\mathcal{F}}^{\prime })\to (S,\mathcal{F})$ are given by $f:S^{\prime }\to S$ and $ϵ:{\mathcal{F}}^{\prime }\to f^{\ast }\mathcal{F}$.

In the above examples, as in Remark 3.1.0.3, we see $\mathsf{QCoh}(S)$ is exactly the category of quasi-coherent sheaves on $S$, and $M_g(S)$ is exactly the category of genus $g$ curves on $S$.

Indeed, $\mathsf{Qcoh}(S)$ is by definition the category with objects being $(X,\mathcal{F})\in \mathsf{Qcoh}$ such that $p(X,\mathcal{F})=S$ and morphisms being $f:(X,\mathcal{F})\to (Y,\mathcal{G})$ such that $p(f)={\mathrm{Id}}_S$. Well, $p(f,ϵ)={\mathrm{Id}}_S$ means we need to have $f:X\to Y$ is the identity, i.e. $X=Y=S$ and $ϵ$ is just a morphism between quasi-coherent sheaves on $S$.

Similarly $M_g(S)$ is category of genus $g$ curves because the projection forces any object to be live over $S$.

定义 3.1.0.8. If $p_{\mathcal{F}}:\mathcal{F}\to \mathcal{C}$ and $p_{\mathcal{G}}:\mathcal{G}\to \mathcal{C}$ be two fibered categories, then a morphism of fibered categories is a functor $g:\mathcal{F}\to \mathcal{G}$ such thatand $g$ sends Cartesian arrows to Cartesian arrows.

We note, for all $U\in \mathcal{C}$, we get $g_U:\mathcal{F}(U)\to \mathcal{G}(U)$. Indeed, since $\mathcal{F}(U),\mathcal{G}(U)$ are categories, $g_U$ is a functor between categories. It is just the same as $g$, i.e. $g_U(x)=g(x)$. We can check this is indeed a functor. Indeed, let $x_1\to x_2\in \mathcal{F}(U),y_1\to y_2\in \mathcal{G}(U)$ be two arrows, we need to show we get a diagramHowever, note we get the following diagramwhere the two triples of vertical arrows (i.e. $(g_U,p_{\mathcal{G}},p_{\mathcal{F}})$) commutes, and the outer square and inner square both commutes, which forces the upper square to commute as desired.

定义 3.1.0.9. If $g,g^{\prime }:\mathcal{F}\to \mathcal{G}$ are two morphisms of fibered categories, then a base-preserving natural transformation $\alpha :g\to g^{\prime }$ is a natural transformation of functors such that for all $U\in \mathcal{F}$, the map ${\alpha }_U:g(U)\to g^{\prime }(U)$ satisfies $p_{\mathcal{G}}({\alpha }_U)={\mathrm{Id}}_{p_{\mathcal{F}}(U)}$, i.e. if we have the following diagramthen we must have $p_{\mathcal{G}}({\alpha }_U)=\mathrm{Id}$. In other word, we want ${\alpha }_U$ to be a morphism in $\mathcal{G}(p_{\mathcal{F}}(U))$.

We note this gives ${\mathrm{HOM}}_{\mathcal{C}}(\mathcal{F},\mathcal{G})$ the category with objects being morphisms of fibered category $g:\mathcal{F}\to \mathcal{G}$ and morphisms being base-preserving natural transformations.

Now, suppose we have $g^{\prime }:\mathcal{F}\to \mathcal{G}$, then we get $g_U:\mathcal{F}(U)\to \mathcal{G}(U)$ for all $U\in \mathcal{C}$. This is kind of looks like a map of presheaves.

Last time we defined a fibered category, which is a functor $p:\mathcal{F}\to \mathcal{C}$ such that pullbacks exist, i.e. for all $z$ and for all $z\to y$, there exists unique $z\to x$ so the following commutesThe morphisms of fibered categories are given by functors such that $g$ sends Cartesian arrows to Cartesian arrows.

Then, for $p:\mathcal{F}\to \mathcal{C}$ fibered category, for all $U\in \mathcal{C}$, we let $\mathcal{F}(U)$ be the category with objects being $x\in \mathcal{F}$ such that $p(x)=U$, and morphisms being $\varphi :x\to y$ such that $p(\varphi )={\mathrm{Id}}_U$.

Thus, say $g:\mathcal{F}\to \mathcal{G}$ be maps of fibered categories over $\mathcal{C}$, then we get $g_U:\mathcal{F}(U)\to \mathcal{G}(U)$ for all $U\in \mathcal{C}$. At the end of last class, we mentioned this implies fibered categories look like presheaves but instead of $\mathcal{F}(U)$ being sets, we have $\mathcal{F}(U)$ is category.

例 3.1.0.10. Consider ${\mathcal{M}}_{g,n}\to \mathsf{Sch}$ be the fibered category of genus $g$ curves with $n$ marked points. In this category, objects of ${\mathcal{M}}_{g,n}$ are $C\stackrel{\pi }{\to }S$ with $\pi$ smooth and on geometric fibers of $S$, $C$ is genus $g$ curves. Next, we need to explain what marked points are. Those are given by sections of $p$, say $p_i:S\to C$, which are distinct points on the geometric fibers. Then the morphisms are diagramssuch that $\pi f=g{\pi }^{\prime }$ and $f{p}_i^{\prime }=p_{i}g$. Then, the projection $p:{\mathcal{M}}_{g,n}\to \mathsf{Sch}$ is given by $(C\stackrel{\pi }{\to }S)\mapsto S$.

One should check that pullbacks exist.

In particular, ${\mathcal{M}}_{1,1}$ is the moduli space of genus $1$ curves with one marked point, i.e. they are exactly elliptic curves.

Next, note we have the fibered category $\mathsf{QCoh}$ and we get $F_i:{\mathcal{M}}_{g,n}\to \mathsf{QCoh}$ map of fibered category, for $1\le i\le n$. The map is given by$$F_i(C\stackrel{\pi }{\to }S):=(S,p_i^{\ast }{\Omega }_{C/S}^1)$$where we take the pullback of relative differential via $p_i$. We also need to define what $F_i$ does on morphisms. Well, suppose we have morphismWe need a map between $(p_i^{\prime }{)}^{\ast }{\Omega }_{C^{\prime }/S^{\prime }}^1\to g^{\ast }{P}_i^{\ast }{\Omega }_{C/S}^1$. Well, we do have a canonical morphism (and in fact its isomorphism), as we will show next. First, note $g^{\ast }{p}_i^{\ast }{\Omega }_{C/S}^1$ is equal to $(p_i^{\prime }{)}^{\ast }{f}^{\ast }{\Omega }_{C/S}^1$ as the diagram commutes, and we also have $(p_i^{\prime }{)}^{\ast }{f}^{\ast }{\Omega }_{C/S}^1=(p_i^{\prime }{)}^{\ast }{\Omega }_{C^{\prime }/S^{\prime }}^1$. Hence we get the desired canonical (iso)morphism as desired.

引理 3.1.0.11. Suppose $g:\mathcal{F}\to \mathcal{G}$ is a map of fibered category. Then $g$ is fully faithful as a map of categories (not fibered category) if and only if $\forall U\in \mathcal{C}$, $g_U:\mathcal{F}(U)\to \mathcal{G}(U)$ is fully faithful.

This result should remind you of: if $\mathcal{F}\to \mathcal{G}$ is a map of presheaves then it is injective if and only if for all $U$, $\mathcal{F}(U)\to \mathcal{G}(U)$ is injective.

证明. Recall fully faithful means we have a bijection between Hom sets (full means surjection between hom sets and faithful means injection between hom sets). Thus, let $x,y\in \mathcal{F}$, we getThen $g$ is fully faithful if and only if for all $h:p_{\mathcal{F}}(x)\to p_{\mathcal{F}}(y)$ in ${\mathrm{Hom}}_{\mathcal{C}}(p_{\mathcal{F}}(x),p_{\mathcal{F}}(y))$, $g$ induces a bijection$$\{x\stackrel{\varphi }{\to }y:p_{\mathcal{F}}(\varphi )=h\}\stackrel{\sim }{\to }\{g(x)\stackrel{\psi }{\to }g(y):p_{\mathcal{G}}(\psi )=h\}$$This is sort of like we show bijection on each of the fibers.

Thus, we fix $h$ downstairsand let $y^{\prime }$ be a fibered product/pullbackThen we see for all $x\to y$, we get unique arrow $x\to y^{\prime }$, i.e. we getso, we have a bijection$$\{x\stackrel{\varphi }{\to }y:p_{\mathcal{F}}(\varphi )=h\}=\{x\stackrel{{\varphi }^{\prime }}{\to }{y}^{\prime }:p_{\mathcal{F}}({\varphi }^{\prime })=\mathrm{Id}\}$$This is because when we actually using definition of pullback, what we get is a diagramNow, since $g(\tilde{h})$ is Cartesian arrow as $\tilde{h}$ is Cartesian (by def of morphism of fibered cat), we see we getWe also have the $g(x)$ floating around, and we getHence we get bijection$$\{g(x)\to g(y)\text{ lying over }h\}=\{g(x)\to g(y^{\prime })\text{ lying over }\mathrm{Id}\}$$with the similar reasoning as above. Hence,$$\{x\to y\text{ lying over }h\}\stackrel{g}{\to }\{g(x)\to g(y)\text{ lying over }h\}$$is a bijection if and only if $g_{p_{\mathcal{F}}(x)}:\mathcal{F}(p_{\mathcal{F}}(x))\to \mathcal{G}(p_{\mathcal{F}}(x))$ is fully faithful. This concludes the proof.

$\square$

定义 3.1.0.12. We say $g:\mathcal{F}\to \mathcal{G}$ a map of fibered categories is an equivalence if there exists $h:\mathcal{G}\to \mathcal{F}$ map of fibered categories and exists a base preserving isomorphism $\alpha :g\circ h\stackrel{\sim }{\to }{\mathrm{Id}}_{\mathcal{G}}$ and $\beta :h\circ g\stackrel{\sim }{\to }{\mathrm{Id}}_{\mathcal{F}}$.

命题 3.1.0.13. For a map of fibered categories $g:\mathcal{F}\to \mathcal{G}$, $g$ is equivalence iff $\forall U\in \mathcal{C}$, $g_U$ is an equivalence (in category theory sense) iff $\forall U\in \mathcal{C}$, $g_U$ is fully faithful and essentially surjective.

证明. We already showed $g$ is fully faithful if and only if all $g_U$ are. Thus we just need to show the claims about essentially surjective (recall essentially surjective for $g:\mathcal{F}\to \mathcal{G}$ means each object $y\in \mathcal{G}$ is isomorphic to an object of the form $g(x)$ where $x\in \mathcal{F}$).

$(\Rightarrow ):$ if $g$ is equivalence, then we want $g_U$ to be essentially surjective. Given $y\in \mathcal{G}(U)$, we have $gh(y)\stackrel{\sim }{\to }y$ and $h$ is a morphism of fibered cats¹, so $h(y)\in \mathcal{F}(U)$.

$(\Leftarrow )$: now assume $g_U$ is essentially surjective for all $U$. We need to construct an equivalence of fibered categories $h:\mathcal{G}\to \mathcal{F}$. Given $y\in \mathcal{G}(U)$, since $g_U$ is equivalence, we know there exists $h(y)\in \mathcal{F}(U)$ such that ${\alpha }_y:y\stackrel{\sim }{\to }{g}_U(h(y))$. Given any $y\stackrel{\varphi }{\to }{y}^{\prime }$ in $\mathcal{G}(U)$, there exists unique $h(\varphi ):h(y)\to h(y^{\prime })$ in $\mathcal{F}(U)$ such thatbecause $g_U$ is fully faithful.

This gives functor $h:\mathcal{G}\to \mathcal{F}$ and also $\alpha :{\mathrm{Id}}_{\mathcal{G}}\stackrel{\sim }{\to }g\circ h$. We need $h$ sends Cartesian arrows to Cartesian arrows. If $y\stackrel{\varphi }{\to }{y}^{\prime }$ is Cartesian in $\mathcal{G}$, then we getand suppose we are given arbitrary $w$ with the following diagramwhere we want to show there exists unique arrow $w\to h(y)$. Since $g$ is fully faithful, there exists dotted arrow $w\to h(y)$ if and only if it holds after we apply $g$. Thus we want to show there exists unique dotted arrow in the following:However, note we can complete the diagram withIn other word, we getbut $y\to y^{\prime }$ is Cartesian, hence we indeed have the dotted arrow as desired.

Lastly, we need $\beta :{\mathrm{Id}}_{\mathcal{F}}\stackrel{\sim }{\to }h\circ g$. If $x\in \mathcal{F}$, we want $x\underset{{\beta }_x}{\overset{\sim }{\to }}h(g(x))$. By full faithfulness of $g$, we just need to show $y(x)\stackrel{\sim }{\to }g(h(g(x)))$. But we do have an isomorphism, ${\alpha }_{g(x)}$. Then, we left as an exercise that ${\beta }_x$ is a natural transformation.

$\square$

Now we have seen fibered categories are analogous to presheaves (over $\mathsf{Sets}$), we ask a natural question: what is analogue of sheaf? The answer is stacks.

Next, we ask if fibered categories are analogous to presheaves, is there a type of Yoneda lemma? Well, there is, and its called $2$-Yoneda lemma.

Before we do this, let’s recall if $X\in \mathcal{C}$, we have fibered category maps $\mathcal{C}/X\to \mathcal{C}$ with morphism $(Y\to X)\mapsto Y$. The analogy is that, $\mathcal{C}/X$ should correspond to $h_X$.

定理 3.1.0.14 ($2$-Yoneda Lemma). For any fibered category $\mathcal{F}\to \mathcal{C}$, and all $X\in \mathcal{C}$, we have a category ${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{F})$ with morphisms being base-preserving natural transformations. Then, we have a equivalence of categories$$\begin{array}{c}\zeta :{\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{F})\to \mathcal{F}(X)\\ g\mapsto g(X\stackrel{{\mathrm{Id}}_X}{\to }X)\end{array}$$

We will prove the $2$-Yoneda lemma next time, and we first prove an corollary of this.

推论 3.1.0.15. For $X,Y\in \mathcal{C}$, we have$$\begin{array}{c}{\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{C}/Y)\to {\mathrm{Hom}}_{\mathcal{C}}(X,Y)\\ f\mapsto f({\mathrm{Id}}_X)\end{array}$$is an equivalence of categories.

Note in the above, ${\mathrm{Hom}}_{\mathcal{C}}(X,Y)$ is a set viewed as a category with objects equal set elements, and morphisms being only identity maps.

证明. Well, apply $2$-Yoneda lemma to $\mathcal{F}=\mathcal{C}/Y$, we get$${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{C}/Y)=(\mathcal{C}/Y)(X)$$where the objects are $X\stackrel{\varphi }{\to }Y$ and morphisms areliving over ${\mathrm{Id}}_X$, i.e. $\psi ={\mathrm{Id}}_X$. Thus, the only morphism we have in $(\mathcal{C}/Y)(X)$ are identity maps, i.e. $(\mathcal{C}/Y)(X)$ is exactly the category ${\mathrm{Hom}}_{\mathcal{C}}(X,Y)$.

$\square$

Thus, we will introduce some notations: we will frequently write $X\to \mathcal{F}$ in place of $\mathcal{F}(X)$ if $\mathcal{F}$ is fibered category. This is justified by $2$-Yoneda because $\mathcal{C}/X\to \mathcal{F}$ is the same as $\mathcal{F}(X)$. So it is just a convenience to write $X$ in place of $\mathcal{C}/X$. The corollary shows ${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{C}/Y)={\mathrm{Hom}}_{\mathcal{C}}(X,Y)$, so $X\to Y$ in place of $\mathcal{C}/X\to \mathcal{C}/Y$ is unambiguous.

Last time, we have the $2$-Yoneda lemma

定理 3.1.0.16 ($2$-Yoneda Lemma). For any fibered category $\mathcal{F}\to \mathcal{C}$, and all $X\in \mathcal{C}$, we have a category ${\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{F})$ with morphisms being base-preserving natural transformations. Then, we have a equivalence of categories$$\begin{array}{c}\zeta :{\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{F})\to \mathcal{F}(X)\\ g\mapsto g(X\stackrel{{\mathrm{Id}}_X}{\to }X)\end{array}$$

Just like the proof of Yoneda lemma, this is not involved, but also not a short proof.

证明. We need to construct $\eta :\mathcal{F}(X)\to {\mathrm{Hom}}_{\mathcal{C}}(\mathcal{C}/X,\mathcal{F})$ that maps $x\in \mathcal{F}(X)$ to a fibered category $({\eta }_x:\mathcal{C}/X\to \mathcal{F})$.

First, we define ${\eta }_x$ on objects: an object in $\mathcal{C}/X$ is a map $Y\stackrel{\varphi }{\to }X$. In particular, we have diagramWell, the natural thing to do is just take a pullback ${\varphi }^{\ast }x$, i.e. we make a choice and get the following diagramand define ${\eta }_x(\varphi ):={\varphi }^{\ast }x\in \mathcal{F}(Y)$.

Next, we define ${\eta }_x$ on morphisms. Suppose we are given morphismin $\mathcal{C}$. We want to know what ${\eta }_x(\xi )$ is. Well, we get the following diagrambut then we get a dotted arrow between $({\varphi }^{\prime }{)}^{\ast }x$ to ${\varphi }^{\ast }x$ as the squares are Cartesian. Hence we haveThis unique dotted arrow gives the desired map on morphisms (i.e. ${\eta }_x(\xi ):({\varphi }^{\prime }{)}^{\ast }x\to {\varphi }^{\ast }x$).

Now we know $\eta$ as a functor $\mathcal{F}\to \mathcal{C}$, why is $\eta$ a morphism of fibered categories. We have to check two things: the first thing is that it respect fibers. However, we checked that already, i.e. $(Y\stackrel{\varphi }{\to }X)\mapsto \text{something in }\mathcal{F}(Y)$.

The second thing we need to show is that $\eta$ takes Cartesian arrows to Cartesian. First, in $\mathcal{C}/X$ all arrows are Cartesian, so we need ${\eta }_x(\text{any arrow})=\text{Cartesian}$. However, note by basic category theory, since the inner square and outer squares are both Cartesian, the dotted arrow must also be Cartesian.

So, we now know $\eta$ on objects. What about morphisms?

Given $f:x^{\prime }\to x$ in $\mathcal{F}(X)$, we want ${\eta }_f:{\eta }_{x^{\prime }}\to {\eta }_x$, i.e. we need ${\eta }_f$ to be base-preserving natural trans.

Given $\varphi :Y\to X$ in $\mathcal{C}/X$ in $\mathcal{C}/X$, we need ${\eta }_f(\varphi ):{\eta }_{x^{\prime }}(\varphi )={\varphi }^{\ast }{x}^{\prime }\to {\eta }_x(\varphi )={\varphi }^{\ast }x$. Lets draw out the diagramsbut then we get a unique dotted arrowthis is our definition of ${\eta }_f(\varphi )$. We still need to show this is base-preserving, i.e. it lives over the identity. Indeed, note $f:x^{\prime }\to x$ lives over the identity (as $f$ is a morphism in $\mathcal{F}(X)$), we see the pullback, i.e. the dotted arrow, also lives over identity. Hence ${\eta }_f$ is base-preserving natural transformation as desired.

We have now defined $\eta$. Next we need to show this is an equivalence.

First, we show $\zeta \eta =\mathrm{Id}$. Note we have$$\zeta \eta (x)=\zeta ({\eta }_x)={\eta }_x(X\stackrel{{\mathrm{Id}}_X}{\to }X)$$but note$${\eta }_x({\mathrm{Id}}_x):\mathcal{C}/X\to \mathcal{F}(X)$$is given by the following squarebut $({\mathrm{Id}}_X{)}^{\ast }x=x$ and hence $\zeta \eta (x)=x$ as desired.

Next, we show $\eta \zeta \cong \mathrm{Id}$. We see we get$$\eta \zeta (f:\mathcal{C}/X\to \mathcal{F})=\eta (f({\mathrm{Id}}_X))={\eta }_{f({\mathrm{Id}}_X)}$$where we see$${\eta }_{f({\mathrm{Id}}_X)}:\mathcal{C}/X\to \mathcal{F}$$and by definition we get the following diagramand taking pullback we getand we want to show ${\eta }_{f({\mathrm{Id}}_X)}(\varphi )={\varphi }^{\ast }(f({\mathrm{Id}}_X))\cong f(\varphi )$ because then ${\eta }_{f({\mathrm{Id}}_X)}(\varphi )=f(\varphi )$, i.e. $\eta \zeta (f)\cong f$.

To show this, it is enough to show the arrow $f(\varphi )\to f({\mathrm{Id}}_X)$ is Cartesian because if we have Cartesian diagramsthen since $2$ pullbacks are canonically isomorphic we get the desired isomorphism.

Now, why is the arrow $f(\varphi )\to f({\mathrm{Id}}_X)$ Cartesian? We always have unique canonical map $\varphi \to {\mathrm{Id}}_X$, as recall morphism of morpihsmes in our case is just try to fill the following diagram:but there is only a unique way to do this, which isThis gives an arrow $\varphi \to {\mathrm{Id}}_X$ which is automatically Cartesian. Thus $f(\varphi )\to f({\mathrm{Id}}_X)$ is Cartesian because $f$ preserves Cartesian arrows. This concludes the proof.

$\square$