The Grothendieck Construction for Cat-Valued Functors

 

Let

$$F:\mathcal{C}\to \mathbf{Cat}$$

be a strict functor. One may think of $F$$F$ as a family of categories varying over $\mathcal{C}$$\mathcal C$: for every object $c\in \mathcal{C}$$c\in\mathcal C$ there is a category $F(c)$$F(c)$, and for every morphism

$$f:c\to d$$

there is a functor

$$F(f):F(c)\to F(d).$$

The Grothendieck construction packages these categories and the transport between them into a single category, denoted by

$${\int }_{\mathcal{C}}F.$$

An object of ${\int }_{\mathcal{C}}F$$\int_{\mathcal C}F$ is a pair

$$(c,x),x\in \mathrm{Ob}F(c).$$

The interesting part is the morphisms. A morphism

$$(c,x)⟶(d,y)$$

consists of a morphism

$$f:c\to d$$

in $\mathcal{C}$$\mathcal C$, together with a morphism

$$\alpha :F(f)(x)\to y$$

in $F(d)$$F(d)$. We write such a morphism as

$$(f,\alpha ):(c,x)\to (d,y).$$

Equivalently,

$${\mathrm{Hom}}_{{\int }_{\mathcal{C}}F}((c,x),(d,y))=\coprod _{f:c\to d}{\mathrm{Hom}}_{F(d)}(F(f)x,y).$$

This should be compared with the category of elements of a functor

$$F:\mathcal{C}\to \mathbf{Set}.$$

If a set is regarded as a discrete category, then a morphism

$$F(f)x\to y$$

exists if and only if

$$F(f)x=y.$$

Thus the ordinary category of elements is simply the special case of the Grothendieck construction in which every fibre is discrete.

The composition law is where the categorical nature of the fibres becomes visible. Suppose that we have

$$(c,x)\stackrel{(f,\alpha )}{\to }(d,y)\stackrel{(g,\beta )}{\to }(e,z).$$

The types of the two fibre morphisms are

$$\alpha :F(f)x\to y\text{in }F(d),$$

and

$$\beta :F(g)y\to z\text{in }F(e).$$

In particular, the expression

$$\beta \circ \alpha$$

does not make sense. The two morphisms do not even lie in the same category, and the target of $\alpha$$\alpha$ is $y$$y$, whereas the source of $\beta$$\beta$ is $F(g)y$$F(g)y$.

However, since

$$F(g):F(d)\to F(e)$$

is a functor, we may apply it to $\alpha$$\alpha$. This gives

$$F(g)(\alpha ):F(g)F(f)x\to F(g)y.$$

Now the types match:

$$F(g)F(f)x\stackrel{F(g)(\alpha )}{\to }F(g)y\stackrel{\beta }{\to }z.$$

Since $F$$F$ is strict,

$$F(g)F(f)=F(gf),$$

and therefore

$$F(gf)x\stackrel{F(g)(\alpha )}{\to }F(g)y\stackrel{\beta }{\to }z.$$

Hence the composite is

$$(g,\beta )\circ (f,\alpha )=(gf,\beta \circ F(g)(\alpha )).$$

In this sense the composition formula is almost forced by type checking.

There is also a useful way to decompose every morphism. Given

$$(f,\alpha ):(c,x)\to (d,y),$$

we may factor it as

$$(c,x)\stackrel{(f,\mathrm{id})}{\to }(d,F(f)x)\stackrel{({\mathrm{id}}_d,\alpha )}{\to }(d,y).$$

The first morphism moves along the base category, while the second stays inside the fibre $F(d)$$F(d)$. It is therefore natural to call them horizontal and vertical morphisms.

Suppose now that

$$\alpha :u\to v$$

is a morphism in $F(d)$$F(d)$ and

$$g:d\to e.$$

We can first move vertically and then transport along $g$$g$:

$$(d,u)\stackrel{(\mathrm{id},\alpha )}{\to }(d,v)\stackrel{(g,\mathrm{id})}{\to }(e,F(g)v).$$

Alternatively, we may first transport $u$$u$ and then apply the transported morphism

$$F(g)(\alpha ):F(g)u\to F(g)v.$$

Thus

$$(d,u)\stackrel{(g,\mathrm{id})}{\to }(e,F(g)u)\stackrel{(\mathrm{id},F(g)(\alpha ))}{\to }(e,F(g)v).$$

The two composites agree. Symbolically, a vertical morphism can be moved past a horizontal transport, provided that it is replaced by its image under $F(g)$$F(g)$:

$$VH⇝HV.$$

This explains the composition law geometrically. Two morphisms, each written in horizontal-vertical form, initially give

$$HVHV.$$

The middle $VH$$VH$ may be exchanged to obtain

$$HHVV,$$

after which the two horizontal morphisms and the two vertical morphisms can be composed:

$$HVHV⇝HHVV⇝HV.$$

The formula

$$(g,\beta )\circ (f,\alpha )=(gf,\beta \circ F(g)(\alpha ))$$

is exactly the algebraic expression of this rewriting.

A number of familiar constructions arise in this way.

If a group $G$$G$ acts on a set $X$$X$, the action is a functor

$$BG\to \mathbf{Set}\subset \mathbf{Cat}.$$

The objects of the Grothendieck construction are the elements of $X$$X$, and

$$\mathrm{Hom}(x,y)=\{g\in G\mid gx=y\}.$$

Thus the Grothendieck construction is the action groupoid

$$X//G.$$

If $G$$G$ acts by order automorphisms on a poset $P$$P$, regarded as a category, then a morphism from $x$$x$ to $y$$y$ consists of a group element $g$$g$ together with a morphism

$$gx\to y$$

in $P$$P$. Since morphisms in a poset correspond to inequalities,

$$\mathrm{Hom}(x,y)=\{g\in G\mid gx\le y\}.$$

This gives the transporter category associated to the $G$$G$-poset $P$$P$.

Another important example is the semidirect product. Suppose that $G$$G$ acts on a group $H$$H$ by automorphisms,

$$\phi :G\to \mathrm{Aut}(H).$$

Regard $G$$G$ and $H$$H$ as one-object categories $BG$$BG$ and $BH$$BH$. The action determines a functor

$$F:BG\to \mathbf{Cat},F(\ast )=BH.$$

The Grothendieck construction again has only one object. Its endomorphisms may be identified with pairs

$$(h,g)\in H\times G.$$

The general composition formula becomes

$$(h,g)(h^{\prime },g^{\prime })=(h{\phi }_g(h^{\prime }),g{g}^{\prime }),$$

up to the convention used for ordering the two components. This is precisely the multiplication law of the semidirect product. Hence

$${\int }_{BG}BH\cong B(H{⋊}_{\phi }G).$$

From this point of view, the twisting term

$${\phi }_g(h^{\prime })$$

is simply the transported fibre morphism

$$F(g)(h^{\prime }).$$

There is one point in the preceding discussion where strictness was essential. In defining composition we used the equality

$$F(g)F(f)=F(gf).$$

For a pseudofunctor this equality is replaced by a specified natural isomorphism

$${\mu }_{g,f}:F(g)F(f)\stackrel{\sim }{\Rightarrow }F(gf).$$

Suppose again that

$$(c,x)\stackrel{(f,\alpha )}{\to }(d,y)\stackrel{(g,\beta )}{\to }(e,z).$$

We still obtain

$$F(g)(\alpha ):F(g)F(f)x\to F(g)y$$

and hence

$$\beta \circ F(g)(\alpha ):F(g)F(f)x\to z.$$

But the fibre component of a morphism lying over

$$gf:c\to e$$

must have source

$$F(gf)x.$$

Since $F(g)F(f)x$$F(g)F(f)x$ and $F(gf)x$$F(gf)x$ are no longer literally equal, we insert the coherence isomorphism:

$$F(gf)x\stackrel{{\mu }_{g,f,x}^{-1}}{\to }F(g)F(f)x\stackrel{F(g)(\alpha )}{\to }F(g)y\stackrel{\beta }{\to }z.$$

Thus the composition becomes

$$(g,\beta )\circ (f,\alpha )=(gf,\beta \circ F(g)(\alpha )\circ {\mu }_{g,f,x}^{-1}).$$

Similarly, the strict equality

$$F({\mathrm{id}}_c)={\mathrm{Id}}_{F(c)}$$

must be replaced by a unit isomorphism. The coherence axioms of a pseudofunctor ensure that the resulting composition is associative and unital.

Thus the passage from strict functors to pseudofunctors amounts to replacing equalities of transport by coherent natural isomorphisms:

$$\text{strict equalities}⟶\text{coherent isomorphisms}.$$

Finally, the Grothendieck construction comes with a natural projection

$$p:{\int }_{\mathcal{C}}F\to \mathcal{C},$$

defined by

$$(c,x)\mapsto c$$

and

$$(f,\alpha )\mapsto f.$$

For a covariant functor

$$F:\mathcal{C}\to \mathbf{Cat},$$

this projection is naturally an opfibration. Conversely, after choosing suitable opcartesian lifts, an opfibration over $\mathcal{C}$$\mathcal C$ determines a pseudofunctor from $\mathcal{C}$$\mathcal C$ to $\mathbf{Cat}$$\mathbf{Cat}$.

This is the Grothendieck correspondence:

$$\boxed{{\displaystyle \text{pseudofunctors }\mathcal{C}\to \mathbf{Cat}⟷\text{opfibrations over }\mathcal{C}.}}$$