I'm trying to learn about Grothendieck fibrations in category theory and was wondering if the following is an example of one:

Let $C$ be a category and let $\tilde{C}$ denote the category whose objects are functors $X:J\rightarrow C$ ($J$ is a small diagram category) and whose morphisms $(J\xrightarrow{X} C)\rightarrow (K\xrightarrow{Y} C)$ are pairs $(f,g)$ where $f:J\rightarrow K$ is a functor and $g:X\Rightarrow Y\circ f$ is a natural transformation.

Is the forgetful functor $F:\tilde{C}\rightarrow Cat$ ($Cat$ denotes the 1-category of small categories) that sends $(J\xrightarrow{X} C)\mapsto J$ a fibration? I'm sure this is the case when the domains of the objects in $\tilde{C}$ are discrete small categories (sets).


1 Answer 1


A functor is a fibration if every morphism in the base to the image of an object in the total category lies below a cartesian morphism to the object in the total category. Thus, it is enough to understand the cartesian morphisms of the functor.

Recall that given a functor $\mathcal E\xrightarrow{p}\mathcal C$, a morphism $Y\xrightarrow{\alpha} Z$ in $\mathcal E$ is $p$-cartesian if whenever $X\xrightarrow{\gamma}Z$ in $\mathcal E$ is such that $p(X)\xrightarrow{p(\gamma)}p(Z)$ in $\mathcal C$ factors as $p(X)\xrightarrow{b}B\xrightarrow{p(\alpha)}p(Z)$ in $\mathcal C$, there is a unique morphism $X\xrightarrow{\beta}Y$ in $\mathcal E$ with $p(\beta)=b$ and $X\xrightarrow{\gamma}Z$ factoring as $X\xrightarrow{\beta}Y\xrightarrow{\alpha}Z$.

In your case, a cartesian morphism above a functor $J\xrightarrow{a}K$ given a functor $K\xrightarrow{Z}C$ will be a natural transformation $Y\overset{\alpha}\Rightarrow Z\circ a$ for functors $\mathcal C\xleftarrow{Y}\mathcal J\xrightarrow{a}\mathcal K\xrightarrow{Z}\mathcal C$ so that for each $X\overset\gamma\Rightarrow Z\circ a\circ b$ for functors $\mathcal C\xleftarrow{X}\mathcal I\xrightarrow{b}\mathcal J\xrightarrow{a}\mathcal K\xrightarrow{Z}\mathcal C$, there is a unique $X\overset\beta\Rightarrow Y\circ b$ for $\mathcal C\xleftarrow{X}I\xrightarrow{b}\mathcal J\xrightarrow{Y}\mathcal C$ so that $X\overset\gamma\Rightarrow Z\circ c$ whiskers as $X\overset\beta\Rightarrow Y\circ b\overset{\alpha . b}\Rightarrow (Z\circ a)\circ b=Z\circ c$.

Choosing $Y=Z\circ a$ and $Y\overset{\mathrm{id}}\Rightarrow Z\circ a$, the unique $X\overset\beta\Rightarrow Y\circ b$ is given by $\beta=\gamma$. In other words, the fibration has the usual splitting of the domain fibration and the $2$-categorical structure plays no role.


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