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Topological Grothendieck Construction

Let $C$ be a small category and $F\colon C^{op}\rightarrow Set$ a functor. The Grothendieck construction is the category $F\wr C$ with objects being pairs $(c,x)$ where $c$ is a object of $C$ and $x\in F(c)$. An arrow from $(c,x)\rightarrow (c',x')$ is an arrow $f\colon c\rightarrow c'$ with $F(f)(x')=x$. This comes with a natural functor $F\wr C\rightarrow C$ forgetting the second coordinate.

In other words it is a category $F\wr C$ together with a functor $F\wr C\rightarrow C$, such that the diagram of nerves

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} N_1(F\wr C) & \ra{d_0} & N_o(F\wr C) \\ \da{} & & \da{} \\ N_1(C) & \ra{d_0} & N_o(C) \end{array} $$ is cartesian.

Now topologize the whole situation, i.e. let $C$ be a topological category (a category internal to $Top$) and $F\colon C^{op}\rightarrow Top$ a functor. The grothendieck construction of this situation should give me a topological category $F\wr C$ with a continuous functor $F\wr C\rightarrow C$, such that the diagram above is cartesian. However, I ran into difficulties making this precise. Taking the non-topological definition for the underlying sets $ob(F\wr C)$ and $mor(F\wr C)$, one can topologize the set of morphisms as subspace of $mor(C)$, but what is the right topology on $ob(F\wr C)$?

In some situations, the way to go is obvious :

  • If $F$ takes values in subspaces of a fixed topological space $X$, one could topologize $ob(F \wr C)$ as subspace of $ob(C)\times X$.
  • If $C$ has a discrete set of objects, $ob(F\wr C)$ can be topologized as a subspace of $\coprod\limits_{c\in ob(C)}\{c\}\times F(c)$.