Mapping cone of $BF:B\mathcal C\to B\mathcal D$ is a classifying space?

Question

Let $F:\mathcal C\to \mathcal D$ be a functor. Then we get an induced map on the classifying spaces $BF:B\mathcal C\to B\mathcal D$. A famous theorem by Quillen says that, under mild assumptions, the homotopy fiber of $BF$ can be realized as the classifying space $B(Y\setminus F)$ of the comma category $Y\setminus F$ where $Y\in \mathcal D$.

Can the homotopy cofiber, i.e. the mapping cone, of $BF$ also be realized as a classifying space of some category? Does the classifying space of the other comma category give something interesting?

Answer 1

I am not sure about the result for categories but there are results on the case of the mapping cone $C(Bf)$ of $Bf: BG \to BH $ where $f: G \to H$ is a morphism of groups. These results are covered in Chapter 5 of the book partially titled Nonabelian Algebraic Topology (EMS Tracts, vol 15, 2011) (pdf available). Here the homotopy $2$-type of $C(Bf)$ is completely described by an "induced" crossed module $$\partial: f_*(G) \to H.$$ This result is one consequence of a 2-dimensional Seifert-van Kampen type theorem, for crossed modules. If $G,H$ are finite so also is $f_*(G)$, and various calculations are given.

More generally, the crossed module $$\partial: \pi_2(X \cup_g CA,X,x) \to \pi_1(X,x) $$ is analogously described in the case $A$ is connected and $g: (A,a) \to (X,x)$.

It is however unclear how to formulate an analogous result for monoids, let alone categories.

Answer 2

Yes, the mapping cone can be realized as the classifying space of a category via Thomason's homotopy colimit theorem. Recall that the mapping cone of a map $ f \colon A \to X$ is the homotopy colimit of the diagram

$$ \require{AMScd} \begin{CD} A @>f>> X \\ @VVV \\ * \\ \end{CD} $$

where $*$ is the space with only point.

Thomason's homotopy colimit theorem says that if $J$ is a small category and $G \colon J \to \text{Cat}$ is a covariant functor to the category of small categories, then

$$ \underset{J}{\text{hocolim}} \, BG \simeq B(\text{Gr}_J G) $$

where $\text{Gr}_J G$ is the Grothendieck construction of $G$. This is a category whose objects are pairs $(x,a)$ with $x$ an object of $J$ and $a$ an object of $G(x)$. And a morphism from $(x,a)$ to $(y,b)$ is pair $(\alpha,f)$ where $\alpha \colon x \to y$ is a morphism in $J$ and $f \colon G(\alpha)(a) \to b$ is a morphism in $G(y)$. Composition is not hard to figure out.

Now use this with the category $J$ given by

$$ \require{AMScd} \begin{CD} a @>f>> b \\ @VgVV \\ c \\ \end{CD} $$

and the functor $G \colon J \to \text{Cat}$ that sends $a$ to $C$, $b$ to $D$, $c$ to $*$ (the category with only one object and one morphism), $f$ to $F$ and $g$ to the constant functor. Then you have

$$ B(\text{Gr}_J G) \simeq \underset{J}{\text{hocolim}} \, BG \simeq C(BF) $$

So you just need to work out what $\text{Gr}_J G$ is in this case.

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There is an important case where things are bit easier (and cool), when you have the constant functor $C \to *$. In this case the mapping cone is the cone of $BC$ and can be realized as follows. Take the category $M$ whose objects are the objects of $C$ and another object, let us call it $\infty$. The morphisms between two objects in $C$ are the same as in the category $C$. There is an unique morphism from any object to $\infty$. And there are no more morphisms. Composition between morphisms in $C$ is still the same. And composition of a morphism in $C$ with a morphism to $\infty$ is the corresponding unique morphism to $\infty$. Then $BM \simeq C(BC)$. This construction can be found in "Topological, simplicial and categorical joins" by R. Fritsch and M. Golasinski.