[I'd be very happy for a better question title, if anyone has any suggestions.]

I have a category $C$, two functors $F,G : C \to \mbox{Cat}$, a natural transformation $\alpha : F \to G$, and a section $\sigma : G \to F$ of $\alpha$. I am interested in the colimits of the composites $C \xrightarrow{F,G} \mbox{Cat} \xrightarrow{N} \mbox{sSet}$. Specifically, calling these $X = \mbox{colim}_C(NF)$ and $Y = \mbox{colim}_C(NG)$, and calling the induced maps again

$$ \alpha : X \stackrel{\curvearrowleft}{\rightarrow} Y: \sigma ,$$

I would like to show that these last maps are inverse weak homotopy equivalences of simplicial sets.

Here are some additional things that I know how to prove about the situation.

  1. Both $NF$ and $NG$ are cofibrant in a model structure on $\mbox{Fun}(C,\mbox{sSet}$), so these are also homotopy colimits.
  2. The above diagram induces an isomorphism $\pi_0 X \cong \pi_0 Y$.
  3. For any $c \in C$ and any map $K \to N(F(c))$, the composite $K \to N(F(c)) \hookrightarrow X$ is homotopic to a map of the form $K \to Y \xrightarrow{\sigma} X$.
  4. $N(C)$ is weakly contractible.

By item 1, we can also study these homotopy types instead via their Grothendieck constructions, namely the opfibrations $\mbox{Gr}(F) \to C$ and $\mbox{Gr}(G) \to C$; more precisely, Thomason's homotopy colimit theorem says that we have weak homotopy equivalences $X' = N(\mbox{Gr}(F)) \xrightarrow{\approx} X$ and $Y' = N(\mbox{Gr}(G)) \xrightarrow{\approx} Y$, under which identifications the above section diagram becomes equivalent to one

$$ \alpha' : X' \stackrel{\curvearrowleft}{\rightarrow} Y' : \sigma' $$

of (nerves of) opfibrations over $N(C)$. Now, a section diagram induces one upon application of any functor, so in particular $\alpha'$ is surjective on $\pi_n$ for all $n$ (and all basepoints). So it would suffice to show that $\sigma'$ is also surjective on $\pi_n$, and by item 2 we only need to check $n \geq 1$. Note too that it suffices to check on free homotopy classes of maps. So a precise statement of my question could take the form:

Do items 3 and 4 suffice to show that any map $K \to X'$ factors through $\sigma' : Y' \to X'$ up to homotopy? (If it helps, we can take $K$ to be finite and connected, and it even suffices to obtain a factorization-up-to-homotopy $K' \to Y' \to X'$ for some replacement $K' \xrightarrow{\approx} K$.)

Of course, I'd also be happy if someone suggested an alternative way to prove this weak equivalence.

In general, I don't think we should expect "spheres in $X'$" to be detectable by the fibers of the opfibration alone; a toy example illustrating this is the torus fibering over the circle by projection onto one coordinate. But then there's item 4: I'd really like it to be true that somehow the weak contractibility of $N(C)$ forces it to be the case that any map $K \to X'$ factors up to homotopy through the inclusion of some fiber $N(F(c)) \hookrightarrow X'$ (which would solve the problem). But this intuition is coming from the undirected case (i.e. the homotopy theory of spaces, not of categories), which means at the very least that it should be taken with a grain of salt. And unfortunately, $C$ isn't filtered (otherwise this would be easy), and so I don't think I have much control over "why" spheres in $N(C)$ are always nullhomotopic.


1 Answer 1


Fix path-connected spaces $A \subset B$ with a retraction $r: B \to A$. Let ${\cal C}$ be the category $* \leftarrow * \rightarrow *$, whose nerve is weakly contractible, and consider the map of diagrams $$ (CB \leftarrow B \rightarrow CB) \to (CA \leftarrow A \to CA). $$ Here $C(-)$ denotes the cone and the maps are all induced by $r$. The induced map of colimits is the map of unreduced suspensions $\Sigma r: \Sigma B \to \Sigma A$. The claim is that this always satisfies your hypotheses. Properties (1) and (4) are both straightforward to check, and property (2) follows because of the path-connectivity constraint.

So we just have to check property (3). This translates to the following: any map $K \to B$ or $K \to CB$ with composite $K \to \Sigma B$ is homotopic to a map $K \to \Sigma A$. However, any map $K \to B$ or $K \to CB$ becomes a nullhomotopic map $K \to \Sigma B$, and so it is automatically homotopic to a map $K \to \Sigma A$.

In particular, if $B = S^1$ and $A = *$, then you do not get an equivalence.

  • 1
    $\begingroup$ Wow, thanks for the quick counterexample Tyler! Thankfully my actual situation is slightly more elaborate than a pushout, so maybe there's still a shot at it being true, but it won't follow directly from the facts I've assembled so far. I guess I need to paint on my wall "Hey, did you try looking at small examples yet?" :-P $\endgroup$ Dec 14, 2014 at 17:44

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