MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathbf{sTop}$ be the functor category $\mathbf{Top}^{{\mathbf{\Delta}}^{\textit{op}}}$, and let $\mathbf{sCat}$ be the functor category $\mathbf{Cat}^{{\mathbf{\Delta}}^{\textit{op}}}$, and let $B:\mathbf{Cat}\rightarrow\mathbf{Top}$ be the classifying space functor (take nerve then realize). How do $B\underline{\mathbf{sCat}}(\mathcal{C},\mathcal{D})$ and $\underline{\mathbf{sTop}}(B\mathcal{C},B\mathcal{D})$ compare?

I think they are weakly equivalent (in the Reedy model structure), and I'm hoping that there might be a trivial cofibration between them. Anyone know a reference for something like this?

share|cite|improve this question
up vote 6 down vote accepted

Here's something easier. Let $C$ and $D$ be categories, and ask: is there an equivalence between $B\underline{\mathrm{Cat}}(C,D)$ (the classifying space of the functor category) and $\underline{\mathrm{Top}}(BC,BD)$ (the mapping space between two classifying spaces)?

In general, the answer is no. For instance, let $C$ be the category $(x\rightrightarrows y)$ with two objects and two non-identity arrows from $x$ to $y$. Then $BC$ is a circle, and $\underline{\mathrm{Top}}(BC,BC)$ is homotopy equivalent to $\mathbb{Z}\times S^1$, whereas $B\underline{\mathrm{Cat}}(C,C)$ is not (it has only finitely many components, for instance).

Taking constant simplicial objects should provide a counterexample to your claim, unless I've misunderstood what you are asking for.

share|cite|improve this answer
Ah, thank you! I confess that I didn't ask that easier question because I thought I had checked that case. Since in the analogous case for nerve you get an isomorphism, the problem has to be with realization. My 'proof' relied on the simplicial map $$ \mathbf{sSet}(S_{\bullet}\times\delta(k),T_{\bullet}) \rightarrow \mathbf{Top}(|S_{\bullet}\times\delta(k)|,|T_{\bullet}|) $$ being a weak equivalence, which is clearly not true. The issue that your example highlights seems to be that $B$ doesn't respect the direction of arrows. So might the (simpler) version be true for groupoids? – Alan Wilder Dec 4 '10 at 5:28
Yep, the simpler version will be true if $C$ and $D$ are groupoids (in fact, even if only $D$ is a groupoid). – Charles Rezk Dec 4 '10 at 5:34
Great; thanks again! – Alan Wilder Dec 4 '10 at 5:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.