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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?

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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.

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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
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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
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