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Let $F,G:C\to D$ be naturally isomorphic functors. Taking the nerve, is $NF,NG:NC\to ND$ homotopy equivalent? Conversely, given a simplicial map $f:NC\to ND$, does there exists a functor $F:C\to D$ such that $NF=f$?

flag	asked Dec 14 2011 at 6:11 Gao 2Man 413●8

Just one comment. The homotopy class of f may be realized by non-equivalent functors $C\rightarrow D$. – Fernando Muro Dec 14 2011 at 7:45

... unless $D$ is a groupoid. – Fernando Muro Dec 14 2011 at 7:46

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@Gao 2Man: Although this is an interesting question, it can be answered just using the definitions. You should tell us what you have tried so far. See also mathoverflow.net/faq#whatnot – Martin Brandenburg Dec 14 2011 at 11:38

As Fernando notes, the content of your title is different from the content of your question: in particular the functor is not full... you cannot recover the category from the homotopy type of its nerve, in general. However, if you take Rezk's version of the nerve to get a bisimplicial set, the homotopy type of this remembers the original category. – Dylan Wilson Dec 14 2011 at 16:02

David Roberts · Accepted Answer · 2011-12-14 06:37:35Z UTC

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Yes to both. A natural transformation is a functor $C\times 2 \to D$, the nerve preserves products, and the nerve of $2$ is the 1-simplex. Geometric realisation gives homotopic maps. The second is elementary by the definition of nerve.

answered Dec 14 2011 at 6:37

David Roberts
11.9k●2●18●56

Is the nerve of a category a fully faithfully functor up to homotopy?

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