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 $NF=f$?

$\begingroup$ Just one comment. The homotopy class of f may be realized by nonequivalent functors $C\rightarrow D$. $\endgroup$– Fernando MuroDec 14, 2011 at 7:45

$\begingroup$ ... unless $D$ is a groupoid. $\endgroup$– Fernando MuroDec 14, 2011 at 7:46

1$\begingroup$ @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 $\endgroup$– Martin BrandenburgDec 14, 2011 at 11:38

$\begingroup$ 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. $\endgroup$– Dylan WilsonDec 14, 2011 at 16:02
1 Answer
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 1simplex. Geometric realisation gives homotopic maps. The second is elementary by the definition of nerve.