most recent 30 from http://mathoverflow.net 2013-06-18T06:14:03Z http://mathoverflow.net/feeds/question/94036 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94036/infty-1-yoneda-embedding-via-the-grothendieck-construction Akhil Mathew 2012-04-14T15:36:10Z 2012-04-15T11:36:13Z

<p>Let $C$ be a quasi-category. Then there is an imbedding $$ C^{op} \times C \to \mathrm{Kan}$$ where $\mathrm{Kan}$ is the quasi-category of Kan complexes. This is essentially constructed in Lurie's book by choosing a model for $C$ as the nerve of a fibrant simplicial category $\mathfrak{C}$ and then taking the nerve of the ordinary Yoneda imbedding $\mathfrak{C} \times \mathfrak{C}^{op} \to \mathrm{Kan}$. </p> <p>However, by the Grothendieck construction, the Yoneda imbedding should correspond to a left fibration (i.e., a quasi-category cofibered in Kan complexes) over $C \times C^{op}$. Is there a direct way to construct such a left fibration? I'm wondering if there is a way to do this without appealing to the theory of simplicial categories. I'd even be happy with something weakly equivalent to this left fibration in the covariant model structure. </p> <p>In Lurie's book, a notion of <em>bifibration</em> (as Mike Shulman observes, this is elsewhere called a <em>two-sided fibration</em>) over a product $S \times T$ of simplicial sets is introduced, to correspond to the idea of a Kan complex functorial in two variables, but covariantly in one and contravariantly in another. I'm not very familiar with this theory, but a more general question would be how to turn a bifibration into a right or left fibration. </p>

http://mathoverflow.net/questions/94036/infty-1-yoneda-embedding-via-the-grothendieck-construction/94102#94102 Jacob Lurie 2012-04-15T11:36:13Z 2012-04-15T11:36:13Z

<p>Let $\mathcal{M}$ be the simplicial set defined by the formula $Hom( \Delta^{J}, \mathcal{M} ) =Hom( \Delta^{J^{op} } \star \Delta^{J}, \mathcal{C} )$, so that an $n$-simplex of $\mathcal{M}$ is a $(2n+1)$-simplex of $\mathcal{C}$. The inclusions of $\Delta^{J^{op} }$ and $\Delta^{J}$ into $\Delta^{J^{op} } \times \Delta^{J}$ induce a left fibration $\mathcal{M} \rightarrow \mathcal{C}^{op} \times \mathcal{C}$, which is the left fibration you are looking for. For details see section 4.2 of my paper "Derived Algebraic Geometry X", entitled "Twisted Arrow $\infty$-Categories".</p> <p>Your more general question can be phrased as follows: given a coCartesian fibration $q: X \rightarrow S$ classified by a functor from $S$ into $\infty$-categories, how can you explicitly construct a Cartesian fibration classified by the same functor? First, construct a simplicial set $Y$ with a map $Y \rightarrow S$, such that maps $T \rightarrow Y$ classify maps $T \rightarrow S$ together with maps from $X \times_{S} T$ to the $\infty$-category of spaces. Then $Y \rightarrow S$ is a coCartesian fibration, whose fibers over a vertex $s \in S$ is the $\infty$-category of presheaves on $X_{s}^{op}$. Restricting to representable presheaves determines a full simplicial subset of $Y_0 \subseteq Y$, and the map $Y_0^{op} \rightarrow S^{op}$ is the Cartesian fibration you're looking for. </p>