# $(\infty, 1)$-Yoneda embedding via the Grothendieck construction

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 $\mathfrak{C} \times \mathfrak{C}^{op} \to \mathrm{Kan}$.

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.

In Lurie's book, a notion of bifibration (as Mike Shulman observes, this is elsewhere called a two-sided fibration) 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.

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Why doesn't this work? Take $\text{Fun}(\Delta^1, C) \rightarrow C \times C^{op}$ defined by sending a morphism to it's source and target (on objects). The fibers seem correct. –  Dylan Wilson Apr 14 '12 at 17:21
And actually it's not hard to make the definition of that map rigorous as simplicial sets (I think). Probably you would send a map $\Delta^n \times \Delta^1 \rightarrow C$ to the pair $\Delta^n \rightarrow C$ (after evaluating on the source), and $\Delta^n \rightarrow C^{op}$ (after evaluating on the target... and maybe reversing the order of something). Anyway, I clearly haven't thought about this very much, so I'll stop here before further revealing my ignorance. –  Dylan Wilson Apr 14 '12 at 17:24
Wait, how does $\mathrm{Fun}(\Delta^1, C)$ map to $C^{op}$? All I can see is that it maps to $C \times C$. –  Akhil Mathew Apr 14 '12 at 18:47
Good point... I guess we need morphisms in the big category to look like commutative squares where the first vertical arrow is flipped... (Side note: stylistically it would make more sense to write $C^{op} \times C \rightarrow Kan$, right?) –  Dylan Wilson Apr 14 '12 at 19:37
(Since, unless I'm mistaken, this should just be the Hom-functor?) –  Dylan Wilson Apr 14 '12 at 19:38

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 $\Delta^{J^{op} } \star \Delta^{J}$ 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".
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.