The weak equivalences in the covariant model structure - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:48:29Z http://mathoverflow.net/feeds/question/73877 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73877/the-weak-equivalences-in-the-covariant-model-structure The weak equivalences in the covariant model structure Akhil Mathew 2011-08-28T01:09:18Z 2011-08-28T12:28:01Z <p>Let $S$ be a simplicial set. Recall that there is a model structure, called the covariant model structure (see HTT ch. 2 and <a href="http://mathoverflow.net/questions/16342/motivation-for-the-covariant-model-structure-on-sset-s" rel="nofollow">this question</a>), on $\mathbf{SSet}/S$ such that:</p> <ol> <li>The cofibrations are the monomorphisms.</li> <li>A map $X \to Y$ of simplicial sets over $S$ is a weak equivalence if $X^\vartriangleleft \sqcup_X S \to Y^\vartriangleleft \sqcup_Y S$ is a categorical equivalence (i.e., a weak equivalence in the Joyal model structure -- that is, one such that when applying the simplicial category functor $\mathfrak{C}$ gives an equivalence of simplicial categories). Here the triangle denotes the left cone.</li> <li>The fibrations are determined; the fibrant objects are the left fibrations $Y \to S$.</li> </ol> <p>I think I understand the motivation for most of this: as Lurie explains, left fibrations are the $\infty$-categorical version of categories cofibered in groupoids (so the fibrant objects model a reasonable concept), and the cofibrations are as nice as can be. But I fail to understand the motivation for the weak equivalences -- not least because I don't have a particularly good picture of what these "left cones" are supposed to model. Why should the weak equivalences be what they are?</p> http://mathoverflow.net/questions/73877/the-weak-equivalences-in-the-covariant-model-structure/73893#73893 Answer by Jacob Lurie for The weak equivalences in the covariant model structure Jacob Lurie 2011-08-28T12:20:57Z 2011-08-28T12:28:01Z <p>Maybe it would be helpful to think about the analogous situation in ordinary category theory. Suppose you are given a category $\mathcal{E}$ and a functor $F$ from $\mathcal{E}$ to the category of sets. There are several ways to encode this functor:</p> <p>$(a)$: Via the Grothendieck construction, $F$ determines a category $\mathcal{C}$ cofibered in sets over $\mathcal{E}$, so that for each object $E \in \mathcal{E}$ you can identify $F(E)$ with the fiber $\mathcal{C}_E$ of the map $\mathcal{C} \rightarrow \mathcal{E}$ over $E$.</p> <p>$(b)$: Using the functor $F$, you can construct an enlargement $\mathcal{E}_F$ of the category $\mathcal{E}$, adding a single object $v$ with $$Hom(E,v) = \emptyset \quad \quad Hom(v,E) = F(E) \quad \quad Hom(v,v) = { id } $$</p> <p>Now suppose we are given another functor $G$ from $\mathcal{E}$ to the category of sets, and a natural transformation $F \rightarrow G$. Then $G$ determines a category $\mathcal{D}$ cofibered in sets over $\mathcal{E}$, and an enlargement $\mathcal{E}_G$ of $\mathcal{E}$. The natural transformation $F \rightarrow G$ determines functors $$ \alpha: \mathcal{C} \rightarrow \mathcal{D} \quad \quad \beta: \mathcal{E}_F \rightarrow \mathcal{E}_G$$ In this situation, the following conditions are equivalent:</p> <p>$(i)$: The natural transformation $F \rightarrow G$ is an isomorphism (that is, for each object $E \in \mathcal{E}$, the induced map $F(E) \rightarrow G(E)$ is bijective.</p> <p>$(ii)$: The functor $\alpha$ is an equivalence of categories.</p> <p>$(iii)$: The functor $\beta$ is an equivalence of categories.</p> <p>Now observe that the category $\mathcal{E}_F$ can be described as the pushout (and also homotopy pushout) of the diagram $$\mathcal{E} \leftarrow \mathcal{C} \rightarrow \mathcal{C}^{\triangleleft},$$ where $\mathcal{C}^{\triangleleft}$ is the category obtained from $\mathcal{E}$ by adjoining a new initial object.</p> <p>Let's now forget the original functors $F$ and $G$, and think only about the categories $\mathcal{C}$ and $\mathcal{D}$ cofibered in sets over $\mathcal{E}$. The equivalence of conditions $(ii)$ and $(iii)$ shows that functor $\alpha: \mathcal{C} \rightarrow \mathcal{D}$ of categories cofibered over $\mathcal{E}$ is an equivalence of categories if and only if the induced map $$ \mathcal{E} \amalg_{ \mathcal{C} } \mathcal{C}^{\triangleleft} \rightarrow \mathcal{E} \amalg_{ \mathcal{D} } \mathcal{D}^{\triangleleft}$$ is an equivalence of categories.</p> <p>Now go to the setting of quasi-categories. Assume for simplicity that $S$ is a quasi-category, and let $f: X \rightarrow Y$ be a map of simplicial sets over $S$. If $X$ and $Y$ are left-fibered over $S$, then we would like to say that $f$ is a covariant equivalence if and only if it an equivalence of quasi-categories. However, we would like to formulate this condition in a way that will behave well also when $X$ and $Y$ are not fibrant. Motivated by the discussion above, we declare that $f$ is a covariant equivalence if and only if it induces a categorical equivalence $$ S \amalg_{X} X^{\triangleleft} \rightarrow S \amalg_{Y} Y^{\triangleleft}.$$ You can then prove that this is a good definition (it gives you a model structure with the cofibrations and fibrant objects that you described, and when $X$ and $Y$ are fibrant a map $f: X \rightarrow Y$ is a covariant equivalence if and only if it induces a homotopy equivalence of fibers $X_s \rightarrow Y_s$ for each vertex $s \in S$).</p>