12
$\begingroup$

Let $S$ be a simplicial set. Recall that there is a model structure, called the covariant model structure (see HTT ch. 2 and this question), on $\mathbf{SSet}/S$ such that:

  1. The cofibrations are the monomorphisms.
  2. 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.
  3. The fibrations are determined; the fibrant objects are the left fibrations $Y \to S$.

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?

$\endgroup$
18
$\begingroup$

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:

$(a)$: 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 $E$.

$(b)$: Using the functor $F$, you can construct an enlargement $\mathcal{E}_F$ of the category $\mathcal{E}$, adding a single object $v$ with $$Hom(E,v) = \emptyset \quad \quad Hom(v,E) = F(E) \quad \quad Hom(v,v) = \{ id \} $$

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:

$(i)$: 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.

$(ii)$: The functor $\alpha$ is an equivalence of categories.

$(iii)$: The functor $\beta$ is an equivalence of categories.

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.

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.

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 $s \in S$).

$\endgroup$
  • $\begingroup$ This is what I was looking for. Thanks. $\endgroup$ – Akhil Mathew Aug 28 '11 at 13:14
  • $\begingroup$ Perhaps it's worth mentioning that the functor in $(\operatorname{Set}_{\Delta})_{/S} \to (\operatorname{Set}_{\Delta})_{/S^{\triangleright}}$ that sends an object $(X,p : X \to S)$ to the object $(S \sqcup_X X^{\triangleright},\dots)$ is fully faithful, and its essential image contains precisely those objects $(Y, q : Y \to S^{\triangleright})$ that have the property that $q$ is an isomorphism over $S \sqcup \{*\}$. $\endgroup$ – Daniel Gerigk Sep 3 '17 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.