5
$\begingroup$

Has anybody attempted to define the notion of a factorization system "in" a derivator?

Something on the lines of this: let $\mathbb{D}$ be a (strong) derivator. We define the orthogonality relation in $\mathbb{D}$ to be the following, if we denote

  1. $\square$ the category $[1]\times[1]$ (the "universal lifting problem"), and $\not\!\square$ the category $[3]$ (the "universal solution");
  2. $u\colon \square\to \not \!\square$ the obvious functor;
  3. $p_L, p_R\colon [1]\to \square$ the embeddings of the left and right side of the square (idem for $q_L,q_R\colon [1]\to \not\!\square$)

Given $f,g\in \mathbb{D}([1])$, and motivated by the fact that this result seems to be true when $\mathbb{D}$ is the represented (pre)derivator $J\mapsto{\cal C}^J$, we say that "$f\perp g$" iff $u^*\colon \mathbb{D}(\not \!\square)\to \mathbb{D}(\square)$ induces an equivalence between the subcategory $\mathbf{S}(f,g)$ of solutions $X\in \mathbb{D}(\not\!\square)$ such that $q_L^*X = f, q_R^* X = g$ and the subcategory $\mathbf{P}(f,g)$ of problems $Y\in \mathbb{D}(\square)$ such that $p_L^*Y = f, p_R^* Y = g$.

The request that every lifting problem has a solution seems to be an essential surjectivity request for $u^*|_{\mathbf{S}(f,g)}$; the orthogonality relation generates the obvious Galois connection between (full, seen as set of objects?) subcategories ${\cal K}\subseteq \mathbb{D}([1])$; a factorization system on $\mathbb D$ is, now, a pair of classes ${\cal E,M}\subseteq \mathbb{D}([1])$ such that $\mathcal E = {}^\perp\mathcal M$, $\mathcal M = \mathcal E^\perp$, and such that the functor $\Phi\colon \mathbb{D}([2])\to \mathbb D([1])$ is an equivalence, when restricted to the subcategory of "commutative triangles" $T$ such that $d_2^*T\in\cal E$, $d_0^*T\in\cal M$, where $d_i\colon [1]\to [2]$ are the obvious faces.

There are several reasons why this notion seems unsatisfying (it only uses a small piece of information provided by $\mathbb{D}$), but I don't want to bias you for the moment. Instead, a bit of speculation: since any sensible definition must re-enact the classical theory, I'm led to hope that this, or another better definition would entail that: $\def\E{\mathcal E} \def\M{\mathcal M} \def\D{\mathbb{D}} \def\op{\text{op}}$

  1. ${\cal E}\cap \cal M$ is the class of isomorphisms (but what is, precisely, an isomorphism in a derivator? An object $f\in\mathbb{D}([1])$ such that $\text{dia}_{[1]}(f)$ is an isomorphism in $\mathbb{D}(e)$? Or an object in the essential image of $\mathbb{D}(\{0\cong 1\})\to \mathbb{D}([1])$? Are these classes the same?)

  2. The following conditions are equivalent for $f\in\mathbb{D}([1])$: (i) $f\perp f$ (ii) $f\perp \mathbb{D}([1])$ (iii) $\mathbb{D}([1])\perp f$; (iv) $f$ is an isomorphism (same question as above).

  3. If $(\E,\M)$ is a FS on $\D$, then $(\M^\op, \E^\op)$ is a FS on $\D^\op$.
  4. Let $\mathbf A$ be a bicomplete category; a FS on $\mathbf A$ induces a FS on $\D_\mathbf A$ (the represented prederivator $J\mapsto {\mathbf A}^J$).
  5. $\mathbf{M}$ a model category, define $J\mapsto \textsf{Ho}(\mathbf{M}^J) = \D_{\mathbf{M}}(J)$; then a Homotopy Factorization System $(\E, \M)$ (in the sense of Joyal) on $\mathbf{M}$, induces a FS on $\D_\mathbf{M}$.
  6. $\mathcal{C}$ an $\infty$-category; $J\mapsto \tau_1(\mathcal{C}^{NJ}) = \D_{\infty,\cal C}(J)$ is a derivator. If $(\E,\M)$ is a quasicategorical FS on $\cal C$, then $???$ is a FS on $\D_{\infty,\cal C}$.

...and many more other things. Have you ever heard of something similar? Do you have any advice on how to improve the definition?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.