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I have the following situation. $M$ is a combinatorial model category, or if you like a locally presentable $(\infty,1)$-category. I have a set of maps $S$ and I let $C$ be the class of maps generated from maps in $S$ via directed colimits. This means every $f\in C$ can be written as a directed colimit of $f_i \in S$, and because $M$ is (co)complete this means the domain of $f$ and is the colimit of the domains of the $f_i$ and similarly for the codomain of $f$.

Is it true that $f$ is in $S$-cell (i.e. it's a retract of some transfinite compositions of pushouts of maps in $S$)? More generally, do we have $S$-cell $=$ $C$-cell?

A word of caution. Various facts about model categories will tell you that $S$-cell is closed under directed colimits and maybe even filtered colimits. But it seems to me that's for colimits taken in $M$ not in $Arr(M)$, i.e. it's saying that if we have a chain $X_0\to X_1\to \dots$ and each map $X_i \to X_{i+1}$ is in $S$-cell then the colimit $X_0\to X_\lambda$ is in $S$-cell. This is not what I'm asking for. I need a ladder $f_0\to f_1\to \dots$ and to know the colimit $f$ is in the same class $S$-cell which all the elements of the directed system are in. I'd like this for any directed colimit of the $f_i$ but I'm willing to accept it for a colimit of the sort just drawn, i.e. a sequential colimit. At least then I'd have an intuition for whether or not this is true.

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    $\begingroup$ To clarify notation, in Dylan's answer below $S$ is the set of split mono's between finite posets, $S$-cell is the set of split mono's (which he calls $C$), what I call $C$ is the closure of $S$ under directed colimits, which Rosicky comments is the set of pure monomorphisms. What Dylan shows is that my $C$ is not contained in $S$-cell, i.e. there is pure mono $\omega \to \omega \cup \infty$ which is not an element of $S$-cell, because it's not split. It's pure because Dylan explicitly writes it as a sequential colimit in $Arr(Pos)$ of elements in $S$. $\endgroup$
    – David White
    May 21, 2013 at 11:59

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The answer is 'no' (sorry I didn't realize this before!)

I remembered someone showing me a weird counterexample in model categories a couple weeks ago (elaborated on from this paper by Rosicky: http://www.math.muni.cz/~rosicky/papers/comb2.pdf)

and it seems to do the trick here:

Consider the category ${\bf Pos}$ of posets. This is certainly presentable. Let $C$ denote the collection of split monomorphisms (which is weakly saturated and generated by split monomorphisms between finite posets). Then we can define a combinatorial model structure on $\{\bf Pos}$ by taking all morphisms as weak equivalences (this satisfies the conditions of, say, Higher Topos Theory A.2.6.7. or one can see this directly). However, $C$ is not closed under sequential colimits because, for example, the non-split monomorphism arrow $\omega \rightarrow \omega \cup \infty$ is the sequential limit of the arrows $[n] \hookrightarrow [n] \cup \infty$, each of which is split.

P.S. What is your distinction between directed colimits and filtered colimits (this terminology has always confused me...)? The way you're using the word makes it seem like "directed" is the same as "finitely filtered".

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    $\begingroup$ Directed colimits are indexed by directed posets, filtered colimits are indexed by filtered categories. $\endgroup$
    – Zhen Lin
    May 18, 2013 at 12:33
  • $\begingroup$ Ok, finally had a chance to read this carefully. It seems like the lemma in HTT is there to complete the construction of a combinatorial model structure by giving a set $J$ of generating trivial cofibrations to match the given set of generating cofibrations which makes $C$ into the cofibrations of Pos, though it´s difficult to get your hands the elements of $J$. This is a great example! It actually answers a harder question than the one I asked, namely restricting to the case where $S$ and $C$ consist entirely of cofibrations $\endgroup$
    – David White
    May 21, 2013 at 11:22
  • $\begingroup$ It seems to me that all these splittings $[n] \cup \infty \to [n] \subset \omega$ should assemble to a map $f:\omega \cup \infty \to \omega$ because the domain is a colimit. Obviously the problem is that there's no largest element in $\omega$ to send the extra point $\infty$ to so $\omega \to \omega \cup \infty$ can't be split. Still, it would be interesting to see exactly why $f$ is not a splitting. After all, for all $n$ we have $f_n \circ i_n = id_{[n]}$ so you'd think the colimit would be $id_\omega$. Can anyone clarify why this argument is wrong? I feel like my intuition is broken here $\endgroup$
    – David White
    May 21, 2013 at 11:38
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    $\begingroup$ @David: The splittings (there is a unique splitting) are not compatible so don't give a map to the colimit. Indeed, the composition $[n] \cup \infty \rightarrow [n] \rightarrow [n+1]$ sends $\infty$ to $n$ whereas the composition $[n] \cup \infty \rightarrow [n+1] \cup \infty \rightarrow [n+1]$ sends $\infty$ to $n+1$. $\endgroup$
    – Dylan Wilson
    May 21, 2013 at 13:09
  • $\begingroup$ @Dylan: that's a good point. Thanks. It makes sense now $\endgroup$
    – David White
    May 21, 2013 at 19:31

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