3
$\begingroup$

I am reading the paper by Jardine and Goerss, Localization theories for simplicial presheaves and having troubles with understand an argument. In this paper, the two authors considered $\mathcal{C}$ to be a small Grothendieck site (so in particular the underlying class of objects is a set and morphisms between two objects form a set). They proposed a list of axiom, called localization theories, so that whenever the site $\mathcal{C}$ satisfies this list, the category $\mathbf{PreSh}(\mathcal{C},\Delta^{op}\mathbf{Sets})$ of simplicial presheavse is a model category. For simplicity, we can think of weak equivalences are sectionwise equivalences and cofibrations are monomorphisms.

One of the axioms is:

Suppose that $\gamma$ is a limit ordinal and there is functor $X: \gamma \longrightarrow \mathbf{PreSh}(\mathcal{C},\Delta^{op}\mathbf{Sets})$ such that for each morphism $i \leq j$ in $\gamma$, $X(i) \longrightarrow X(j)$ is a trivial cofibration, then every canonical morphism $$X(i) \longrightarrow \operatorname{lim}_{j \in \gamma}X(j)$$ is a trivial cofibration.

As fas as I understand, this axiom was listed in order to make the small object argument work. Let me first pick a cardinal $\alpha$ bouding the set of morphisms of $\mathcal{C}$ and $\beta$ be some cardinal greater than $2^{\alpha}$ (this is where I do not understand the choice). Then the small object argument comes into the account: it factors any morphism $f: X \longrightarrow Y$ into $$X \longrightarrow X_{\beta} \longrightarrow Y$$ where $X_{\beta}$ is defined by a transfinite induction argument (basically, it is like what we should do for simplicial sets, we should fill all the holes $\partial \Delta[n] \subset \Delta[n]$), i.e. $X_{\beta}$ is some $\beta$-transfinite composition. What I do not understand is why $X_{\beta} \longrightarrow Y$ has to be a fibration. The authors proved this by using the lifting property. They proved that $X_{\beta} \longrightarrow Y$ has the right lifting property w.r.t all trivial cofibration $i:U \longrightarrow V$ with $V$ being $\alpha$-bounded (meaning that $\left|V_n(A) \right| \leq \alpha$ for every $n\geq 0$ and $A \in \mathcal{C}$).

Their argument is:

A morphism $U \longrightarrow X_{\beta}$ must factor through some stage $X_{\gamma}$ for some $\gamma < \beta$ because otherwise $U$ would have too many subobjects. Once the factorization is found, the lifting problem is solved.

I do not see why? Why $U \longrightarrow X_{\beta}$ does not factor implies that $U$ would have too many subobjects? And precisely, how many is too many?

$\endgroup$

1 Answer 1

1
$\begingroup$

To answer the question as it is stated: $U$ is an object in a locally presentable category, therefore $U$ is a small object, hence the corepresentable functor of $U$ preserves $α$-filtered colimits for some regular cardinal $α$. More precisely, the image of $U$ in such a colimit has a cardinality bounded by the product of cardinalities of $U$ and $\cal C$. Therefore, it must factor through some fixed stage of an $α$-filtered colimit if $α$ is greater than this cardinality.

However, that being said, the result discussed in the main post has been superseded by a considerably more general and easier to use theorem of Smith: for any left proper combinatorial model category, the left Bousfield localization at any set of morphisms $S$, and, more generally, at any class of morphisms $S$ that forms an accessible subcategory of $C$, exists. The conditions of Goerss and Jardine (especially, condition E7) simply guarantee that there is a set $S$ of such morphisms. For a reference, see, for example, Theorem 4.7 in Barwick's On left and right model categories and left and right Bousfield localizations.

$\endgroup$
7
  • $\begingroup$ I understand what you commented, making an use of smallness, but what I am still not over is the choice of $\beta$, how can I prove that $U$ is $\beta$-small (let's ignore regularity of cardinals for simplicity). $\endgroup$
    – Alexey Do
    Commented May 18, 2023 at 21:06
  • 1
    $\begingroup$ @AlexeyDo: Take $β$ to be any regular cardinal larger than the total number of elements in all sets $U(X)$, $X∈C$. Now if $U→\mathop{\rm colim}_i Y_i$ is a morphism of simplicial presheaves, then the image of $U$ is a simplicial subpresheaf of the colimit, which has at most as many elements as $U$. Since colimits of simplicial presheaves are computed objectwise, every individual element in the image factors through some stage $Y_i$. Also, if restrictions of two sections of $U$ to the same object in $\cal C$ are equal, then the relevant elements become equal in some $Y_j$. $\endgroup$ Commented May 18, 2023 at 21:22
  • $\begingroup$ @AlexeyDo: Since $β$ exceeds the cardinality of the set of $i$ and $j$, all elements in the image $U$ as well as any relations between them factor through some fixed stage $Y_k$, which proves $U$ is $β$-small. $\endgroup$ Commented May 18, 2023 at 21:23
  • $\begingroup$ Thank you, but I do not see Jardine and Goerss assume the regularity of cardinals, is this a mistake in their paper? $\endgroup$
    – Alexey Do
    Commented May 19, 2023 at 12:20
  • $\begingroup$ @AlexeyDo: If $β$ is a nonregular cardinal, then a category has $β$-filtered colimits if and only if it has $β^+$-filtered colimits, where $β^+$ is the successor cardinal of $β$, which is regular. Therefore, $β$-filtered colimits for nonregular cardinals $β$ are a redundant concept. Of course, if you only want to show an object to be $β$-small, there is no need to assume regularity. $\endgroup$ Commented May 19, 2023 at 16:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .