5
$\begingroup$

I'm reading a few papers on reflective factorization systems and I've just noticed they're all mentioning a procedure which seems very similar to the small object argument.

First of all, some background. My first encounter with the small object argument was Garner's paper about which I've asked several questions and learned a lot from. Lately I've been revisiting the basic theory of strong factorization systems in the context of categorical Galois theory and its connection to semi-left-exact reflections. I do not have a good understanding of the transfinite subtleties of small object arguments, but as far as I understand, what they do is freely construct cofibrantly generated factorization systems (weak or strong).

This question is about an apparent similarity between the goal of the small object argument, and an important procedure proving the finite well-completeness implies that reflective prefactorization systems are actually factorization systems.

Let $F\dashv G$ be an adjunction between $\mathsf C\overset{F}\rightleftarrows \mathsf B$. The prefactorization system on $\mathsf C$ fibrantly generated by the image of $G$ is $(\mathcal L,\mathcal R)=(\operatorname{\!^\perp Im}G, (\operatorname{\!^\perp Im}G)^\perp)$. Generally speaking this is not a strong factorization system. Let $f$ be an arrow. Seemingly the only way to get any interesting factorization of it is to factor the naturality square below through the pullback of $B\rightarrow GFB\leftarrow GFA$. This yields a factorization $f=v\circ w$, where $v$ is the pullback of $GFf$ along $\eta_B$ and $w$ is the unique arrow induced by the universal property of the pullback $P$. By definition $GFf\in \mathcal R$, and since $\mathcal R$ is always universal, so is $v$. Hence, in a sense, we already have "half" the $(\mathcal L,\mathcal R)$-factorization of $f$.

$$\require{AMScd} \begin{CD} A @>{\eta_A}>> GFA\\ @V{f}VV @VV{GFf}V\\ B @>>{\eta_B}> GFB \end{CD}$$

$w$ though is not generally in $\mathcal L$, which is always the class of arrows inverted by $F$. However, since $\mathcal R$ is closed under composition, finding an $(\mathcal L,\mathcal R)$-factorization of $w$ itself would suffice. The tentative strategy now is to take the intersection $m$ of all (strong) subobjects of the pullback object $P$ that are also contained in $\mathcal R$. If this creature exists, its universal property implies it must factor $v$ as $v=mg$ for some $g$, while limit closedness implies it's also in $\mathcal R$. Some careful arguments then show $g\in \mathcal L$.

The condition on a finitely complete category asking for intersections of strong subobjects to exist is called finite well-completeness. Adámek and Rosicky's book mentions locally $\lambda$-presentable categories as those in which every objects is a directed colimit of $\lambda$-presentable objects of a certain set, which seems at the very least analogous to finite well-completeness. Since locally presentable categories are basically "ones which admit a small object argument", I'd like to understand precisely the relation of the picture I sketched above with small object arguments in general.

Also, what can be done in the situation described if instead of an adjunction one just starts with some class of arrows $\Sigma$?

$\endgroup$
1

1 Answer 1

3
$\begingroup$

This isn't a full answer because it isn't completely precise, but I would say that the relationship is between "predicative" and "impredicative" constructions of universal objects.

Suppose $P$ is a poset and $f:P\to P$ is monotone and inflationary, i.e. $x\le y \Rightarrow f(x)\le f(y)$ and $x\le f(x)$. And say we have $a\in P$ and we want to construct the least fixed point of $f$ above $a$. (Since $f$ is inflationary, we have $f(x)=x$ iff $f(x)\le x$.)

One way to do it is to take the infimum of all fixed points of $f$ above $a$, if it exists. This is clearly still above $a$, and below all fixed points of $f$ above $a$, while it is itself a fixed point of $f$ since $f(\bigwedge x) \le f(x) \le x$ for each such $x$, hence $f(\bigwedge x) \le \bigwedge x$. This in an "impredicative" construction, analogous to the well-completeness argument.

Another way to do it is to consider the sequence $a \le f(a) \le f(f(a))\le f(f(f(a))) \le \cdots$ and take its supremum. This is also clearly above $a$, and below all fixed points above $a$ since if $a\le x$ and $f(x)=x$ then $f(a) \le f(x) = x$ too (and so on). And if $f$ preserves this supremum, then it is a fixed point of $f$, since $f(\bigvee_n f^n(a)) = \bigvee_n f(f^n(a)) = \bigvee_n f^{n+1}(a) = \bigvee_n f^n(a)$. (And if $f$ doesn't preserve this supremum, then we can continue into transfinite ordinals in hopes of finding some supremum that $f$ does preserve.) This is a "predicative" construction, analogous to the small object argument.

Universal properties in posets often have two constructions like this. However, when generalizing to categories, predicative constructions tend to fare better, because they involve (co)limits whose size is at least potentially smaller than that of the ambient category, while impredicative constructions tend to involve (co)limits of roughly the same size as the ambient category. Such "large limits" do sometimes exist --- well-completeness is one example, another is total categories --- but they are the exception rather than the rule; indeed a classical result of Freyd says that if a category has all limits of the same size of itself, then it must be a poset.

Locally presentable categories are those that are "determined by a small amount of data" in a certain precise sense. This ensures that the (co)limits involved in predicative constructions are (usually) in fact small, so that the small object argument works. And it also ensures that certain large limits exist, since they can also be "determined by a small amount of data"; for instance, every locally presentable category is total. So that gives at least an intuitive reason why local presentability seems to make both of these sorts of arguments work in some cases.

$\endgroup$

Your Answer

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

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