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We say that an object $X$ of a category $C$ is $\kappa$-compact (also $\kappa$-presentable and $\kappa$-accessible) for a cardinal $\kappa$ if $h^X(\cdot):=Hom(X,\cdot)$ commutes with all $\kappa$-filtered colimits.

In Makkai-Pare, a different but equivalent definition is given, that $X$ is $\kappa$-compact if for any $\kappa$-filtered functor $F:I\to C$ (i.e. the category $I$ is $\kappa$-filtered), any morphism $X\to \varinjlim F$ factors as $X\to F(i) \to \varinjlim F$, and any two factorizations $X\to F(i)\to \varinjlim F$ and $X\to F(i')\to \varinjlim F$ admit a majorant factorization $X\to F(i'')\to \varinjlim F$, that is, a factorization $X\to F(i'')\to \varinjlim F$ along with a commutative diagram

$$\begin{matrix} X&\to&F(i)\\ \downarrow&&\downarrow\\ F(i')&\to&F(i'')\\ \end{matrix}$$

such that $$F(i)\to \varinjlim F=F(i)\to F(i'')\to \varinjlim F$$ and $$F(i')\to \varinjlim F=F(i')\to F(i'')\to \varinjlim F$$ are the natural maps.

Let's try to show that they are equivalent:

Proof. '$\Rightarrow$' The existence of a factorization follows from the fact that the factorization exists in the category of sets, since giving a map $X\to \varinjlim F$ is the same as giving a map $$*\to h^X\varinjlim_i F(i)\cong\varinjlim_i h^XF(i),$$ and since we can give the colimit as a quotient of the disjoint union, the point $*\to \varinjlim_i h^XF(i)$ maps through some $h^XF(i)$, which gives us the necessary commutative triangle upon inspection.

Question:

In the definition of Makkai-Pare, it seems (to me) that we should need the existence of a majorant factorization for any family of factorizations indexed by a $\kappa$-small set. Why is this not the case?

Also, how do we prove the existence of the majorant factorization? I would guess that it follows from some property of filtered colimits in the category of sets, but I'm not familiar with what that is.

Edit: I would look in Makkai-Pare, but the only version I can find is on google books, and it cuts off literally right after the statement of the second definition. If anyone has it and is feeling generous, I'd very much appreciate any "assistance" you can give in finding it ;) (I'm currently not at school, so the library is not an option).

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Most of the formulas here aren't rendering for me — they give “Unknown control sequence ‘\underrightarrow’”. –  Peter LeFanu Lumsdaine Aug 2 '10 at 19:06
    
It sounds strange, but when I switched of "Preview Math" all the formulas became visible! –  Sasha Aug 2 '10 at 19:09
    
That is \varliminf, which is a macro for \underrightarrow{\operatorname{lim}}. I don't know what browser you're using, but it displays fine for me Opera and Firefox. –  Harry Gindi Aug 2 '10 at 19:17
    
A possible replacement is \underset{\rightarrow}{lim} and it would give the result as in faketestsite.stackexchange.com/questions/96/varinjlim-display . –  Anweshi Aug 2 '10 at 19:47
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It works fine after one refresh(at every visit). –  Anweshi Aug 2 '10 at 20:30

1 Answer 1

up vote 2 down vote accepted

The definitions are equivalent as stands; no extra conditions (eg majorants for infinite sets of factorisations) are needed. This is essentially because colimits in $\mathbf{Sets}$ are computed finitarily.

One way to present the colimit $\underset{i}{\varinjlim} Hom(X,F(i))$ (see eg Mac Lane CWM) is as

  1. the coproduct of all the sets $Hom(X,F(i))$, quotiented by

  2. the relation $\sim$ defined by: for $f \colon X \to F(i)$ and $g \colon X \to F(i')$, take $f \sim g$ if there is some $i''$ and a commutative square as in your original post.*

Looking at it this way, the two conditions in Makkai and Paré’s definition say that the canonical map $\underset{i}{\varinjlim} Hom(X,F(i)) \to Hom(X,\underset{i}{\varinjlim} F(i))$ is

  1. surjective;

  2. injective;

so together they say exactly that it’s an iso, which is what the usual definition says.

* For a general colimit, we’d need to use “the equivalence relation generated by $\sim$” (which is still something finitary), but if the colimit is filtered, so a fortiori if it’s $\kappa$-filtered, then $\sim$ is already an equivalence relation.

share|improve this answer
    
@Peter: You might want to try out \underset{i}{\varinjlim} in place of \varinjlim_i . –  Anweshi Aug 2 '10 at 22:14
    
thanks, good suggestion! –  Peter LeFanu Lumsdaine Aug 2 '10 at 23:51

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