# K-good trees and K-compactness of colimits over K-small downwards-closed subposets (500 point bounty if answered by Midnight EST))

## Question:

Let $D:A\to (X\downarrow C)$ be a $\kappa$-good $S$-tree rooted at $X$ for a collection of morphisms $S$ in $C$, where $\kappa$ is a fixed uncountable regular cardinal. Then according to the proof of Lemma A.1.5.8 of Higher Topos Theory by Lurie, for any $\kappa$-small downward-closed $B\subseteq A$, the colimit of the restricted diagram, $\varinjlim D|_B$ is $\kappa$-compact in $(X\downarrow C)$.

Why is this true? (It is stated without proof.)

## Definitions:

For your convenience, here are the definitions:

Recall that an object $X$ in $C$ is called $\kappa$-compact if $h^X(\cdot):=\hom(X,\cdot)$ preserves all $\kappa$-filtered colimits (where $\kappa$-filtered means "$<\kappa$"-filtered, since the terminology is different depending on the source).

Recall that an $S$-tree rooted at $X$ for a collection of morphisms $S$ in $C$ consists of the following data:

• An object $X$ in C (the root)
• A partially ordered set $A$ whose order structure is well-founded (the index)
• A diagram $D:A\to (X\downarrow C)$ such that given any element $\alpha\in A$, the canonical map $$\varinjlim D|_{\{\beta:\beta<\alpha\}}\to D(\alpha)$$ is the pushout of some map $U_\alpha\to V_\alpha\in S$.

We say that an $S$-tree is $\kappa$-good if for all of the morphisms $U_\alpha\to V_\alpha$ above, $U_\alpha$ and $V_\alpha$ are $\kappa$-compact, and such that for any $\alpha\in A$, the subset $\{\beta: \beta < \alpha \}\subseteq A$ is $\kappa$-small.

Edit: It's easy to reduce the proof to showing that $D(\alpha)$ is $\kappa$-compact, since projective limits of diagrams $B\to Set$ are $|Arr(B)|$-accessible (and therefore $\kappa$-accessible since $B$ is $\kappa$-small), we perform the computation for $I$ a $\kappa$-filtered poset, and $F:I\to C$, assuming that $D(\alpha)$ is $\kappa$-compact for all $\alpha\in B$:

$$\begin{matrix}\ \varinjlim_I Hom_C(\varinjlim_B D, F)&\cong&\varinjlim_I\varprojlim_{B^{op}} Hom_C(D,F)\\ &\cong& \varprojlim_{B^{op}}\varinjlim_I Hom_C(D,F)\\ &\cong& \varprojlim_{B^{op}} Hom_C(D,\varinjlim_IF)\\ &\cong& Hom_C(\varinjlim_B D,\varinjlim_IF) \end{matrix}$$

Edit 2: I think the above reduction actually won't work, since it doesn't use the hypothesis that B is downward-closed.

• If someone answers this in the next 3 hours, I will gladly award him or her a 500 point bounty as soon as it becomes possible (one must wait two days. It is forced by the system). Jun 20, 2010 at 22:17
• $\kappa$-compact objects are stable under $\kappa$-small colimits (Cor. 5.3.4.15). Reduction is valid, and $D(\alpha) = D_B$ for $B = \{\beta \leq \alpha\}$, but not helpful. Want transfinite induction on downward-closed subsets. Choose an increasing sequence $B(\gamma)$ as in proof of Lemma A.1.5.11. Base: $B(0) = \emptyset$ and $D_{\emptyset} = X$. Next: $D_{B(\gamma)} \to D_{B(\gamma+1)}$ is a pushout of a pushout of $U_{\alpha_{\gamma+1}} \to V_{\alpha_{\gamma+1}}$. Limit ordinal $\lambda$: $D_{B(\lambda)}$ is a $\lambda < \kappa$-small colim of earlier. (c.f., pf of Lemma A.1.5.6) Jun 21, 2010 at 3:41

Does this work?

To prove $D(\alpha)$ is $\kappa$-compact for all $\alpha$ in $A$, assume otherwise, that there exists some counterexample. Then, by the fact $A$ is well-ordered, there is a minimal counterexample (i.e., there is a minimal element $\alpha$ in the set of $\gamma \in A$ such that $D(\gamma)$ is not $\kappa$-compact). This means $D_\beta$ is $\kappa$-compact for all $\beta \lt \alpha$. Since $\{\beta: \beta \lt \alpha\}$ has cardinality less than $\kappa$, we have that

$$colim_{\beta: \beta \lt \alpha} D(\beta)$$

is $\kappa$-compact. Now, given a diagram of the form

$$V_\alpha \leftarrow U_\alpha \to colim_{\{\beta: \beta \lt \alpha\}} D(\beta)$$

in the category of $\kappa$-compact objects, its pushout is also $\kappa$-compact. But the hypothesis is that $D(\alpha)$ is the pushout for some such diagram, so $D(\alpha)$ is $\kappa$-compact, and we have reached a contradiction.

So $D(\alpha)$ is $\kappa$-small for all $\alpha \in A$. It follows that $colim_{\beta \in B} D(\beta)$ is $\kappa$-compact for any subposet $B \subseteq A$ whenever this is a $\kappa$-small colimit. (The restriction to downward-closed $B$ is not much loss of generality, because if $B \subseteq A$ is full, then the colimit over such a $B$ is isomorphic to the colimit over its downward closure, since $B$ is cofinal in its downward closure.)

• This does seem to work, but I'll wait for somebody to confirm before I accept it. Jun 21, 2010 at 9:33
• Yeah, I can't find anything wrong with it, so I'm accepting it. Jun 21, 2010 at 11:12
• I'll give you a 250 point bounty just because you're such a good sport =). Jun 21, 2010 at 16:34
• That's very, uh, bountiful of you! Thx. Jun 21, 2010 at 17:03
• The deed is done! (I gave you 300)! Jun 22, 2010 at 20:08

Let me add a remark on a subtle point which doesn't seem to be addressed in Todd's answer. (I'm sorry I'm digging up a decade-old post!) We wanted to show that, given a $$\kappa$$-good $$S$$-tree $$D:A \to (X\downarrow C)$$ and a donward-closed $$\kappa$$-small subset $$B\subset A$$, the colimit $$D_B=\operatorname{colim}_BD\vert_B$$ (colimit taken in $$X\downarrow C$$) is $$\kappa$$-compact in $$X\downarrow C$$. And this boils down to the assertion that each $$D(\alpha)$$ is $$\kappa$$-small. And for this, as is observed in Todd's answer, it suffices to show that assuming $$D_{\{\beta \in A\mid \beta <\alpha \}}$$ is $$\kappa$$-compact in $$X\downarrow C$$, the object $$D_\alpha$$ is $$\kappa$$-compact in $$X\downarrow C$$ also. By definition of $$\kappa$$-good $$S$$-trees, we have a pushout diagram in $$C$$ of the form

$$\require{AMScd} \begin{CD} U @>>> V\\ @VVV @VVV \\ D_{\{\beta \in A\mid \beta <\alpha \}} @>>> D(\alpha) \end{CD}$$

where $$C$$ and $$D$$ are $$\kappa$$-compact objects of $$C$$. But this is just a pushout square in $$C$$, and there is no guarantee that $$U$$ and $$V$$ are $$\kappa$$-compact objects in $$X\downarrow C$$ (what if there isn't any arrow $$X\to U$$, say?). So we cannot conclude by using the fact that $$\kappa$$-small colimits of $$\kappa$$-compact objects are $$\kappa$$-compact.

Nevertheless, we can still say that $$D(\alpha)$$ is $$\kappa$$-compact as an object of $$X\downarrow C$$. Indeed, the commutative square

$$\require{AMScd} \begin{CD} (X\downarrow C)(D(\alpha),-) @>>> (X\downarrow C)(D_{\{\beta \in A\mid \beta <\alpha \}},-)\\ @VVV @VVV \\ C(V,-) @>>> C(U,-) \end{CD}$$

of functors $$(X\downarrow C)\to \mathsf{Set}$$ is cartesian (by direct verification), and all the vertices except for the upper left one preserve $$\kappa$$-filtered colimits by hypothesis; note that $$\kappa$$-filtered colimits in $$X\downarrow C$$ can be computed in $$C$$ because filtered diagrams are connected. Thus, as a $$\kappa$$-small limit of functors preserving $$\kappa$$-filtered, we conclude that $$(X\downarrow C) (D(\alpha),-)$$ also preserves $$\kappa$$-filtered colimits, i.e., that $$D(\alpha)$$ is $$\kappa$$-compact.