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.)


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.

  • $\begingroup$ 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). $\endgroup$ Commented Jun 20, 2010 at 22:17
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    $\begingroup$ $\kappa$-compact objects are stable under $\kappa$-small colimits (Cor. 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) $\endgroup$ Commented Jun 21, 2010 at 3:41

2 Answers 2


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.)

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


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