MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a site $C$, there are various standard notions for an object $X \in C$ being compact. For instance:

  1. Every covering family $\lbrace U_i \to X \rbrace$ has a finite subfamily that is still covering.

  2. The functor $C(X,-)$ commutes with filtered colimits.

  3. After Yoneda-embedding, the functor $Sh_C(X, -)$ commutes with filtered colimits.

  4. After $\infty$-Yoneda-embedding, the functor $\infty Sh_C(X, -)$ commutes with filtered $\infty$-colimits.

These notions are closely related but subtly different. For instance for $C = Top$ it is well known that the first two are not equivalent without further fine-tuning.

What can one say about the relation of 1. to 3. and 4. ?

It seems to me that one can say for instance: compactness in the first sense implies that $Sh_C(X,-)$ commutes with mono-filtered colimits, and this should generalize to the $\infty$-case in the suitable sense.

What else can one say?

share|cite|improve this question
Some of these general questions are treated in Exposé VI of SGA 4. – Martin Brandenburg Apr 30 '12 at 8:55

The question that you ask might be better phrased intrinsically without referring to sites. Fix an $0 \leq n \leq \infty$ and let $\mathbf{X}$ be an $n$-topos. Recall that an object $X \in \mathbf{X}$ is quasi-compact if for every effective epimorphism of the form $\coprod_{i \in I} U_i \to X$ there is a finite subset $I_0 \subseteq I$ such that $\coprod_{i \in I_0} U_i \to X$ is an effective epimorphism. On the other hand, let us say that $X \in \mathbf{X}$ is small if the functor $Map_{\mathbf{X}}(X,-)$ preserves filtered colimits. Your question can then be phrased as follows:

What is the relation of between the property of being quasi-compact and the property of being small?

Using the fact that colimits in an $n$-topos are universal one can easily verify that $X \in \mathbf{X}$ is quasi-compact or small if and only if the terminal object in $\mathbf{X}_{/X}$ is quasi-compact or small respectively. We can hence assume that $X$ itself was the terminal object $\ast \in \mathbf{X}$. Now it is not hard to prove that $\ast \in \mathbf{X}$ is quasi-compact if and only if $\ast$ is small when regarded as an object of the underlying $0$-topos of $\mathbf{X}$ (i.e. the full subcategory spanned by $(-1)$-truncated objects). Hence we see that $\ast$ being small implies $\ast$ being quasi-compact. However, when $n > 0$ the inverse implication is false in general.

Here is a counterexample that works for every $0 < n \leq \infty$. Let $G$ be a free group on an infinite set of generators and let $\mathbf{X}$ be the $n$-topos of $(n-1)$-types equipped with a $G$-action. Then the terminal object $\ast$ is quasi-compact but the functor represented by $\ast$ is the homotopy fixed point functor, which does not commute with filtered colimits for such $G$.

share|cite|improve this answer
Hi Yonatan! Thanks for reviving this 2 year old question! What you write is closely related to the issue of commuting over mono-filtered diagrams that I mentioned. But your formulation is much nicer and better, thanks. – Urs Schreiber Jun 9 '14 at 19:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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