In Riehl's Category Theory in Context, Sect. 4.6, we find the following:

Definition 4.6.16. Let $\kappa$ be a regular cardinal.¹ A locally small category $\mathsf{C}$ is locally $\kappa$-presentable if it is cocomplete and if it has a set of objects $S$ so that:

  1. Every object in $\mathsf{C}$ can be written as a colimit of a diagram valued in the subcategory spanned by the objects in $S$.
  2. For each object $s \in S$, the functor $\mathsf{C}(s,-): \mathsf{C} \rightarrow\mathsf{Set}$ preserves $\kappa$-filtered colimits.

A functor between locally $\kappa$-presentable categories is accessible if it preserves $\kappa$-filtered colimits.

For example, a large variety of categories whose objects are sets equipped with some sort of "algebraic" structure are locally finitely presentable (meaning locally $\omega$-presentable); see Definition 5.5.5. A locally $\kappa$-presentable category is also locally $\lambda$-presentable for any $\lambda>\kappa$.

¹ A regular cardinal is an infinite cardinal $\kappa$ with the property that every union of fewer than $\kappa$ sets each of cardinality less than $\kappa$ has cardinality less than $\kappa$.

First of all: I ignore the definition of "cardinal." I don't know any set theory, the only intuition of "cardinal" that I have is "isomorphism class in $\mathsf{Set}$."

My only question is: in Riehl's text, is "$\lambda>\kappa$" a typo? Shouldn't it be $\lambda<\kappa$ instead? It is what my intuition tells me, but again, I have no formal background to support this (since I ignore the relevant set-theoretic definitions). The definition of "$\kappa$-filtered colimit" I know is the one from Wikipedia. (I should also say, the naive definition of the order in cardinals I know is: $\kappa<\lambda$ if there is an injection $K\to\Lambda$ for some sets $K$, $\Lambda$ with cardinal $\kappa$, $\lambda$, respectively.)

  • $\begingroup$ (Small remark: there being an injection $K \to \Lambda$ is the definition of $\kappa \le \lambda$. The strict inequality $\kappa < \lambda$ means that $\kappa \le \lambda$ and $\kappa \ne \lambda$ - or equivalently, "there is an injection $K \to \Lambda$ but no injection/no bijection $\Lambda \to K$ for some/any sets $K$ and $\Lambda$ of cardinalities $\kappa, \lambda$".) $\endgroup$ Dec 25, 2023 at 1:12

1 Answer 1


It is not a typo*. When a category is locally $\lambda$-presentable, you can think of its objects being "smaller than $\lambda$" (in a suitable sense, don't take this literally!). When $\lambda \geq \kappa$, every object that is smaller than $\kappa$ is of course also smaller than $\lambda$, and hence locally $\kappa$-presentable implies locally $\lambda$-presentable. This is, of course, not a formal proof. But I hope that it corrects your intution you had which had it vice versa. For example, every locally $\omega$-presentable category is also locally $\lambda$-presentable (since $\omega \leq \lambda$).

Formal proof: It is enough to show that every functor preserving $\kappa$-filtered colimits also preserves $\lambda$-filtered colimits**. For this, it is enough to prove that every $\lambda$-filtered poset is also $\kappa$-filtered. Well, if $T$ is a subset of the poset with $< \kappa$ elements, it also has $< \lambda$ elements, hence has an upper bound.

*Funny enough, I also thought this is a typo when reading the book by Adamek-Rosicky for the first time many years ago.

**By definition, a $\lambda$-filtered colimit is a colimit indexed by a $\lambda$-filtered poset, meaning that every subset with $< \lambda$ elements has an upper bound.

  • $\begingroup$ Thank you so much! To sum it all up: If $\mathsf{J}$ is a $\kappa$-filtered category, then the bigger $\kappa$ is, the "smaller" $\mathsf{J}$ will be, right? $\endgroup$ Dec 24, 2023 at 12:39
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    $\begingroup$ I don't agree with that summary. Also, notice that $\kappa$ is always $\kappa$-filtered (since $\kappa$ is regular). $\endgroup$ Dec 24, 2023 at 12:40
  • $\begingroup$ Thanks again. Then I guess the relevant phenomenon at play may be more subtle. $\endgroup$ Dec 24, 2023 at 12:43
  • $\begingroup$ Is anything unclear? $\endgroup$ Dec 24, 2023 at 15:13
  • $\begingroup$ No, it's just that I've never worked with these ideas : ) $\endgroup$ Dec 24, 2023 at 15:31

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