In Adamek and Rosickys' Locally Presentable and Accessible Categories, I came across the following statement (I'm paraphrasing),

every $\mu$-presentable object in a locally $\lambda$-presentable category (for regular cardinals $\mu\geq\lambda$) is a $\mu$-small colimit of $\lambda$- presentable objects. The proof is rather technical, however the following weaker statement is trivial: each $\mu$-presentable object is a split quotient of a $\mu$-small colimit of $\lambda$-presentable objects.

The book then goes on to prove the second statement. But, doesn't this statement directly prove the first as a split quotient can be expressed as a coequaliser between the corresponding split idempotent and the identity map, making it a coequaliser of a $\mu$-small colimit of $\lambda$-presentable objects and hence a $\mu$-small colimit of $\lambda$-presentable objects itself?

  • $\begingroup$ Suppose $x$ is the colimit at issue and $y$ is its split quotient, the given $\mu$-presentable object. What's your proposed diagram colimiting to $y$ consisting only of $\lambda$-presentable objects? The diagram defining $y$ uses $x$ twice, and only one of those occurrences can be removed via $x$'s universal property. $\endgroup$ – Kevin Arlin Jan 11 '17 at 18:47
  • $\begingroup$ Look at Makkai-Paré book, 2.3.11, 2.3.3, 2.1.2. $\endgroup$ – Ivan Di Liberti Jan 17 '18 at 0:20

Suppose $X_j$ is a diagram with shape having $\Omega$ many $\lambda$-presentable objects and $\Omega\leq \kappa<\mu$ many morphisms, and $X_j\xrightarrow{f_j}X$ its colimit. When $\lambda\leq\mu$, the $\lambda$-presentable $X_j$ are also $\mu$-presentable, so $X$ is $\mu$-small colimit of $\mu$-presentable objects, and hence $\mu$-presentable.

As a special case, any (split) quotient $X\rightrightarrows X\to Z$, $Z$ is also $\mu$-presentable since it is the colimit of a finite diagram of $\mu$-presentable objects.

By combining the two diagrams, you get that $Z$ is the colimit of the diagram $X_j\xrightarrow{f_j}X\rightrightarrows X$, whose shape has size at least $2^{\Omega}+\kappa$, which may be larger than $\mu$ if $2^\Omega>\mu$. In particular, the resulting diagramin does not have to be $\mu$-small. For this reason, the converse of the first statement does not follow directly from the converse of the second.

  • $\begingroup$ Well, and your diagram also fails to consist of $\lambda-$presentable objects, no? $\endgroup$ – Kevin Arlin Jan 11 '17 at 22:12
  • $\begingroup$ ... yes, of course. $\endgroup$ – Vladimir Sotirov Jan 11 '17 at 22:48
  • $\begingroup$ Just wondering why to bother with the counting of arrows. $\endgroup$ – Kevin Arlin Jan 11 '17 at 23:19
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    $\begingroup$ because sometimes I miss the obvious... $\endgroup$ – Vladimir Sotirov Jan 11 '17 at 23:25
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    $\begingroup$ Ah, sure. I thought you were ...-ing at me, not at yourself. That's a nice argument, anyway, not sure I've ever seen it used outside of proving that a large-complete category is a preorder. $\endgroup$ – Kevin Arlin Jan 11 '17 at 23:29

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