After introducing the notion of finitely presentable object as an object $A$ such that ${\sf Hom}(A, -)$ preserves directed colimits, an "explicit" form of it is given:

$A$ is finitely presentable iff $(B, \bar{b}_i)$ is the colimiting cone for a directed diagram $(B_i, b_{ij}$, then for every arrow $f : A \to B$:

  • There is a $g : A \to B_i$ such that $f = \bar{b}_i \circ g$
  • For any $g', g'': A \to B_i$, such that $f = \bar{b}_i \circ g' = \bar{b}_i \circ g''$, there exists $j \geq i$ such that $b_{ij} \circ g' = b_{ij} \circ g''$.

(They can be found in page 2 of http://arxiv.org/pdf/1312.0432v1.pdf, on in Adamek and Rosicky's book, for example)

How is it proved that both conditions are equivalent? Is it that immediate? When I draw the diagram for the preserved colimit, I find that I should prove that for any $f : A \to B$ there must exist a $g : A \to D_i$ for some $i$ such that ${\sf Hom}(A, b_i)(g) = f$. Should I show that otherwise ${\sf Hom}(A, B) \setminus \{ f \}$ would be a "smaller" colimiting cone?


The point is that the directed colimit of $\hom(A,B_i)$'s in $\Bbb{Set}$ is the following quotient set: $${\coprod_i\hom(A,B_i)}\ \ /\sim$$ where $f_i\sim f_j$ for some $f_i\in\hom(A,B_i)$ and $f_j\in\hom(A,B_j)$, if $$\exists k\ge i,j:\ b_{ik}\circ f_i=b_{jk}\circ f_j\,.$$ Try to use this to get to the explicit form.

  • $\begingroup$ Thanks, I derived it using your hint! However, I still wonder if there's other way of showing the result which is more immediate, or one that doesn't use the form of colimits in ${\sf Set}$. $\endgroup$ – learer May 10 '14 at 2:24
  • $\begingroup$ Well, the proposition about directed colimits in the original category and in $\Bbb{Set}$. $\endgroup$ – Berci May 10 '14 at 11:42

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