1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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