2
$\begingroup$

Let $(\Lambda,\le )$ be a directed set, which we can understand as a small category: The set of all objects is $\Lambda$ and for $\lambda,\lambda '\in \Lambda$ there exists an unique morphism $i_{\lambda,\lambda '}:\lambda\to \lambda '$ iff $\lambda\le \lambda '$. Otherwise we don't have any morphisms.

Let $R-$MOD be the category of $R$-modules, $F:(\Lambda,\le )\to R$-MOD a functor. The directed colimit of $F$ is

$$\operatorname{colim}\limits_{\lambda} F:=\coprod\limits_{\lambda\in \Lambda}F(\lambda)/\sim ,$$where $v\sim F( i_{\lambda,\lambda '})(v)$ for $v\in F(\lambda),\; \lambda\le \lambda '$.

My question: is $\sim$ really an equivalence relation? This realtion isn't symmetric, or am I wrong?

Best

$\endgroup$

1 Answer 1

1
$\begingroup$

EDIT : [I will assume that by $\coprod\limits_{\lambda\in \Lambda}F(\lambda)$ you mean $\bigoplus\limits_{\lambda\in \Lambda}F(\lambda)$ using the general notation for coproduct (and not the disjoint union, which would be wrong).] Actually for directed colimits it doesn't matter.

Then you have to quotient by the equivalence relation that is generated by $v\sim F( i_{\lambda,\lambda '})(v)$ (which you have to show is a congruence).

$\endgroup$
4
  • $\begingroup$ ok, the professor in lecture states that this relation is already an equvalence relation, but this seems to be wrong, right? And we definited the colimit with the disjoint union. I don't know, why this should be wrong $\endgroup$
    – alg
    Mar 5, 2016 at 13:47
  • $\begingroup$ Strictly speaking, I guess you could call it wrong. But generally people implicitly symmetrize their relations in the definitions, so I wouldn't consider it wrong, just maybe somewhat misleading. $\endgroup$ Mar 5, 2016 at 14:26
  • 1
    $\begingroup$ As for the "disjoint union" part, it's my mistake : for directed colimit it doesn't matter. It's just bad practive, because for general colimits you will take the direct sum. $\endgroup$ Mar 6, 2016 at 17:51
  • $\begingroup$ thank you, your answer is very helpful! $\endgroup$
    – alg
    Mar 9, 2016 at 8:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .