0

$X_i\in\mathcal{A},i\in I$,where $I$ is an index set.

Let $\Lambda=\{X_i,X\}$ containing all $X_i$,$X=\coprod_{i\in I} X_i$.

Then $\Lambda$ is a directed set,with preoder $id_i:X_i\to X_i,\iota_i:X_i\to \coprod X_i$.

Consider the colimit of the system $\{j\}_{j\in\Lambda}$:

Claim:$\{X=\coprod X_i,\iota_i:X_i\to X,id_X\}$ is the colimit of the system.

Proof:If $M$ commutes with the system with $m_i:X_i\to M,m:\coprod X_i\to M$,then $m$ must be the morphism determined by the universal property of coproduct.

Hence we can define $f:X\to M$ as $m$.By the universal property of coproduct,$f$ is unique.Therefore $X$ is the colimit of the system.

Is the proof above correct?Thanks in advance!

shdvt
  • 639
  • 5
    Yes, but this doesn't let you construct the coproduct unless you already have the coproduct, and it is not what people mean when they say that coproducts are filtered colimits. Typically I assume what is meant is the observation that arbitrary coproducts are filtered limits of finite coproducts; that is, you take the directed set to consist of the coproducts over all finite subsets of $I$, ordered by inclusion. – Qiaochu Yuan Jan 31 '23 at 04:33
  • @QiaochuYuan Thanks.So if we have a functor that preserves filtered colimit,can we state that it preserves coproduct?Since I found it here that Free-R Module functor commutes with filtered colimits,but not preserves direct sum.Does this mean that we cannot find a 'naturual way'(maybe without using coproduct in the system) to construct coproduct as a filtered colimit? – shdvt Jan 31 '23 at 05:09
  • here in the comment and the first answer:https://math.stackexchange.com/q/3761216 – shdvt Jan 31 '23 at 05:10
  • 3
    No. It means that if a functor preserves finite coproducts and filtered colimits, then it preserves coproducts. The free $R$-module functor from $\text{Set}$ to $\text{Mod}(R)$ preserves all colimits. – Qiaochu Yuan Jan 31 '23 at 05:14
  • 1
    Note that a directed limit over a directed poset with a maximum element is always isomorphic to the object indexed by the maximum. – Arturo Magidin Jan 31 '23 at 06:21

1 Answers1

1

$\def\C{\mathcal{C}} \def\colim{\operatorname{colim}} \def\Hom{\operatorname{Hom}}$I will answer the question in the title:

Yes.

Lemma. A category with finite coproducts and filtered colimits has all coproducts. Specifically, if $\C$ is such a category and $\{X_i\}_{i\in I}$ is a set of objects of $\C$, then $$ \label{1}\tag{1} \coprod_{i\in I}X_i=\underset{F\in T}{\colim}\coprod_{j\in F}X_j, $$ where $T$ is the set of finite subsets of $I$, which is a directed set. The inclusion map of $X_i$ is given by the structure map of $X_i=\coprod_{j\in\{i\}}X_j$ into the colimit in $T$.

Proof. Since $$ \label{2}\tag{2} \Hom\left(\underset{F\in T}{\colim}\coprod_{j\in F}X_j,Y\right) \cong\underset{F\in T}{\lim}\prod_{j\in F}\Hom\left(X_j,Y\right), $$ it suffices to verify formula \eqref{1} in the category of sets (because then from \eqref{2} we'll conclude that $\underset{F\in T}{\colim}\coprod_{j\in F}X_j$ satisfies the universal property of the coproduct of $\{X_i\}_{i\in I}$).

If $\{X_i\}_{i\in I}$ are sets, then $$ \underset{F\in T}{\colim}\coprod_{j\in F}X_j =\frac{\coprod_{F\in T}\coprod_{j\in F}X_j} {x\sim \iota(x),\; \iota:\coprod_{j\in F}X_j\hookrightarrow \coprod_{j'\in F'}X_{j'},\; F\subset F'}. $$ It is not difficult to verify that this is the same as the disjoint union of the $X_i$'s. $\square$

Elías Guisado Villalgordo
  • 4,758