# Existence of morphisms in a free completion under directed colimits,$\lambda$-accessible category

Let $\cal K$ be a $\lambda$-accessible category with directed colimits and $\cal C$ be its representative full subcategory consiting of $\lambda$-presentable objects. Let $\cal L$ be free completion of $\cal C$ under directed colimits.There is a functor $F:$Ind$(\cal C)\to \cal K$ preserving directed colimits. We want to show that this $F$ is faithful. An unclear proof follows. Consider two distinct morphisms $f$,$g$ in Ind$(\cal C)=L$.Why there are morphisms $f',g':A\to B$ in $\cal C$ and morphisms $u:A\to K$ and $v:B\to L$ such that $fu=vf'$ and $gu=vg'$. From now on I do understand. Since $F$ is faithful on $\cal C$ (because it is identity on $\cal C$), $Ff',Fg'$ are distinct.Since $Fv$ is a monomorphism,$Ff,Fg$ are distinct. See also the page 7 here.I also do understand that for any $\lambda$-presentable object $K$ any morphism $h$ from $K\to D$ to a $\lambda$-directed colimit of objects $D_i$, $D$=colim$D_i$, essentially uniquely factorizes through some $D_i$.

• Do you know a construction of the free cocompletion under filtered colimits? – Kevin Carlson Apr 17 '18 at 2:15
• They are just $\lambda$-preserving functors from $\cal K$ into $\mathbb{Set}$, right? – user122424 Apr 17 '18 at 11:20
• Well, there's no $\lambda$ involved. If $C$ had finite colimits then the completion under finite colimits $L$ would be the functors $F:C^{op}\to Set$ sending finite colimits to limits. If $C$ is general, then instead one says $F$ is flat: its category of elements is cofiltered. But the way to get your hands on $L$ is to view its objects as filtered colimits of representables. This makes it relatively easy to compute the hom-sets in $L$ in terms of those in $C$. – Kevin Carlson Apr 17 '18 at 17:14
• Notice I'm using filtered colimits where Rosicky uses directed ones-there's no important difference for this purpose. – Kevin Carlson Apr 17 '18 at 17:15

## 1 Answer

Given $f,g\colon K\to L$ you can express $K$ and $L$ as directed colimits $K=\mathrm{colim}_i A_i$ and $L=\mathrm{colim}_j B_j$ with $A_i$ and $B_j$ in $\cal C$. Let $u_i\colon A_i\to K$ and $v_j\colon B_j\to L$ be the maps into the colimits.

(1) First take some $A=A_i$ and $u=u_i\colon A_i\to K$.

(2) Now observe that in $\mathrm{Ind}({\cal C})$ the objects from $\cal C$ are finitely presentable (not just $\lambda$-presentable). Therefore the two composite maps $fu\colon A\to K \to L$ and $gu\colon A\to K \to L$ factor through some $v_j\colon B_j\to L$ and $v_k\colon B_k\to L$.

(3) Because the colimit diagram for $L$ is directed, there is a common upper bound $B_j \rightarrow B_l \leftarrow B_k$. Set $B=B_l$ and $v=v_l$. Then $fu$ and $gu$ both factor through $v\colon B\to L$