A special case of amalgamation in $\mathrm{Ind}(\mathcal{C})$ Let $\newcommand{\C}{\mathcal{C}}\C$ be a category with amalgamation, i.e. such that every span has a cocone. In this MO question, it is said that one can prove by induction that if $B ← A → C$ is a span in $\DeclareMathOperator{\Ind}{Ind}\Ind(\C)$ with $A ∈ \C$, then it has a cocone. I can only show this if $B$ and $C$ are colimits of chains of objects in $\C$ (to continue the induction, I need to drop the hypothesis that $A$ is in $\C$), or if amalgamation in $\C$ is functorial without the condition on $A$. How does one proceed? If it is false, is there something else similar which is true? The case of interest to me is when both $A$ and $B$ are in $\C$.
 A: Lemma: any object $X$ in $Ind(\mathcal C)$ which is not finitely presentable is a retract of the colimit of a smooth chain of objects $X = \varinjlim_{\alpha < \lambda} X_\alpha$ of lower presentability rank. Chain means the colimit is indexed by and ordinal $\lambda$ and smooth means that the functor $\alpha \mapsto X_\alpha$ preserves colimits (i.e. $X_\beta = \varinjlim_{\alpha<\beta} X_\alpha$ when $\beta < \lambda$ is limit).

Now to prove the claim. Let $B \leftarrow A \to C$ be an amalgamation problem in $Ind(\mathcal C)$ with $A \in \mathcal C$ and assume by induction that $Ind(\mathcal C)$ has amalgamation for spans $B'\leftarrow A \to C'$ where $rk(B'),rk(C') \leq rk(B),rk(C)$ and moreover that the amalgam $D'$ may be chosen to have $rk(D') \leq max(rk(B'),rk(C'))$ where $rk$ means presentability rank. Write $A = \varinjlim A_\alpha$, $C = \varinjlim C_\alpha$ as colimits of smooth chains of objects of lower presentability rank. We may assume without loss of generality that the maps $B \leftarrow A \to C$ factor through $B_0 \leftarrow A \to C_0$. We build a chain of amalgamations $D_\alpha$ of $B_\alpha \leftarrow A \to C_\alpha$, functorial in $\alpha$ [even though amalgamation in the whole category is not assumed to be functorial], by induction on $\alpha$. When $\alpha = 0$, let $D_0$ be any amalgamation. When $\alpha$ is limit, take a colimit of over the amalgams already constructed.
To amalgamate $B_{\alpha+1} \leftarrow A \to C_{\alpha+1}$, first choose an arbitrary amalgam $D_{\alpha+1}^0$. Then we have an amalgamation problem $D_\alpha \leftarrow A \to D_{\alpha+1}^0$, which we amalgamate to get $D_{\alpha+1}$. One must check that all the requiste diagrams commute (since we need this amalgam to be functorial in $\alpha$ in order to take a colimit at the limit stages).
