Let $F: \mathcal{K} \to \mathcal{C}$ be a functor between $\lambda$-accessible categories, you can assume $\mathcal{C}$ to be Set if needed.


Is it true that $F$ is $\lambda$-accessible if and only if it preserves $\lambda$-directed colimits of $\lambda$-presentable objects?



Can I check the accessibility of a functor on directed colimits of presentable objects?
Yes, and this doesn't require any assumption on $\mathcal C$. This follows from the following basic fact: if $I$ is filtered and $(A\_i)\_{i\in I}$ is a diagram of categories with colimit $A$, then the colimit of any functor $F: A\to\mathcal C$ may be computed as
$$
colim\_{a\in A} F(a) = colim\_{i\in I}(colim\_{a\in A\_i}F\_i(a)),
$$
where $F\_i$ is the restriction of $F$ to $A\_i$ (provided the inner colimits all exist).


Write $\mathcal K=Ind\_\lambda(\mathcal K\_0)$. Every $X\in \mathcal K$ is then the colimit of the forgetful functor $\mathcal K\_0/X \to \mathcal K\_0\subset\mathcal K$, and $\mathcal K\_0/X$ is $\lambda$-filtered. Let $I$ be $\lambda$-filtered and let $(X\_i)\_{i\in I}$ be a diagram in $\mathcal K$ with colimit $X$. Then $colim\_i \mathcal K\_0/X\_i = \mathcal K\_0/X$ and so
 $$
 colim\_{i} F(X\_i) = colim\_{i} (F(colim\_{Y\to X\_i} Y)) = colim\_i (colim\_{Y\to X\_i} F(Y)) = colim\_{Y \to X} F(Y) = F(colim\_{Y\to X} Y)=F(X).
 $$


