This question will fade from very specific to very generic.


Consider an $\lambda$-accessible category $\mathcal{K}$ and let's call $\text{Pres}\_{\lambda} \mathcal{K}$ its $\lambda$-presentables.
Is the following statement true?



> 
> Given a category with directed colimits $ \mathcal{C}$ and a functor $F: \text{Pres}\_{\lambda} \mathcal{K}\to \mathcal{C}$, there is an extension $\bar{F} : \mathcal{K} \to \mathcal{C}.$
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> 
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I do have an idea to prove it.


Consider objects $K, K' \in \mathcal{K}$ and a map $K \stackrel{f}{\to} K'$. I need to give a definition for $\bar{F}(K), \bar{F}(K'), \bar{F}(f)$.


This is my attempt: Choose a directed diagram of $\lambda$-presentables $K\_i$ such that $K = \text{colim }K\_i$ and same for a diagram $K'\_j$ whose colimit will be $K'$.


The map $K\_i \to K \to K'$ factors through $K\_i \to K \to K'\_{\bar{i}}$ because of presentability of $K\_i$. We call this map $f\_i$.
Moreover there is a natural map $$\text{colim}F(K\_{\bar i}) \stackrel{i}{\to} \text{colim}F(K\_{ i})$$ 


I pose $$\bar{F}(K) = \text{colim}F(K\_i) $$ $$\bar{F}(f) = i \circ \text{colim}(F(f\_i)).$$ 


Probably with this definition composition will not work, but maybe one can make it work.




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Now question gets more vague. Consider a category $\mathcal{K}$ and $\mathcal{G}$ the subcategory generated by a strong generator.


Can the following statement be true with some additional hypotesis?



> 
> Given a category with directed colimits $ \mathcal{C}$ and a functor $F: \mathcal{G}\to \mathcal{C}$, there is an extension $\bar{F} : \mathcal{K} \to \mathcal{C}.$
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> 
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Extending functors defined on dense subcategories
$\mathcal K$ is a free cocompletion of its $\lambda$-presentables under $\lambda$-filtered colimits. Thus $\bar{F}$ exists and preserves $\lambda$-filtered colimits.


