image of generator in filtered colimit in grothendieck category

Question

Suppose $\mathscr{A}$ is a grothendieck abelian category with generator $R$, is it true that $$\varinjlim \mathrm{Hom}(R,M_i) =\mathrm{Hom}(R,\varinjlim M_i)$$ if $M_i$ is a filtered system of objects.

Answer 1

No. For instance, take $\mathscr{A}=Ab$ with $R=\mathbb{Z}^{\oplus\mathbb{N}}$. We can write $R$ as the filtered colimit of the subgroups $\mathbb{Z}^n$, but the identity map $R\to R$ does not factor through any $\mathbb{Z}^n$.

Even worse, it is possible that there does not exist any generator for which this is true. For instance, let $\mathscr{A}=Ab^\mathbb{N}$ be the category of graded abelian groups. It is clear that any generator $R$ must be nonzero in each degree. But then we can write $R$ as a filtered colimit of objects that are nonzero in only finitely many degrees, and again the identity map $R\to R$ will not factor through any of them.