When does $\mathrm{Ext}^\ast(K,{-})$ preserve filtered colimits? Fix an abelian category $\mathcal{A}$ that admits filtered colimits, and for $K,H\in\mathcal{A}$, write $\mathcal{A}^n(K,H) = \mathrm{Ext}^n(K,H)$ for the group of Yoneda $n$-extensions of $K$ by $H$, definable in any abelian category. My question is:


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*Under what conditions on $\mathcal{A}$ and $K$ does $\mathcal{A}^n(K,{-})\colon \mathcal{A}\rightarrow\mathrm{Set}$ preserve filtered colimits?


I would be happy with an answer for $n=1$. In the case $n=0$, an object $K$ such that $\mathcal{A}(K,{-})$ preserves filtered colimits is called compact. However, it does not appear that this is sufficient for $\mathcal{A}^n(K,{-})$ to preserve filtered colimits, as the former is concerned with maps out of $K$ and the latter with maps into $K$.
I can prove that $\mathcal{A}^1(K,{-})$ does preserve filtered colimits in the somewhat contrived situation where $\mathcal{A}$ is a compactly-generated abelian category and $K\in\mathcal{A}$ a compact object such that:


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*If $G\in\mathcal{A}$ is compact, then every subobject of $G$ is compact,

*If $P\subset\mathrm{Sub}(K)$ is a filtered subposet such that $\bigvee P = K$, then $K\in P$.
The first condition fails in case $\mathcal{A} = \mathrm{Mod}_R$ and $K=R$ for a ring $R$ which is not Noetherian, so these conditions are rather suboptimal. I do not know an example where the second condition fails.
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Here is a sketch of the proof under these conditions. We must show $\mathrm{colim}_j \mathcal{A}^1(K,H_j)\rightarrow\mathcal{A}^1(K,H)$ is an isomorphism for all filtered colimits $H=\mathrm{colim}_j H$. 
For surjectivity, it is sufficient by compact generation to show that every element $x\in \mathcal{A}^1(K,H)$ can be written as $x=f_\ast(y)$, where $f\colon H'\rightarrow H$ with $H'$ compact and $y\in\mathcal{A}^1(K,H')$. Given an extension $0\rightarrow H\rightarrow G\rightarrow K\rightarrow 0$, we may write $G=\mathrm{colim}_{j\in\mathcal{J}} G_j$ with $\mathcal{J}$ filtered and $G_j$ compact, and by condition (ii) there is some $j$ such that $G_j\rightarrow K$ is epi. If $H' = H\times_G G_j$, then $H'$ is compact by condition (i), and the extension is the image of $0\rightarrow H'\rightarrow G_j\rightarrow K\rightarrow 0$.
For injectivity, say $H=\mathrm{colim}_j H_j$ as a filtered colimit. The sequences $0\rightarrow H_j\rightarrow H\rightarrow H/H_j\rightarrow 0$ give an exact sequence $\mathrm{colim}_j \mathcal{A}(K,H/H_j)\rightarrow\mathrm{colim}_j\mathcal{A}^1(K,H_j)\rightarrow\mathrm{colim}_j\mathcal{A}^1(K,H)$, and the first term is zero as $K$ is compact and $\mathrm{colim}_j H/H_j = 0$.
 A: If your abelian category is a module category, then this is well understood. To get arbitrary exts to preserve filtered colimits, you want your module to be quasi-isomorphic to a complex of finitely generated projectives. Brown proved this in his paper "A homological criterion for finiteness." Over a Noetherian ring, this is just asking for your module to be finitely generated, which generalizes the special case of your result that is he case of modules over a Noetherian ring.
The assumptions in your argument aren't too insane: that subobjects of compacts be compact is an abstract Noetherian condition, while the condition on unions of filtered families of subobjects always holds. Indeed, since $K$ is compact, its identity morphism factors through some subobject in the family! However, an abelian category in which every object is a filtered colimit of compacts is always a reflective subcategory, close down under filtered colimits, of the category of modules over a "ring with many objects", that is, a small preadditive category. So your version and Brown's have surprisingly close to the same scope, and I'm sure Brown's theorem generalizes to your setting.
In more general abelian categories, there's no hope, because in general there are not even any objects whose non-derived homs preserve filtered colimits. For instance, such an object x of the opposite of a module category would satisfy, in the module category, $Hom(\prod_i y_i,x)\cong \prod_i Hom(y_i,x).$ This is absurd, as follows for $x, y_i$ all the same finite cyclic group of prime order by a dimension count and for $x, y_i$ all the integers by Specker's theorem that the dual of a countable direct product of copies of the integers is the countable direct sum.
