Significance of adjoint relationship with Ext instead of Hom

Question

For categories $\mathcal{C}$ and $\mathcal{D}$, a pair of functors $\mathcal{C} \xrightarrow{\;R\;} \mathcal{D}$ and $\mathcal{C} \xleftarrow{\;L\;} \mathcal{D}\,$ are an adjoint pair if for any objects $X$ of $\mathcal{C}$ and $Y$ of $\mathcal{D}$ we have a bijection

$$ \operatorname{Hom}_\mathcal{C}(LY,X) \;\simeq\; \operatorname{Hom}_\mathcal{D}(Y,RX) \,. $$

Recently, I've come across an exercise^* asking to prove that a pair of functors are an adjoint pair, and then additionally to prove that

$$ \operatorname{Ext}^1_\mathcal{C}(X,LY) \;\simeq\; \operatorname{Ext}^1_\mathcal{D}(RX,Y) \,. $$

What is the significance of this adjoint-like relationship with $\operatorname{Ext}$ instead of $\operatorname{Hom}$? The exercise I'm looking at didn't provide any motivation for proving this.

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* In the exercise, $X$ and $Y$ are categories of quiver representations, and $R$ and $L$ are reflection functors across a sink and source vertex respectively. This category is hereditary, so $\operatorname{Ext}^i$ vanish for $i > 1$.

Answer 1

$\operatorname{Ext}$ is the derived functor of $\operatorname{Hom}$. So you're basically looking at a "derived" adjunction. Or rather an adjunction which is compatible with the dg-structure, I guess.

Answer 2

The isomorphism is not true for an arbitrary choice of adjoint functors $L$ and $R$, even when the abelian categories $\mathcal{C}$ and $\mathcal{D}$ are hereditary. Thus, a possible motivation for the the authors might be to show that the functors they are considering (reflection functors) are particularly nice.

To see a counter-example, take $\mathcal{C}$ to be the category of modules over the path algebra $A$ of the quiver $1\to 2$ over a field $k$, and let $\mathcal{D}$ be the category of vector spaces over $k$. Take $L = ?\otimes_k A$ and $R= \operatorname{Hom}_A(A, ?)$ to be the adjoint pair. Finally, take $X$ to be the representation $k\to 0$ and $Y=k$.

Then $LY = A$, so $\operatorname{Ext}_A^1 (X, LY) \cong k$, but $\operatorname{Ext}_k^1 (RX, Y) = 0$, since all $\operatorname{Ext}^1$ groups vanish in the category of vector spaces. Therefore, the two extension groups are not isomorphic.