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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.

* 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 $i > 1$.

Mike Pierce
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2 Answers2

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$\operatorname{Ext}$ 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.

Najib Idrissi
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    I think it might be better to see it as an adjunction in the derived category. – Tobias Kildetoft Feb 20 '18 at 07:20
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    I'm not seeing exactly how the adjunction in the derived category relates back to the original category though. Or at least I'm not familiar with the dg-structure in the category I'm looking at (quiver representations). I don't think the authors of book I'm looking at are thinking about the derived category. So maybe they're thinking about it and not being explicit about it, or they have something else in mind? – Mike Pierce Feb 20 '18 at 20:11
  • @MikePierce I think there are two separate things here: Why would we consider something like this (or rather, why would we ever expect something like it to hold), and why is it nice to have something like this. For the latter, it should be "clear" that it can be very handy for computing Ext groups, in the same way that the usual version is usuful for computing Hom's. – Tobias Kildetoft Feb 21 '18 at 08:00
  • @MikePierce Given an abelian category, you can always consider its derived category. The original category embed in its derived category by considering chain complexes concentrated in degree zero. An additive functor induces a functor on the derived category. Then what you're asked to prove is that for the functor considered, the adjunction on the base category becomes an adjunction on the derived categories. – Najib Idrissi Feb 21 '18 at 08:03
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    @NajibIdrissi Not exactly since the adjunction is reversed (at the level of $operatorname{Ext}^1$, $L$ becomes the "right adjoint"). This sound more like a duality statement to me. – Roland Feb 21 '18 at 09:48
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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.

Pierre-Guy Plamondon
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