Lemma 13.7.1 (0A8D)—The Stacks project

Lemma 13.7.1. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor between triangulated categories. If $F$ admits a right adjoint $G: \mathcal{D'} \to \mathcal{D}$ , then $G$ is also an exact functor.

Proof. Let $X$ be an object of $\mathcal{D}$ and $A$ an object of $\mathcal{D}'$ . Since $F$ is an exact functor we see that

$\begin{align*} \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X, G(A[1])) & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(X), A[1]) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(X)[-1], A) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(X[-1]), A) \\ & = \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[-1], G(A)) \\ & = \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X, G(A)[1]) \end{align*}$

By Yoneda's lemma (Categories, Lemma 4.3.5) we obtain a canonical isomorphism $G(A)[1] = G(A[1])$ . Let $A \to B \to C \to A[1]$ be a distinguished triangle in $\mathcal{D}'$ . Choose a distinguished triangle

$G(A) \to G(B) \to X \to G(A)[1]$

in $\mathcal{D}$ . Then $F(G(A)) \to F(G(B)) \to F(X) \to F(G(A))[1]$ is a distinguished triangle in $\mathcal{D}'$ . By TR3 we can choose a morphism of distinguished triangles

$\xymatrix{ F(G(A)) \ar[r] \ar[d] & F(G(B)) \ar[r] \ar[d] & F(X) \ar[r] \ar[d] & F(G(A))[1] \ar[d] \\ A \ar[r] & B \ar[r] & C \ar[r] & A[1] }$

Since $G$ is the adjoint the new morphism determines a morphism $X \to G(C)$ such that the diagram

$\xymatrix{ G(A) \ar[r] \ar[d] & G(B) \ar[r] \ar[d] & X \ar[r] \ar[d] & G(A)[1] \ar[d] \\ G(A) \ar[r] & G(B) \ar[r] & G(C) \ar[r] & G(A)[1] }$

commutes. Applying the homological functor $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(W, -)$ for an object $W$ of $\mathcal{D}'$ we deduce from the $5$ lemma that

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(W, X) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(W, G(C))$

is a bijection and using the Yoneda lemma once more we conclude that $X \to G(C)$ is an isomorphism. Hence we conclude that $G(A) \to G(B) \to G(C) \to G(A)[1]$ is a distinguished triangle which is what we wanted to show. $\square$

Comments (3)

Comment #8615 by nkym on August 13, 2023 at 08:57

Additional ) needed in the most left hand side of the first equation

Comment #9423 by Stacks project on June 05, 2024 at 17:02

Already fixed.

Comment #9801 by Elías Guisado on October 15, 2024 at 04:13

I don't think it is necessarily the case that commutativity of the diagram $(FGA\to FGB\to FX\to (FGA)[1])\to(A\to B\to C\to A[1])$ implies commutativity of the diagram $(GA\to GB\to X\to (GA)[1])\to(GA\to GB\to GC\to (GA)[1])$ . My worry is that when $G$ is regarded as a triangulated functor with the isomorphism $G\circ [1]\cong [1]\circ G$ as built in the proof it may happen that the unit $\eta:1\Rightarrow GF$ and counit $\varepsilon: FG\Rightarrow 1$ won't be trinatural (in the sense of #9797). Only if $\eta,\varepsilon$ are trinatural we will have an adjunction in the $2$ -category of categories with translation (that contains (pre-)triangulated categories as a subcategory), which will induce an adjunction between categories of triangles via the $2$ -functor $\mathrm{t}(-)$ (explained in #9797). From inspection of the current proof, the isomorphism $\xi':G\circ [1]\cong [1]\circ G$ can be seen to equal $\xi'=T[GT^{-1}(\varepsilon_{T}\circ \xi_{T^{-1}GT}^{-1})\circ\eta_{T^{-1}GT}]$ (where $\xi$ is the isomorphism $F\circ[1]\cong [1]\circ F$ ). However, it is not clear that $\eta,\varepsilon$ will be trinatural (i.e., compatible with $\xi,\xi'$ ). In [M, Proposition 47], Murfet chooses a different $\xi'$ , namely (after dualizing his proof by assuming $F\dashv G$ ), $(\xi')^{-1}=GT\varepsilon\cdot G\xi_G\cdot\eta_G.$ With this definition for $\xi'$ , Murfet shows that $\eta$ and $\varepsilon$ turn out to be trinatural.

References

[M] D. Murfet, Triangulated Categories Part I, http://therisingsea.org/notes/TriangulatedCategories.pdf

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2 comment(s) on Section 13.7: Adjoints for exact functors

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