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 $\xi _ X : F(X[1]) \to F(X)[1]$ be as in Definition 13.3.3. Let $\epsilon _ A : F(G(A)) \to A$ be the adjunction map. Consider the composition

\[ F(G(A)[1]) \xrightarrow {\xi _{G(A)}} F(G(A))[1] \xrightarrow {\epsilon _ A[1]} A[1] \]

This map is adjoint to a map $G(A)[1] \to G(A[1])$ which we claim to be an isomorphism. To see this, by the Yoneda lemma it suffices to show that we get a bijection on applying $\mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[1], -)$ for every object $X$ of $\mathcal{D}$. Now, every morphism $f : X[1] \to G(A)[1]$ is of the form $g[1]$ for a unique $g : X \to G(A)$ and every $g$ is adjoint to a unique $h : F(X) \to A$ and in turn $h[1] \circ \xi _ X$ is adjoint to a unique $i : X[1] \to G(A[1])$. The rule sending $f$ to $i$ gives a bijection between $\mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[1], G(A)[1])$ and $\mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[1], G(A[1]))$. That this bijection is given by the morphism above follows from the discussion in Categories, Section 4.24 and the following calculation

\begin{align*} \epsilon _ A[1] \circ \xi _{G(A)} \circ F(f) & = \epsilon _ A[1] \circ \xi _{G(A)} \circ F(g[1]) \\ & = \epsilon _ A[1] \circ F(g)[1] \circ \xi _ X \\ & = (\epsilon _ A \circ F(g))[1] \circ \xi _ X \\ & = h[1] \circ \xi _ X \end{align*}

Some details omitted. We will show that $G$ is an exact functor using the inverse of the isomorphisms $G(A)[1] \to G(A[1])$. These isomorphisms are functorial in $A$ (details omitted).

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

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

Recall that the arrow $F(X) \to F(G(A))[1]$ is the composition of $F(X) \to F(G(A)[1])$ with $\xi _{G(A)}$. Hence using that $G$ is the right adjoint of $F$ and our discussion above, we conclude the existence of 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 and the right vertical arrow is the morphism constructed above. 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 (6)

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

Comment #11494 by Elías Guisado on May 26, 2026 at 05:17

This result in a sharpened form (the adjunction in the $2$ -category $\mathbf{Cat}$ can be promoted to an adjunction in the $2$ -category $\mathbf{Triang}$ ) also appears as Proposition 1.1 in Bondal, Orlov, Reconstruction of a variety from the derived category and groups of autoequivalences.

Comment #11543 by Stacks Project on May 29, 2026 at 16:43

Leaving as is.

Comment #11610 by Stacks Project on July 09, 2026 at 15:59

Sorry for the late realization that you pointed out an actual error in the proof. I've fixed it without adding too much text by choosing the "correct" isomorphisms of $G(A)[1]$ with $G(A[1])$ . See this commit.

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

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