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If we have two triangulated categories, let's say $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$, with translation functors $\Sigma_{1}$ and $\Sigma_{2}$, respectively. A exact functor or triangulated functor between these two categories is defined by an additive functor $F:\mathcal{T}_{1} \to \mathcal{T}_{2}$ such that there is a natural isomorphism

$$\eta:= \lbrace \eta_{X}:F\Sigma_{1}(X) \to \Sigma_{2}F(X) \rbrace_{X \in \operatorname{Obj}(\mathcal{T_{1}})}.$$

In a way that for every distinguished triangle $X \xrightarrow{u} Y \xrightarrow{v} Z \xrightarrow{w} \Sigma_{1}X$ in $\mathcal{T_{1}}$ then $FX \xrightarrow{F(u)} FY \xrightarrow{F(v)} FZ \xrightarrow{\eta_{X} \circ F(w)} \Sigma_{2}FX$ is a distinguished triangle in $\mathcal{T_{2}}$. My question is that if we have that this natural transformation $\eta$ induced by the triangulated functor $F:\mathcal{T}_{1} \to \mathcal{T}_{2}$ we also have a natural isomorphism $\Sigma_{2}^{-i}F \cong F \Sigma_{1}^{-i}$ for every $i \in \mathbb{Z}$?? I guess this is true by checking out on proof of "the adjoint functor of a triangulated functor is also triangulated". I tried to prove this simple case $\Sigma_{2}^{-1}F \cong F \Sigma_{1}^{-1}$.

I was thinking maybe I didn't understand the definition of a triangulated functor correctly and there is a natural isomorphism $\eta$ for each distinguished triangle in $\mathcal{T}_{1}$?? Can anyone tell me if this is true? Otherwise I don't have any idea how to obtain this natural isomorphism :/

I came with this question since I was following for this proof Daniel Huybrechts' book "Fourier-Mukai Transforms in Algebraic Geometry" page 15 Proposition 1.41. As there is one step "using my notation" where he uses an isomorphism:

$$\operatorname{Hom}_{\mathcal{T}_{2}}(\Sigma_{-2}F(X),Y) \cong \operatorname{Hom}_{\mathcal{T}_{2}}(F \Sigma^{-1}(X),Y).$$

This question has been stuck in my head for so many day so I will really appreciate your help. Thanks!

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1 Answer 1

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$\Sigma_2$ is an equivalence, so to construct such an isomorphism it suffices to construct an isomorphism $F\cong\Sigma_2^i F\Sigma_1^{-i}$ (to which you can then apply $\Sigma_2^{-i}$)

Similarly, by using precomposition instead of postcomposition, it suffices to construct an isomorphism $F\Sigma_1^i\cong \Sigma_2^iF$. But this one is given to you by iterating $\eta$, so we are done.

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  • $\begingroup$ Thanks for your help! @MaximeRamzi $\endgroup$
    – Cos
    Commented Nov 15, 2020 at 21:31

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