# Question regarding the natural isomorphism given by a triangulated functor

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!

$$\Sigma_2$$ 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.