Definition of triangulated functor

Question

In Huybrechts' book "Fourier Mukai transforms in algebraic geometry" he defines (def 1.39) an exact (or triangulated) functor between triangulated categories as follows. An exact functor is an additive functor $F: D \rightarrow D'$ such that

i) There exists a natural isomorphism $F \circ T_{D} \simeq T_{D'} \circ F$,

ii) Any distinguished triangle $$ A \rightarrow B \rightarrow C \rightarrow A[1] $$ is mapped to a distinguished triangle $$ F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow F(A)[1] $$ where $F(A[1])$ is identified with $F(A)[1]$ via the previous isomorphism.

I don't really understand this definition. Does it mean that there exists a natural isomorphism as in i) such that ii) holds, or that for every isomorphism such in i) it is true that distinguished triangles are mapped into distinguished triangles?

I tried to prove that by changing the natural isomorphism in i) one gets isomorphic triangles $$ F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow F(A)[1] $$ but I am not able to define the isomorphism of triangles.

Answer 1

Both questions you ask are true - you need the triangulated functor to preserve distinguished triangles, and to do so would need a natural isomorphism $\eta:F\circ T_{D}\rightarrow T_{D'}\circ F$.

Note that $F(A[1]) \simeq F(A)[1]$ by $\eta$.

An isomorphism of triangles is defined in a natural way:

$$X \rightarrow Y \rightarrow Z \rightarrow X[1] \simeq X'\rightarrow Y'\rightarrow Z'\rightarrow X'[1] \iff X \simeq X', Y \simeq Y', Z \simeq Z'$$ I don't personally know Huybrechts' book, but this how I've seen them defined in other references on triangulated categories.

From these pieces of information, we can deduce that $$FX \rightarrow FY \rightarrow FZ \rightarrow F(X[1]) \simeq FX \rightarrow FY \rightarrow FZ \rightarrow F(X)[1]$$