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.


1 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 $$FX \rightarrow FY \rightarrow FZ \rightarrow F(X[1]) \simeq FX \rightarrow FY \rightarrow FZ \rightarrow F(X)[1]$$


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