let $T$ be a triangulated category and $A \to B \to C \to A[1]$ a triangle in $T$ such that for every $A_0 \in T$ the induced long sequence

$... \to \hom(A_0,A) \to \hom(A_0,B) \to \hom(A_0,C) \to \hom(A_0,A[1]) \to \hom(A_0,B[1]) \to ...$

is exact. is then $A \to B \to C \to A[1]$ exact? I'm a beginner, so this could be rather trivial. I've checked it in special cases, but I can't translate the proof into $T$. I would like the result because it would imply that the homological algebra you know for abelian groups takes over to triangulated categories. comparable to the fact that you can work with group objects in arbitrary categories as with ordinary groups. you can do it directly with the axioms, but this is a big mess.

in many answers I've seen here so far, even standard facts are enriched with wonderful, ample insights. so you are also invited to dish durt about these basics of triangulated categories because I'm just beginning to learn them. and it would be cool if there is a online full version of neemans book "Triangulated categories" ;-).

let $T$ be a triangulated category and $A \to B \to C \to A[1]$ a triangle in $T$ such that for every $A_0 \in T$ the induced long sequence

$... \to \hom(A_0,A) \to \hom(A_0,B) \to \hom(A_0,C) \to \hom(A_0,A[1]) \to \hom(A_0,B[1]) \to ...$

is exact. is then $A \to B \to C \to A[1]$ exact? I'm a beginner, so this could be rather trivial. I've checked it in special cases, but I can't translate the proof into $T$. I would like the result because it would imply that the homological algebra you know for abelian groups takes over to triangulated categories. comparable to the fact that you can work with group objects in arbitrary categories as with ordinary groups. you can do it directly with the axioms, but this is a big mess.

in many answers I've seen here so far, even standard facts are enriched with wonderful, ample insights. so you are also invited to dish durt about these basics of triangulated categories because I'm just beginning to learn them.

let $T$ be a triangulated category and $A \to B \to C \to A[1]$ a triangle in $T$ such that for every $A_0 \in T$ the induced long sequence

$... \to \hom(A_0,A) \to \hom(A_0,B) \to \hom(A_0,C) \to \hom(A_0,A[1]) \to \hom(A_0,B[1]) \to ...$

is exact. is then $A \to B \to C \to A[1]$ exact? I'm a beginner, so this could be rather trivial. I've checked it in special cases, but I can't translate the proof into $T$. I would like the result because it would imply that the homological algebra you know for abelian groups takes over to triangulated categories. comparable to the fact that you can work with group objects in arbitrary categories as with ordinary groups.

in many answers I've seen here so far, even standard facts are enriched with wonderful, ample insights. so you are also invited to dish durt about these basics of triangulated categories because I'm just beginning to learn them.

let $T$ be a triangulated category and $A \to B \to C \to A[1]$ a triangle in $T$ such that for every $A_0 \in T$ the induced long sequence

$... \to \hom(A_0,A) \to \hom(A_0,B) \to \hom(A_0,C) \to \hom(A_0,A[1]) \to \hom(A_0,B[1]) \to ...$

is exact. is then $A \to B \to C \to A[1]$ exact? I'm a beginner, so this could be rather trivial. I've checked it in special cases, but I can't translate the proof into $T$. I would like the result because it would imply that the homological algebra you know for abelian groups takes over to triangulated categories. comparable to the fact that you can work with group objects in arbitrary categories as with ordinary groups. you can do it directly with the axioms, but this is a big mess.

in many answers I've seen here so far, even standard facts are enriched with wonderful, ample insights. so you are also invited to dish durt about these basics of triangulated categories because I'm just beginning to learn them. and it would be cool if there is a online full version of neemans book "Triangulated categories" ;-).