I am reading through the appendix of Commutative Algebra - with a View Towards Algebraic Geometry and I came across essentially the following question: Suppose that $\require{AMScd}$ \begin{CD} 0 @>>> F' @>\alpha>> F @>\beta>> F'' @>>> 0 \end{CD} is a short exact sequence of complexes from a category $\mathcal{A}$, show that it induces an exact triangle in $D(\mathcal{A})$, but not necessarily in $K(\mathcal{A})$.

As a hint, it says to prove that $F''$ is quasi-isomorphic to $M(\alpha)$ (the mapping cone of $\alpha$) by showing that $\require{AMScd}$ \begin{CD} 0 @>>>M(\alpha') @>>> M(\alpha) @>>> F'' @>>> 0 \end{CD} is a short exact sequence of complexes, where $\alpha'$ is the isomorphism from $F'$ to $\alpha(F')\subset F$.

I can show that the latter is indeed a short exact sequence, but that's it. I am quite new to derived categories and the concept of exact triangles in general is quite confusing to me.

  • $\begingroup$ I think you need to read some materials about triangulated category and derived category, then you can understand this question. Actually $K(A)$ is a triangulated category, its distinguished triangles are the class of triangules that are isomorphic to the triangles induced by mapping cone. Derived category $D(A)$ is the verdier quotient of $K(A)$, it takes all quasi-isomorphisms in $K(A)$ to isomorphisms in $D(A)$. We can get that for any short exact sequence of complexes of $A$, it induces a distinguished triangle in $D(A)$. $\endgroup$ – Xiaosong Peng Jun 23 '17 at 2:13
  • $\begingroup$ I will do that, thank you. I have two follow-up questions: 1) Is there a source where they explain derived categories with more (non-trivial) examples, I feel like that might help me. 2) Specifically to my question, I figured out how to do it, so should I answer my own question here (answer is quite long, but ultimately not hard) or is there a way to indicate that it doesn't need answering any more? $\endgroup$ – Anton V. Jun 25 '17 at 17:35
  • $\begingroup$ 1) maybe you can look at this book "triangulated categories in the representation theory of finite dimensional algebras" by Happle; 2) Of course you can answer your own question $\endgroup$ – Xiaosong Peng Jun 26 '17 at 1:31

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