# Distinguished triangle induced by short exact sequence

I'm reading about derived categories of abelian categories from Huybrecht's book Fourier-Mukai transforms in algebraic geometry. I'm having a lot of trouble with one of the exercises. In fact, at this point, I'm confident that the result is false.

Claim 1 (Huybrechts Exercise 2.27): Suppose $0 \to A \to B \to C \to 0$ is a short exact sequence in an abelian category $\mathcal{A}$. Then embedding this into the homotopy $K(\mathcal{A})$ (as complexes supported in degree zero), we get a distinguished triangle $A \to B \to C \to A[1]$.

My question is pretty much a repeat of Short exact sequence makes exact triangle in derived category, but there is no real answer there.

Based on answers and comments in short exact sequences of complexes and triangles in the homotopy category and especially A short exact sequence that cannot be made into an exact triangle. (Weibel 10.1.2), it seems like Huybrechts has made a false claim. It seems like the right way to modify this is:

Claim 2: Suppose $0 \to A \to B \to C \to 0$ is a short exact sequence in an abelian category $\mathcal{A}$. Then embedding this into the derived category $D(\mathcal{A})$ (as complexes supported in degree zero), we get a distinguished triangle $A \to B \to C \to A[1]$.

Another thing it seems like people are implying in the threads I mentioned is that this is actually a specific instance of the following more general statement.

Claim 3: Suppose $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ is a short exact sequence in the chain complex category $\operatorname{Kom}(\mathcal{A})$. Then viewing this in the derived category $D(\mathcal{A})$ as $A^\bullet \to B^\bullet \to C^\bullet \to A^\bullet[1]$, it is a distinguished triangle.

My questions are: Am I correct that Claim 1 is false? Am I correct that Claims 2 and 3 are true? Can someone provide details on how to prove Claims 2 and/or 3?

Claim 1 is indeed wrong : the short exact sequence $0\to\mathbb{Z/2Z}\to\mathbb{Z/4Z}\to\mathbb{Z/2Z}\to 0$ does not define a distinguished triangle in the homotopy category. Indeed by definition a triangle $A\to B\to C\to A[1]$ is distinguished iff it is isomorphic to a triangle $A\to B\to\operatorname{Cone}(A\to B)\to A[1]$, but $C=\operatorname{Cone}(\mathbb{Z/2Z\to Z/4Z})$ is not isomorphic to $\mathbb{Z/2Z}$ is the homotopy category.

To see this, note that if you have an isomorphism $C\simeq \mathbb{Z/2Z}$ it will have to be represented by a morphism of complexes $\mathbb{Z/2Z}\to C$ (this is because $Hom_K=Hom_C/\sim_{htp}$), in other words, there will be a morphism of complexes : $$\require{AMScd} \begin{CD} 0@>>>0@>>>\mathbb{Z/2Z}@>>>0\\ @.@VVV@VVV@.\\ 0@>>>\mathbb{Z/2Z}@>>>\mathbb{Z/4Z}@>>>0 \end{CD}$$

which is an isomorphism in the homotopy category.

This is easy so see that there is only two such morphisms. But none of them are a homotopy equivalence : in fact they are not even a quasi-isomorphism !

This argument fail in the derived category because morphisms $\mathbb{Z/2Z}\to C$ does not necessarily comes from a morphism of complexes. We have in fact $$Hom_D(\mathbb{Z/2Z},C)=Hom_D(P,C)=Hom_K(P,C)=Hom_C(P,C)/\sim_{htp}$$ where $P$ is any projective resolution of $\mathbb{Z/2Z}$.

Take your favorite resolution of $\mathbb{Z/2Z}$ (for example $P=\mathbb{Z\overset{\times 2}\to Z}$), you will see that there is a quasi-isomorphism $P\to C$. Even more : there is also a morphism $P\to\mathbb{Z/2Z}[1]$ leading to a triangle : $$\mathbb{Z/2Z\to Z/4Z}\to P\to\mathbb{Z/2Z}[1]$$ which is isomorphic in the derived category to $\mathbb{Z/2Z\to Z/4Z}\to C\to\mathbb{Z/2Z}[1]$.

Claim 2 and Claim 3 are true but require a precision. Claim 2 follows from Claim 3, so let's have a look at Claim 3.

The correct claim is the following : if you have a short exact sequence of complexes $0\to A\to B\to C\to 0$, then there is a map $C\to A[1]$ in the derived category such that $A\to B\to C\to A[1]$ is a distinguished triangle.

In particular :

• this is not true for any map $C\to A[1]$
• the map $C\to A[1]$ only lives in the derived category and not necessarily at the level of complexes (as you can see by taking a short exact sequence of complexes concentrated in a single degree as above).

From your linked thread (Short exact sequence makes exact triangle in derived category) you know that there is a quasi-isomorphism $\varphi:\operatorname{Cone}(A\to B)\to C$.

So the map $C\to A[1]$ is constructed as : take the inverse of $\varphi$, that is $\varphi^{-1}:C\to\operatorname{Cone}(A\to B)$ and compose with $\operatorname{Cone}(A\to B)\to A[1]$.

(Note however that $\varphi^{-1}$ only exists in the derived category.)

With this construction of $C\to A[1]$, the claim is now obvious since by construction : $$\begin{CD} A@>f>> B@>i>>\operatorname{Cone}(A\to B)@>p>> A[1]\\ @|@|@V\varphi VV@|\\ A@>f>> B@>g>> C@>p\circ \varphi^{-1}>> A[1] \end{CD}$$ commutes and so is an isomorphism of triangles in the derived category.

• I took the example of the linked thread, but an even easier short exact sequence which is not a distinguished triangle in the homotopy category is $$0\to\mathbb{Z\to Z \to Z/2Z}\to 0$$ – Roland Mar 27 '18 at 14:31
• Thanks for the extremely thorough answer. Can you say a bit more about the quasi-isomorphism $\varphi$? Is it basically projection onto the $B$ components? How do I see that it is a quasi-isomorphism? – Joshua Ruiter Mar 28 '18 at 0:10
• I mean, is $\varphi$ projection onto the $B$ components followed by $B \to C$. – Joshua Ruiter Mar 28 '18 at 0:42
• There is no map $\operatorname{Cone}(A\to B)\to B$ (the natural map goes the other way). But yes, on each degree take $\operatorname{Cone}(A\to B)^n=B^n\oplus A^{n+1}\to B^n\to C^n$ and note that this defines a morphism of complexes (that is : it is compatible with the differentials). This also shows that $g=\varphi\circ i$ (the middle square of the last diagram commutes) – Roland Mar 28 '18 at 5:13
• That makes sense. Why is $\varphi$ a quasi-isomorphism? – Joshua Ruiter Mar 28 '18 at 12:23