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First of all, let me fix some notation.

  1. Let $\mathcal D$ be a triangulated category in the sense of Verdier-Grothendieck (for example, the homotopy category $\mathbf{K}(k)$ of cochain complexes over a fixed commutative ring $k$). I call cone of a morphism $f : A \rightarrow B$ the object $C(f)$ (uniquely determined up to isomorphism) such that $A \stackrel{f}\rightarrow B \rightarrow C(f) \rightarrow A[1]$ is a distinguished triangle in $\mathcal D$. When $\mathcal D = \mathbf{K}(k)$, $C(f)$ can be identified (up to homotopy equivalence) to the mapping cone of the chain map $f$.

  2. In any category $\mathcal C$, I say that two morphisms $f: A \to B$ and $f' : A' \to B'$ are isomorphic, if they are isomorphic in the category of morphisms $\mathrm{Mor}(\mathcal C)$, that is, there are two isomorphisms $u : A \to A'$ and $v : B \to B'$ such that $vf = f'u$.

Now, my question is the following: is there a triangulated category with a pair of parallel morphisms $f,f' : A \to B$ such that $f$ is not isomorphic to $f'$ but $C(f)$ is isomorphic to $C(f')$? I believe that an example could be found in the category $\mathbf{K}(k)$.

Of course, if we don't require $f$ and $f'$ to be parallel, then we may find examples in any reasonable triangulated category: just set $f=1_0$, the identity of a zero object, and $f' = 1_A$, the identity of a nonzero object. Then, both cones are zero objects (a general fact in triangulated categories), but clearly $f$ is not isomorphic to $f'$.

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3 Answers 3

up vote 5 down vote accepted

Let $R$ be the ring $R=\mathbb C[x,y]$, and let $B$ be the $5$-dimensional $R$-module with shape like a 'W'. That is, basis elements are $a_1,a_2,a_3,b_1,b_2$ and the module structure is given by $$y \cdot a_1=b_1,$$ $$x \cdot a_2=b_1,$$ $$y \cdot a_2=b_2,$$ $$x \cdot a_3=b_2,$$ and all other products of generators and basis elements are zero.

Let $A=\mathbb C$ be the trivial $R$-module and consider the parallel morphisms $f,f' \colon A \rightarrow B$ defined by $f(z)=zb_1$ and $f'(z)=zb_2.$ Now ${\mathrm{coker}} \; f \simeq {\mathrm{coker}} \; f'$ as $R$-modules, but $f$ and $f'$ are non-isomorphic in $\mathrm{Mor}(\mathrm{Mod} \; R)$. This gives an example in the derived category of $\mathrm{Mod} \; R$.

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2  
If you work instead over a full polynomial ring $k[x,y]$, this gives an example where both $A$ and $B$ are perfect complexes. –  Eric Wofsey Jan 22 '13 at 0:07
    
Sure, might be better. I'm used to thinking small $k$-dimension, but here maybe small homological dimension is more relevant :) –  Dag Oskar Madsen Jan 22 '13 at 0:28
    
Changed the ring after Eric's suggestion. –  Dag Oskar Madsen Jan 22 '13 at 0:56
    
I'm checking this example in detail, I'm quite sure it works. I will accept this answer, I think it's the "simplest" one given here. Of course, I would like to thank the other answerers, too: I believe that all examples given here are correct. –  Francesco Genovese Jan 22 '13 at 14:19

Let $C$ be a non-contractible complex. Let $X$ be a direct sum of a countably infinite number of copies of $C$ plus a countable infinite number of copies of $\Sigma C$. Then the inclusion of $C$ into $X$ as a direct summand, and the null map from $C$ to $X$, are non-isomorphic maps with isomorphic mapping cones.

Even if it works, this example feels like a swindle. Is there one with finitely generated modules?

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I'm still unable to see the isomorphism between the cones: could you explain it with a little bit more detail? In any case, thank you! –  Francesco Genovese Jan 21 '13 at 21:56
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$X$ is a direct sum of infinitely many copies of $C$ and infinitely many copies of $\Sigma C$. Consider the map $C\to X$ that is inclusion into the first summand. The cone of this map is obtained by removing one copy of $C$ from $X$. This is isomorphic to $X$, because $\infty -1=\infty$. Now consider the null map from $C$ to $X$. The cone of this map is $X\oplus \Sigma C$. This, again, is isomorphic to $X$, because $\infty +1 =\infty$. Thus the two cones are isomorphic to each other. –  Gregory Arone Jan 21 '13 at 22:11

Yet another example. Take $R$ any ring such that $R\cong R\oplus R$. Consider the following parallel morphisms $f,g\colon R\rightarrow R$: $f=0$ the trivial morphism, and

$$g=\left(\begin{smallmatrix} 1&0\\\0&0 \end{smallmatrix}\right)\colon R\cong R\oplus R\longrightarrow R\oplus R\cong R.$$ Both have isomorphic mapping cone

$$C\colon \cdots\rightarrow0\rightarrow R\stackrel{0}\rightarrow R\rightarrow0\rightarrow\cdots$$

but $f\ncong g$ since $f=0\neq g$.

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