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](http://en.wikipedia.org/wiki/Mapping_cone_(homological_algebra)) 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 $f'$.



Example: a pair of nonisomorphic parallel morphisms with isomorphic cones
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 $\mathrm{Mod} \; R$.


