MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f'$.

share|cite|improve this question
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 $\mathrm{Mod} \; R$.

share|cite|improve this answer
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?

share|cite|improve this answer
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
$X$ 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 $f=0\neq g$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.