Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I set this problem in the framework of (pretriangulated) dg-categories; everything can probably be translated in the world of stable $(\infty,1)$-categories.

Let $\mathcal A$ be a pretriangulated dg-category. It is known that the homotopy category $H^0(\mathcal A)$ is triangulated, with exact triangles coming from the "pre-triangles" in $\mathcal A$. Assume we have a diagram in $\mathcal A$:

,

where the rows are pretriangles in $\mathcal A$, the vertical arrows are closed and of degree $0$, and everything is commutative in $H^0(\mathcal A)$. This diagram, in other words, induces a morphism of exact triangles in $H^0(\mathcal A)$.

Now, let $h : A \to B'$ be a degree $-1$ morphism such that \begin{equation} dh = vf - f'u, \end{equation} which exists by hypothesis. In the dg-category $\mathcal A$, there is a canonical (closed, degree $0$) morphism $C(u,v,h) : C(f) \to C(f')$ induced (functorially!) between the cones. This morphism also makes the diagram

commute in $H^0(\mathcal A)$. My question is the following: is it true or false, in general, that $[w] = [C(u,v,h)]$ as morphisms in $H^0(\mathcal A)$? Actually, there are some subtleties. Better said: can I choose $u': A \to A'$, $v': B \to B'$ closed of degree $0$ such that $[u]=[u']$, $[v]=[v']$, and $h' : A \to B'$ closed of degree $-1$ with $dh' = v'f - f'u'$, such that $[w] = [C(u',v',h')]$? I believe the answer of the above question is false, but perhaps it is true with some hypothesis on $\mathcal A$?

share|improve this question
    
It seems to me that in the $\infty$-stable world your question has affirmative answer, in view of the way you obtain the arrow $C(f) \to C(f')$: it is simply the arrow induced between two homotopy pushouts, and the simplicial set of the choices you have to solve this problem is contractible (it is an initial object "somewhere"). Tell me if you want me to expand this argument (indeed too sketchy to be convincing). –  tetrapharmakon Aug 5 at 14:56
1  
The answer to both questions is no in general, as you guess. Counterexamples are not that obvious though. If I have time (and can get a better connection where I am) I'll come back with one. –  Fernando Muro Aug 5 at 15:55
    
@tetrapharmakon, the problem is the other direction: given a map $w : C(f) \to C(f')$ as above, it is not necessarily induced by some $u$ and $v$. –  Adeel Aug 5 at 19:35
    
Actually, it is induced by some $u$ and $v$; the problem is that we are not assured that it is induced by the given $u$ and $v$. –  Francesco Genovese Aug 5 at 20:05
    
Sorry, I was being silly. I think tetrapharmakon is right actually, the space of maps $w$ as above should be contractible. –  Adeel Aug 5 at 21:19

1 Answer 1

up vote 1 down vote accepted

The following example comes from Neeman's ``Some new axioms\dots" (J. Algebra, 1991). A triangle is contractible if it is a direct sum of (translations of) triangles of the form $0\rightarrow X\rightarrow X \rightarrow 0$. Contractible triangles are exact, but exact triangles are seldom contractible. Assume that $X\rightarrow Y\rightarrow Z\rightarrow\Sigma X$ is not a contractible exact triangle. Then $$\begin{array}{ccccccc} X&\stackrel{f}\rightarrow&Y&\stackrel{i}\rightarrow&Z&\stackrel{q}\rightarrow&\Sigma X\\ {\scriptstyle 0}\downarrow&&{\scriptstyle 0}\downarrow&&{\scriptstyle q}\downarrow&&{\scriptstyle 0}\downarrow\\ Y&\stackrel{i}\rightarrow&Z&\stackrel{q}\rightarrow&\Sigma X&\stackrel{-\Sigma f}\rightarrow &\Sigma Y \end{array}$$ is a morphism of triangles and $q$ cannot be obtained in the way you described. Otherwise, $h$ would be simply a morphism $h\colon \Sigma X\rightarrow Z$ and, up to sign, $qhq=q$ necessarily. This leads to contractibility of the initial exact triangle.

This argument is very general, so we rather see it `in action' in $D(\mathbb Z)$. We can take the exact triangle associated to the short exact sequence $\mathbb Z/2\hookrightarrow \mathbb Z/4\twoheadrightarrow \mathbb Z/2$

$$\begin{array}{ccccccc} \mathbb Z/2&\stackrel{f}\rightarrow&\mathbb Z/4&\stackrel{i}\rightarrow&\mathbb Z/2&\stackrel{q}\rightarrow&\Sigma \mathbb Z/2\\ {\scriptstyle 0}\downarrow&&{\scriptstyle 0}\downarrow&&{\scriptstyle q}\downarrow&&{\scriptstyle 0}\downarrow\\ \mathbb Z/4&\stackrel{i}\rightarrow&\mathbb Z/2&\stackrel{q}\rightarrow&\Sigma \mathbb Z/2&\stackrel{-\Sigma f}\rightarrow &\Sigma \mathbb Z/4 \end{array}$$ Here $q\in \operatorname{Ext}_{\mathbb Z}^1(\mathbb Z/2,\mathbb Z/2)\cong \mathbb Z/2$ is the generator, but there are not degree $-1$ maps $\mathbb Z/2\rightarrow \mathbb Z/2$, i.e. the only map $\Sigma\mathbb Z/2\rightarrow\mathbb Z/2$ is the trivial map. Therefore, the standard completion of the first commutative square would be the trivial one, not the former.

The last explicit counterexample turned out to be more obvious than I remembered.

share|improve this answer
    
Thanks! What if I work with dg-categories over a field? Should I be able to construct a counterexample with this technique? –  Francesco Genovese Aug 5 at 19:21
1  
Of course, take $k[x]$ and $k[x]/(x)\hookrightarrow k[x]/(x^2)\twoheadrightarrow k[x]/(x)$ –  Fernando Muro Aug 6 at 8:16

Your Answer

 
discard

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.