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Fix a stable $\infty$-category $\mathcal{C}$ and two (co)fibre sequences $a \rightarrow b \rightarrow c$ and $x \rightarrow y \rightarrow z$ in $\mathcal{C}$. Now suppose we are given a map $a \rightarrow x$ such that the composite $c[-1] \rightarrow a \rightarrow x \rightarrow y$ is homotopic to zero.

From this data, I would like to extend the map $a \rightarrow x$ to a map of (co)fibre sequences

$\begin{array}{cccccc} & a & \rightarrow & b & \rightarrow & c\\\ \\\ & \downarrow & & \downarrow & & \downarrow\\\ \\\ & x & \rightarrow & y & \rightarrow & z\\\ \end{array}$

and I would like to know exactly what choices are involved in such an extension.

I believe that such an extension exists: the universal property of cofibre sequences should give a map $b \rightarrow y$ so that we have the first two vertical arrows in a map of (co)fibre sequences, and then the argument that the homotopy category of $\mathcal{C}$ is triangulated allows us to construct the third vertical arrow $c \rightarrow z$.

Now the question is the following: What choices are involved in the construction of the vertical arrows?

For the second vertical arrow, $b \rightarrow y$, I think that the universal property of cofibre sequences should say that I need to actually specify a homotopy from the composite $c[-1] \rightarrow a \rightarrow x \rightarrow y$ to zero. Is that correct?

To construct the third vertical arrow, $c \rightarrow z$, we would likewise need to specify a homotopy from the composite $a \rightarrow b \rightarrow y \rightarrow z$ to zero, but it seems to me that this is already given by the previous choices. Is that correct?

If so, then we can fill in the third vertical arrow more or less canonically. Is this one of the ways in which stable $\infty$-categories are richer than triangulated categories, in which we have only the existence of some third vertical arrow $c \rightarrow z$?

(EDIT: Thanks to David White for providing the above diagram for the map of cofibre sequences.)

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Your guesses sound right to me. –  Mike Shulman Jun 19 '12 at 16:58

1 Answer 1

Hi dhagbert,

Everything you say seems right. In particular the map $c \to z$ is determined up to contractible choice by a choice of homotopy from the composite $$ c[-1] \to a \to x \to y $$ to the zero map. (Depending on how you diagram-chase, you can rephrase this as: "the map $b \to y$ is determined up to contractible choice by a choice of homotopy..." and you can also replace the above composite by another one.) As you seem to intuit, one uses the universal property of both push-outs and pull-backs to see this (hence the necessity of being stable).

And yes, I think you point out a richness. In the Verdier axiom (TR3), you only remember that a certain homotopy commutative square exists (without specifying the homotopy) so you know that the vertical map $c \to z$ exists, but nothing more. In the stable oo-category setting, however, we know from the above paragraph that the space of certain homotopies is equivalent to the space of vertical maps. So once you specify the homotopy commutative square, then you specify the vertical map (up to contractible choice).

You asked what choices are involved in the extension you're after, and that choice is precisely the homotopy of the above composite to the 0 map. As you and I have both pointed out, there are various equivalent ways to phrase this choice (such as filling in a homotopy from a different composite to the 0 map) but it seems you're familiar enough -- and I'm sleepy enough -- that I won't need to spell out all these equivalent phrasings.

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Thanks. Could you suggest an illustrative place in the literature where such things are discussed? –  dhagbert Jun 20 '12 at 12:35
    
A great place to start is probably David Ben-Zvi's answer to this post: mathoverflow.net/questions/815/triangulated-vs-dg-a-infinity. It doesn't go into the details of (TR3) and its deficiencies, but it definitely highlights why you want to remember higher coherence data (like which homotopy commutative square you take in your map of fiber sequences), which triangulated categories don't do. –  Hiro Lee Tanaka Jun 20 '12 at 14:07

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