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Let $f\colon X \rightarrow Z$ and $g\colon Y \rightarrow Z$ be two morphisms in a stable infinity category $\mathcal{C}$. How does one show that the $\infty$-categorical fiber product $X \times_Z Y$ can be canonically identified with the fiber of the map $f - g \colon X \oplus Y \rightarrow Z$?

I am sure that the question is quite basic (for instance, something quite similar is used implicitly in the proof of Proposition 1.1.3.4 of Lurie's "Higher algebra"), but I would nevertheless appreciate an explanation. The type of answer I am not looking for is "this is trivial consequence of the definitions" because I believe that there is something to check: $X \times_Z Y$ makes sense in any $\infty$-category that has fiber products, whereas the other description uses the stability assumption.

While we are at it, let me ask a related bonus question: with $f$ and $g$ as above, suppose that we also have maps $a\colon F \rightarrow X$ and $b\colon F \rightarrow Y$ that make the evident diagram commute and such that $\mathrm{fib}(a) \simeq \mathrm{fib}(g)$ via the induced map (a five lemma type of setting where one of the vertical maps is an isomorphism). Do these conditions imply that the map $F \rightarrow X \times_Z Y$ is an isomorphism?

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In fact what you need is that your ∞-category is additive (i.e. that it has direct sums and that the canonical commutative monoid structure on the mapping spaces is group-like). All stable categories are additive, so I'll just prove it in this case.

Step 1: The pullback $X\times_Z Y$ is equivalent to the pullback $(X\times Y)\times_{Z\times Z}Z$. This fact is true in every ∞-category with finite limits

I'm sure there are many ways of proving this. One way of doing it (thanks to Omar Antolín-Camarena for the idea) is to consider the diagram

$$\require{AMScd} \begin{CD} Y @>>> Z @<<< Z\\ @VVV @VVV @VVV\\ \ast @>>> \ast @<<< Z\\ @AAA @AAA @AAA\\ X @>>> Z @<<< Z \end{CD}\,,$$

where $\ast$ is the terminal object. Taking the limit first in the horizontal direction and the in the vertical direction we obtain $X\times_Z Y$, while taking the limits in the opposite order we obtain $(X\times Y)\times_{Z\times Z}Z$. Since both methods compute the limit of the whole diagram, the two objects must be equivalent.

Step 2: The diagonal $Z\to Z\oplus Z$ is the fiber of the "difference" map $Z\oplus Z\xrightarrow{(1_Z,-1_Z)} Z$

This is where we use the additivity hypothesis. Recall that one way of phrasing that the ∞-category is additive is that the shear map $$Z\oplus Z\xrightarrow{\left(\substack{1\, 1\\ 0\, 1}\right)} Z\oplus Z$$ is an equivalence. Then there is an equivalence of sequences $$\require{AMScd} \begin{CD} Z @>{\left(\substack{0\\1}\right)}>> Z\oplus Z @>{(1\,0)}>> Z\\ @| @V{\left(\substack{1\,1\\0\,1}\right)}VV @|\\ Z @>{\left(\substack{1\\1}\right)}>> Z\oplus Z @>{(1\,-1)}>> Z \end{CD}$$ Since the top one is a fiber sequence, so is the bottom one.

Step 3 Profit!

Let us consider the following diagram $$\require{AMScd} \begin{CD} X\times_Z Y @>>> Z @>>> 0\\ @VVV @V{\left(\substack{1\\1}\right)}VV @VVV\\ X\oplus Y @>{(f\,g)}>> Z\oplus Z @>{(1\,-1)}>> Z \end{CD}$$ Since the left and right squares are cartesian (by step 1 and 2), so is the big one, which was the thesis.


Bonus question: In every $\infty$-category with finite limits, if you have a square of pointed objects $$\require{AMScd} \begin{CD} F @>{a}>> X\\ @V{b}VV @V{f}VV\\ Y @>{g}>> Z \end{CD} $$ there is a canonical equivalence between the fiber of the map $F\to X\times_Z Y$ and the fiber of the induced map $\mathrm{fib}(a)→\mathrm{fib}(g)$. This can be proven as before, by considering the diagram $$\require{AMScd} \begin{CD} \ast @>>> X @<<{a}< F\\ @VVV @V{f}VV @V{b}VV\\ \ast @>>> Z @<<{g}< Y\\ @AAA @A{g}AA @AA{1_Y}A\\ \ast @>>> Y @<{1_Y}<< Y \end{CD}\,,$$ and taking the limit in the two directions (you can find more details in the first section of this paper). Incidentally this object is called the total fiber of the square.

Since in a stable ∞-category a map is an equivalence iff the fiber is trivial, this gives an affermative answer to your query.

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  • $\begingroup$ Thank you, this is very useful. Do you happen to also know the answer to the bonus question? $\endgroup$ – Question Machine Jun 4 at 16:27
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    $\begingroup$ @QuestionMachine Let me add a quick remark $\endgroup$ – Denis Nardin Jun 4 at 16:29
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    $\begingroup$ For step 1 you can also use that pullbacks commute will pullbacks: $(X \times Y) \times_{Z \times Z} Z \simeq$ $(X \times_1 Y) \times_{Z \times_1 Z} (Z \times_Z Z) \simeq^\ast$ $(X \times_Z Z) \times_{1 \times_1 Z} (Y \times_Z Z) \simeq$ $X \times_Z Y$, where $1$ is terminal and the equivalence labelled $\ast$ is the commutation step. $\endgroup$ – Omar Antolín-Camarena Jun 5 at 13:32
  • $\begingroup$ @OmarAntolín-Camarena That's a nice way of seeing it! I'm thinking of adding a variant of that proof to my answer $\endgroup$ – Denis Nardin Jun 5 at 15:10
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    $\begingroup$ @OmarAntolín-Camarena If you are interested I can give a more conceptual explanation: the category $I_n$ is the poset of chains of nonempty subsets of $\{1,..,n\}$. There is a cofinal functor $I_n→P_{\neq\varnothing}\{1,..,n\}$ sending every chain to its maximal element and a right fibration $I_n→P_{\neq\varnothing}\{1,..,n\}$ sending every chain to the subset given by the cardinalities of the elements of the chain. The diagram I used is the case $n=2$. This is helpful when your original cube has a $G$-action for a finite group $G$, because this allows you to remove the action. $\endgroup$ – Denis Nardin Jun 5 at 16:41
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A proof of this in the context of stable derivators can be found in my paper Mayer-Vietoris sequences in stable derivators with Moritz Groth and Kate Ponto. It is rather more verbose than Denis's, but one reason for that is that it's intended to be fully precise. In an $\infty$-category, of course, you can't actually just write down some objects and arrows and have a "diagram"; there are coherences that need to be produced too, and kept track of all the way through. Derivators are a way of forcing oneself to remember that. (With that said, it's quite possible that if Denis's proof were made precise it would still be shorter than ours.)

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    $\begingroup$ If Denis’s answer is to be considered imprecise, then it would be hard to find a single precise statement in the mathematical literature. $\endgroup$ – crystalline Jun 6 at 0:16
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    $\begingroup$ I'll admit that I skimped in specifying the homotopies, but if you look at the diagrams I wrote there are only two kind of homotopies needed to be specified: homotopies between maps to the terminal object (where you have only one choice), and the defining homotopy for the map $-1:Z→Z$ (a nullhomotopy of $1-1$). Note that $-1$ is essentially defined so that the second square in step 2 commutes. I agree that normally one needs to be careful in specifying what you mean, but here it seems superfluous. $\endgroup$ – Denis Nardin Jun 6 at 5:30
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    $\begingroup$ @DenisNardin I would say there is more, For instance, it's not immediately obvious to me what it means to "take the limit in the horizontal/vertical direction" of the first diagram, maybe a right Kan extension; but whatever that is, the process produces a 3- or 4-dimensional diagram with a bunch more coherences in it. One then needs to identify not just the objects appearing in the first sets of pullbacks -- up to equivalence -- but also the morphisms between them as being, e.g., equivalent to the diagonal morphism $Z\to Z\times Z$ or whichever other morphisms one wants. $\endgroup$ – Mike Shulman Jun 6 at 12:33
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    $\begingroup$ Unless I am mistaken all the cohérences follow trivially from what I have described via universal properties (e.g. you can identify the map $Z\to Z\times Z $ with the diagonal by seeing that its composition with the projections are the horizontal arrows $Z\to Z $ on the right column of my first diagram) By taking the limit first in one direction and then in another I just mean that there is a natural equivalence $\lim_{I\times J} F \cong \lim_I \lim_I F $ for every functor $F:I\times J\to C$, no other coherences needed (beyond of course the definition of F) $\endgroup$ – Denis Nardin Jun 6 at 12:51
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    $\begingroup$ @MikeShulman I agree that this is a very tricky thing (and many a mistake were made by careless people forgetting the higher homotopies), but I think we cannot write papers as if they were computer-formalized: they'd be unreadable by actual human beings (and many papers are nigh-unreadable as it is). At some point you'd have to either believe me that I checked that all the issues I'm declaring trivial are actually trivial (and I did check),or check for yourself.If you want to continue this discussion I'd be happy to show you (in chat) in painstaking detail how to formalize every missing detail $\endgroup$ – Denis Nardin Jun 6 at 13:49

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