TITLE: Describing fiber products in stable $\infty$-categories
QUESTION [14 upvotes]: 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?

REPLY [3 votes]: 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.)