Suppose that $\mathcal{A}$ is an additive $\infty$-category. By this I mean that $\mathcal{A}$ is pointed, semi-additive (i.e., admits biproducts, which I will call direct sums and denote by $\oplus$), and that every object is group-like in the sense that the shear map $x \oplus x \to x \oplus x$ is an equivalence. In particular, all mapping spaces in $\mathcal{A}$ inherit a canonical structure of an $\mathbb{E}_{\infty}$-group. Now suppose that $$ x \stackrel{i}{\to} y \stackrel{r}{\to} x $$ is a retract diagram in $\mathcal{A}$, and that the map $r: y \to x$ admits a fiber ${\rm fib}(r) \to y$. I would like to show that the induced map $x \oplus {\rm fib}(r) \to y$ is then an equivalence (and so, in particular, $x \to y$ is a "summand inclusion"). If we were dealing with ordinary categories instead of $\infty$-categories then the proof could go like this: one would consider the "complementary idempotent" $e:{\rm Id} - ir: y \to y$. Then $re: y \to x$ is the zero map and hence $e$ factors as $y \to {\rm fib}(r) \to y$. The composition ${\rm fib}(r) \to y \to {\rm fib}(r)$ is then seen to be the identity, so that we have identified ${\rm fib}(r)$ as the retract of $y$ corresponding to the idempotent $e$. The map $x \oplus {\rm fib}(r) \to y$ is then the inclusion part of a retract diagram whose associated idempotent is $ir + e = {\rm Id}$, and is hence an isomorphism.

When trying to make this proof work $\infty$-categorically all kinds of subtleties arise, and it ends up being an overly complicated proof (which causes an undesirable digression in the text I'm working on, since, as it happens, the paper in question is not about retracts in additive $\infty$-categories). On the other hand, this claim seems to be something of a "standard fact", though I was not able to find a reference. My question is hence:

1) Does anyone know a reference for this claim?

2) If no reference is known, do you see a good way to prove this that gives a short proof in the $\infty$-categorical setting?

  • $\begingroup$ Can you apply the standard proof to the homotopy category? $\endgroup$ – Tyler Lawson Aug 28 at 14:56
  • $\begingroup$ Does the proof of HA. help? $\endgroup$ – Dylan Wilson Aug 28 at 14:59
  • $\begingroup$ @TylerLawson, I don't see exactly how, because ${\rm fib}(r)$ is not (a-priori) the fiber in the homotopy category. $\endgroup$ – Yonatan Harpaz Aug 28 at 15:34
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    $\begingroup$ @YonatanHarpaz Good point. So let's be a little more careful. For a generic object $z$, the composition-with-$r$ map $Map(z,y) \to Map(z,x)$ has homotopy fiber $Map(z,fib(r))$ over the zero map; this is because the Yoneda embedding preserves limits. Upon applying the long exact sequence on homotopy, based at the trivial map, we get a sequence $\pi_1 (Map(z,y),\ast) \to \pi_1 (Map(z,x),\ast) \to [z,fib(r)] \to [z,y] \to [z,x]$. $\endgroup$ – Tyler Lawson Aug 28 at 18:01
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    $\begingroup$ Moreover, the composition-with-$r$ map has a natural (pointed!) section, given by composition-with-$i$, which provides sections of maps in this exact sequence. In particular, $[z,fib(r)]$ maps isomorphically to the set of elements in $[z,y]$ that are sent to the zero map in $[z,x]$. $\endgroup$ – Tyler Lawson Aug 28 at 18:02

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