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Aug 28, 2019 at 18:15 comment added Yonatan Harpaz @TylerLawson, this looks pretty good. It means that ${\rm fib}(r)$ actually is the fiber of $r$ in the homotopy category. In addition, since products and coproducts descend to $Ho(\mathcal{A})$ we have that $Ho(\mathcal{A})$ is also additive, which means that we can run the classical argument there (or find a place to quote it...). Pretty slick!
Aug 28, 2019 at 18:02 comment added Tyler Lawson 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 $[z,x]$.
Aug 28, 2019 at 18:01 comment added Tyler Lawson @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 $\pi_1 (Map(z,y),\ast) \to \pi_1 (Map(z,x),\ast) \to [z,fib(r)] \to [z,y] \to [z,x]$.
Aug 28, 2019 at 15:35 comment added Yonatan Harpaz @DylanWilson, the reference you give is about idempotent completion (in the stable case). Here it's not exactly this, I already have the retract, and I want the direct sum decomposition.
Aug 28, 2019 at 15:34 comment added Yonatan Harpaz @TylerLawson, I don't see exactly how, because ${\rm fib}(r)$ is not (a-priori) the fiber in the homotopy category.
Aug 28, 2019 at 14:59 comment added Dylan Wilson Does the proof of HA.1.2.4.6 help?
Aug 28, 2019 at 14:56 comment added Tyler Lawson Can you apply the standard proof to the homotopy category?
Aug 28, 2019 at 14:14 history asked Yonatan Harpaz CC BY-SA 4.0