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Let $f\colon X\to Y$ be a morphism of spectra. The associated morphism in the stable homotopy category is an epimorphism, if and only if it fits into a distinguished triangle $$ X\xrightarrow{f} Y\xrightarrow{0}Z. $$ (By abuse of notation, I use the same symbol ''$f$'' here for the induced morphism in the stable homotopy category.) From this we get a long exact sequence on stable homotopy groups and the induced morphisms on stable homotopy groups $\pi_k(f)\colon \pi_k(X)\to\pi_k(Y)$ are all surjective. I wonder whether the converse is true:

Is $f$ an epimorphism in the stable homotopy category if it induces surjections $\pi_k(f)\colon \pi_k(X)\to\pi_k(Y)$ on all stable homotopy groups?

(I guess the answer is 'no' but I cannot find a counterexample.)

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No. If $f: X\to Y$ is a non-zero morphism in the stable homotopy category inducing the zero map on all homotopy groups (according to Is this morphism of spectra zero in the stable homotopy category? you already know that such exist), then the fiber $F\to X$ of $f$ induces an epimorphism on all homotopy groups without being a split epimorphism in the stable homotopy category.

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  • $\begingroup$ Thanks Hanno, very nice answer. $\endgroup$ – user8463524 Feb 10 '15 at 10:22
  • $\begingroup$ You're welcome, jeffrey! $\endgroup$ – Hanno Feb 10 '15 at 13:01

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