# Is this morphism of spectra zero in the stable homotopy category?

Let $f\colon A\to B$ a morphism of spectra and suppose that both spectra $A$ and $B$ have only one non-zero stable homotopy group $\pi_n$, more precisely $$\pi_k(A)=0=\pi_k(B)$$ for $k\neq n$. Suppose further that $f$ induces the zero-morphism on $\pi_n$, i.e. $$\pi_n(f)\colon \pi_n(A)\to \pi_n(B)$$ is the zero-morphism of abelian groups.

Is $f$ necessarily the zero-morphism in the stable homotopy category?

I think this may be true because the heart of the usual t-structure on the stable homotopy category is equivalent to the category of abelian groups, but I am not sure about this.

Surprisingly there are non-zero morphisms of spectra (so called non-zero ghost maps) that induce the zero-map on all stable homotopy groups.

## 1 Answer

Yes.

If $A$ and $B$ have only one non-zero stable homotopy group they both are Eilenberg-Maclane spectra (shifted by $n$, if you will).

Maps between E-M spectra correspond to (stable) cohomological operations and $[EM(A),EM(A)]_0=\operatorname{Hom}(A,A)$.

• Dear Grigory, thank you for the answer. Is it also true that if $A$ and $B$ have non-zero stable homotopy groups $\pi_n$ only in a finite range $a\leq n \leq b$ and $f$ is the zero morphism on all those $\pi_n$, then $f=0$? – user8463524 Jan 9 '15 at 8:35
• No, that generalization is not true: take any non-trivial stable cohomological operation — say, Bockstein $EM(n)\to EM(n+1)$. – Grigory M Jan 9 '15 at 10:53
• For future readers: the so-called "Freyd's generating hypothesis" should be an interesting thing to read about. The question seems to be converging to "in which cases is the homotopy functor faithful?" or some variation of that. – Bruno Stonek Jan 27 '17 at 12:38