Let $$f_*:\pi_n(A)\to\pi_n(X)$$ be an isomorphism induced by a map $f:A\to X$ of based topological spaces. Does the suspension $\Sigma f$ induce an isomorphism $$(\Sigma_f)_*:\pi_{n+1}(\Sigma A)\to\pi_{n+1}(\Sigma X)?$$
Is the suspension of a $\pi_n$ isomorphism a $\pi_{n+1}$ isomorphism?
2 Answers
No. Consider $A=X=S^1$ with $n=2$. Let $f$ be a constant map. Then $f$ induces an isomorphism on $\pi_2(S^1)=0$. The suspension is null-homotopic, so also induces the zero map on $\pi_3(S^2)\cong\mathbb Z$. (Generated by the Hopf map.) So $\Sigma f$ does not induce an isomorphism.
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0Yes, thank you. Do you know if the statement holds for $n=0$ and $n=1$? – 2011-10-16
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1I don't know. I don't see an argument off the top of my head. One thing I do know is that the suspension of a homology isomorphism is a homology isomorphism, as you may know. – 2011-10-16
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1@JohnZhao, the statement does hold for $n=0$, essentially by the Hurewicz theorem. The statement also holds for $n=1$ for the same reason provided that you strengthen your assumption to include that the map is an isomorphism for all choices of basepoint; otherwise the inclusion $* \to * \coprod S^1$ is a counterexample. – 2011-10-17
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0@Tyler: Thanks for the addendum. – 2011-10-17
If you add the condition that $f$ is $n$-connected (ie, $\pi_k(f)$ is an isomorphism for all $k\leq n$ and a epimorphism for $k=n+1$), then it is true, applying the
Whitehead Theorem (Theorem 10.28 in [Switzer, Algebraic Topology]): Let X, Y be simply connected spaces.
(1) If $f\colon X\rightarrow Y$ is a $n$-connected map,, then $H_k(f)$ is an isomorphism for all $k\leq n$ and a epimorphism for $k=n+1$.
(2) If $f\colon X\rightarrow Y$ is a map such that $\pi_1(f)$ is an isomorphism and $H_k(f)$ is an isomorphism for all $k\leq n$, and a epimorphism for $k=n+1$, then $f$ is $n$-connected.
Back to the question, since $f$ is $n$-connected, by (1) $H_k(f)$ is also an isomorphism for all $k\leq n$ and an epi for $k=n+1$, hence $H_k(\Sigma f)$ is an isomorphism for all $k\leq n+1$ and an epi for $k=n+2$. In addition, $\pi_1(\Sigma A)$ and $\pi_1(\Sigma X)$ are both abelian and (provided $f$ is at least $0$-connected) $H_1(\Sigma f)$ is an isomorphism, hence $\pi_1(\Sigma f)$ is also an isomorphism. Finally, apply (2) to obtain that $\Sigma f$ is $(n+1)$-connected.
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0Two remarks: 1) Above you can replace $n$-connected by $\pi_k(f)$ is an isomorphism for all $k\leq n$, to get $\pi_{k}(\Sigma f)$ is an isomorphism for all $k\leq n+1$ (ie, forget the epis). – 2011-10-17
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02) Also, using Freudenthal suspension theorem, if $A$ is $i$-connected and $X$ is $j$-connected, the map $\pi_{n+1}(\Sigma A)\rightarrow \pi_{n+1}(\Sigma X)$ is an isomorphism if $\pi_{n}(A)\rightarrow \pi_{n}(X)$ is an isomorphism and $n\leq 2\min(i,j)$. – 2011-10-17