Is the suspension of a $\pi_n$ isomorphism a $\pi_{n+1}$ isomorphism? 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 $$(\Sigma_f)_*:\pi_{n+1}(\Sigma A)\to\pi_{n+1}(\Sigma X)?$$
 A: 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.
A: 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.
