An unstable homotopy class need not yet be stable

By [F1], the loop $[a,b]$ corresponds to the word $ab{a}^{-1}{b}^{-1}$ in the free group on $a,b$. No two adjacent letters in this four-letter word are formal inverses, so it is already reduced and is not the empty word. The reduced-word uniqueness in [F2] therefore proves that $[a,b]\mathrm{\ne }1$. Thus it is a nonzero initial-stage element of

Since $X$ is connected, [F5] with connectivity parameter $1$ and $i=1$ gives the suspension homomorphism $E:{\pi }_1(X)\to {\pi }_2(\mathrm{\Sigma }X)$. This is its surjective endpoint; no injectivity is promised. The target is abelian by [F3]. Every homomorphism from $F(a,b)$ to an abelian group kills the commutator, so $E([a,b])=0$.

Freudenthal uses the two-cone suspension. Collapsing its basepoint meridian gives the reduced suspension $S^1\wedge X$ used by ${\mathrm{\Sigma }}^{\mathrm{\infty }}X$. The meridian is a nonempty contractible CW subcomplex, so [F4] makes the collapse an isomorphism on ${\pi }_2$; the quotient formula is the reduced-suspension formula in the prespectrum definition. Thus the image found in step 1.2 is the prespectrum bonding image.

In the stable colimit, $[0,[a,b]]=[1,E([a,b])]=[1,0]=0$. Thus a genuinely nontrivial unstable class dies before stabilization. No value of ${\pi }_4(S^3)$ or positive stable stem is assumed. $\mathrm{\square }$