The identification $G=\Omega^\infty \Sigma^\infty S_0$ of the stable group of self homotopy equivalences of spheres with the suspension spectrum

My topology professor told me in a discussion that the suspension spectrum $colim \Omega_n \Sigma_n S_0$ is the same as the monoid $G$ where $G=colim G_n$ where $G_n$ are self homotopy equivalences of $S_n$.

I just want to ask if this is correct or whether I heard her wrong.

My first guess is that $G=\pi_0 \Omega^\infty \Sigma^\infty S_0$ is the correct statement. A homotopy equivalence of $S_k$ is an element $\alpha=\pm 1 \in \pi_k(S^k)=\pi_0( \Omega_k \Sigma_k S_0)$.

This gives a map from $G_k \to \pi_0\Omega^\infty \Sigma^\infty S_0$.

But since the connected components of $\Omega^\infty \Sigma^\infty$ are not contractible since the stable k stems $\pi_k^S(S_0)$ are nontrivial for $k>0$, I can't identify $\pi_0 \Omega^\infty \Sigma^\infty S_0$ with $\Omega^\infty \Sigma^\infty S_0$.

Hence I don't see any possible way of even finding a map from $G \to \Omega^\infty \Sigma^\infty S_0$.

Do any of you?

• For sensible spaces, $[\Sigma X,Y]=[X,\Omega Y]$ and, in particular, $[S^n,S^n]=[\Sigma^nS^0,\Sigma^nS^0]=[S^0,\Omega^n\Sigma^nS_0]$, and this last set is the set of path components of $\Omega^n\Sigma^nS^0$. – Mariano Suárez-Álvarez Jun 9 '16 at 23:03
• ok so as I wrote above I get map $colim_n G_n \to \pi_0(\Omega^\infty \Sigma^\infty)$. This is also just what you pointed out, since $G_n=[S^n,S^n]_+$. But there is no identification between $\pi_0(\Omega^\infty \Sigma^\infty)$ and $\Omega^\infty \Sigma^\infty$ because $\Omega^\infty \Sigma^\infty$ doesn't have contractible components. Hence there is no way of identifying $G$ with $\Omega^\infty \Sigma^\infty S_0$. Is this assesment correct? – user062295 Jun 9 '16 at 23:19
• Of course not. $\Omega^\infty S^\infty S^0$ is quite far fro having contractible path components! – Mariano Suárez-Álvarez Jun 9 '16 at 23:28
• Yes. $\pi_k(\Omega \infty S^\infty S_0=pi_k^S(S^0)$ which are not zero. But it seems in any case I can accept that I just heard my topology teacher wrong and not worry about it anymore. Thanks for the help. – user062295 Jun 9 '16 at 23:32
• Ask him what he means! It is not reasonable to ask the Internet™ for what someone you can email has in mind. – Mariano Suárez-Álvarez Jun 10 '16 at 2:27

This sounds correct. Each component of $\Omega^\infty S^\infty S_0$ gives a homotopy equivalence of $S^k$ for all $k$ greater than some $K$. Thus $G= \pi_0(colim S^k \Omega^k S_0)$.