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?
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 $G= \pi_0(colim S^k \Omega^k S_0)$.