$E_{\infty}$ spaces are $A_{\infty}$ spaces

While studying the well-known "Geometry of Iterated Loop Spaces", I found this corollary which is not completely clear to me. (By $\mathcal{M}$ is meant the operad given by $\mathcal{M}(j):=\Sigma_j$, whith the other data suitably defined and such that its algebras are precisely the topological monoids).

More precisely, it is unclear to me why the homotopies witnessing that $\pi_2:\mathcal{C}\times \mathcal{M} \to \mathcal{M}$ is a homotopy equivalence should be $\Sigma$-equivariant. If so, I will certainly agree that any $E_{\infty}$ space is an $A_{\infty}$ space.

• You might want to mention that $\mathcal M$ is the "associative operad", the operad whose algebras are associative monoids. – André Henriques Nov 2 '14 at 21:00
• Sure, I am sorry for that! – Edoardo Lanari Nov 2 '14 at 21:00
• If $E_{\infty}$ and $A_{\infty}$ Operads are cofibrant replacements then the operad map $Ass\rightarrow Com$ gives you a map of operads $A_{\infty}\rightarrow E_{\infty}$, which in particular means than any $E_{\infty}$-algebra (space) is an $A_{\infty}$-algbera (space). – Ilias A. Nov 2 '14 at 21:01
• @Fedotov: I can't understand your comment. Though I know quite well model categories, I have gone through operads only following May's concrete approach, hence I don't know any homotopy theory of operads yet. – Edoardo Lanari Nov 2 '14 at 21:04
• Also asked on math.SE... – Najib Idrissi Nov 15 '14 at 10:15

If $X$ and $Y$ are $G$-spaces for any group $G$, then the projection $\pi\colon X\times Y\to Y$ is a $G$-map, trivially: $\pi (gx,gy) = gy$. The map $\pi_2$ of your question is a very special case; note that $\mathcal M$ as I defined it is an operad as I defined operads, with $\Sigma_j$ acting on $\mathcal M(j)$.
Model categories are extremely important but entirely irrelevant here, and taking cofibrant approximations of operads tends to destroy their relevant individuality: different $E_{\infty}$ operads play seriously different roles in the applications, which are what people should care about.
Also, by my definition, $E_{\infty}$ operads are $\Sigma$-free rather than just locally contractible, so $Com$ is certainly not an example; $E_{\infty}$ spaces are far more general than $Com$ spaces, which of course are just commutative topological monoids.
This edit is to answer Lano's clarification of his question. Lemma 3.7, p. 24, op cit shows that a local equivalence over $\mathcal M$ is necessarily a local $\Sigma$-equivalence. The statement of Corollary 3.11 follows.
• Dear professor May, I have been imprecise. What is not clear is why are the homotopies (witnessing that $\pi_2$ is a homotopy equivalence) equivariant? $\mathcal{C}(j)$ is contractible, though it is explicitely stated that the homotopies involved are not necessarily equivariant. – Edoardo Lanari Nov 2 '14 at 23:03