Spaces homotopy equivalent to $A_{\infty}$-spaces I ask this question after reading Peter May's "Geometry of Iterated Loop Spaces", where the problem is definitely hinted at but I couldn't find a definite answer. 
Recall a symmetric operad $\mathcal{C}$ in spaces is called $A _{\infty}$ if the action of $\Sigma _{n}$ on $\mathcal{C}_{n}$ is free, transitive on path components and moreover each component of $\mathcal{C}_{n}$ is contractible. Thus, an operad $\mathcal{C}$ is $A_{\infty}$ if and only if it admits a map $\mathcal{C}_{n} \rightarrow \mathcal{M}$ of operads that is a homotopy equivalence, where $\mathcal{M} _{n} = \Sigma _{n}$ is the discrete operad such that algebras over it are precisely topological monoids. 
An $A_{\infty}$-space is a space acted on by some $A_{\infty}$ operad $\mathcal{C}$. 
Let $f: X \rightarrow Y$ be a homotopy equivalence of pointed, path-connected spaces and suppose that $Y$ is an $A_{\infty}$-space, say acted on by operad $\mathcal{C}$. Does there exist a structure of an $A_{\infty}$-space on $X$ such that the map $f$ becomes a mapping of $A_{\infty}$-spaces? 
I would like to leave this question a little vague in hope of more general answers, but if I was asked to make it precise, I would say that I am looking for an operad $\mathcal{D}$ together with a map of operads $\mathcal{D} \rightarrow \mathcal{C}$ that is a levelwise a weak equivalence and a $\mathcal{D}$-structure on $X$ such that $f$ is a map of $\mathcal{D}$-spaces. 
More generally, can we do this when $\mathcal{C}$ is not necessarily $A_{\infty}$? Ie. can we transport a structure of an algebra over an operad through a homotopy equivalence, up to replacing our operad by some larger one living over $\mathcal{C}$ (yet weakly equivalent)? What happens when $f$ is not a homotopy equivalence, but only a weak equivalence?
 A: It is true for $A_\infty$ spaces. It's a special case of the Homotopy Transfer Theorem, which you can lookup online. You can find a reference (in the algebraic context) in Algebraic Operads, by Loday and Vallette. There's also "Algebra+Homotopy=Operad" by Vallette.
This theorem is basically the reason that we ask this seemingly strange condition of a free, transitive $\Sigma_n$-action on $\mathcal{C}_n$ and that each path component is contractible, because in this case the operad is cofibrant (a concept from model category theory, for which there are many references - I like Dwyer and Spaliński' Homotopy theories and model categories).
This is not in general true for arbitrary operads, that if $X$ is a $P$-algebra and $Y$ is weakly equivalent to $X$ then $Y$ is a $P$-algebra. But in this case (under some assumptions), then $Y$ is a $P_\infty$-algebra. I won't burden you with the details, but basically this is a "up to homotopy" version of $P$. For example $A_\infty$-algebras are associative algebras "up to homotopy", or $E_\infty$-algebras are commutative algebras "up to homotopy". This is hard to define, as you can see in the various references.
Here's how it works, roughly, in an algebraic context (to which I'm more accustomed, but it works almost exactly the same for topological operads).
Suppose $(A, d_A)$ is a differential graded associative algebra, and that $(B, d_B)$ is a deformation retract of $A$, in the sense that there are maps:
$$\begin{align} p &  : A \to B & i : B \to A \end{align}$$
such that $pi = id_B$ and there is a homotopy $h$ acting on $A$ such that $id_A - ip = h d_A + d_A h$.
Then if $\mu : A \otimes A \to A$ is the product on $A$, you could be tempted to define a new product $m_2 = m : B \otimes B \to B$ on $B$ by:
$$ m(x,y) = p(\mu(i(x), i(y)))$$
The problem is, it's not associative: $(xy)z \neq x(yz)$. But if you define $m_3 : B \otimes B \otimes B \to B$ by:
$$ m_3(u,v,w) = -p \mu(i(u), h\mu(i(v), i(w))) + p \mu(h\mu(i(u), i(v)), i(w))$$
(this is almost the associator except the homotopy $h$ is taken into account), then for all $x,y,z$ in $B$:
$$(xy)z - x(yz) = d_B m_3(x,y,z) + m_3(d_B x, y, z) + m_3(x, d_B y, z) + m_3(x, y, d_B z)$$
Which means that in the homology $H^*B$, the product is associative! So it's associative "up to homotopy". And then you can define higher homotopies $m_4, m_5\dots$ to make it fully coherent. You then get an $A_\infty$-structure on $B$, not an associative algebra structure (where $m_3 = m_4 = \dots = 0$).
