Infinite loop spaces Let $X, Y$ be  infinite loop spaces: $X = QA$ and $Y = QB$, where $A,B$ are connected topological spaces, and $Q$ stands for $\Omega^\infty S^\infty.$ Let $f:X \to Y$ be a continuous map such that $\Omega f: \Omega X \to \Omega Y$ has a left inverse, i.e. there is a map
$g: \Omega Y \to \Omega X$
 such that 
$g \circ \Omega f =$ identity.                      (*)
Is it true that $f$ itself also has a left inverse?
The "proof" would be: Apply to the equality (*) the functor $B$, associating to an $H$-space $Z$ its classifying space $BZ.$
Seemingly we get what we wanted:
The map $Bg$ would be the left inverse of $B\Omega f$, which is equal to $f$.(Is it?)
The problem with this argument is that if $g$ is not an $H$-space homomorphism, then the map $Bg$ makes no sense.
Is it true that any map $g:\Omega QA \to \Omega QB$ is homotopic to an $H$-homomorphism?
(Then the above proof seems to work.)
 A: If I am not mistaken, we get a counter-example to your last question easily.
First of all we have $\Omega Q\Sigma A=QA$ so we only need to find a map from
$QA$ to $QB$ that is not a $H$-map to get a counter-example.
Denote by $i_X$ the standard map $X\rightarrow QX$ for spaces $X$.  Then
$i_QX :QX\rightarrow QQX$ is almost never a loop map. This can be seen by looking at the homology.
However, this doesn't give a counter example to your first question.
A: Take the Hopf bundle, $S^1\to S^3\stackrel{\eta}{\to}S^2$. That the Hopf invariant of $\eta$ is $1$, is that same as saying that the image of $\eta\in\pi_3S^2\simeq\pi_2\Omega S^2$ under second James-Hopf invariant $H_\#:\pi_2\Omega S^2\to \pi_2\Omega S^3\simeq \pi_3S^3$ induced by second James-Hopf map $H:\Omega\Sigma X\to\Omega\Sigma(X\wedge X)$ is $1$ corresponding to the identity of $S^3$. This is equivalent to saying that the composition 
$$\Omega S^3\stackrel{\Omega\eta}{\to} \Omega S^2\stackrel{H}{\to}\Omega S^3$$
is homotopic to the identity. Hence, $\Omega S^2\simeq\Omega S^3\times S^1$ only as spaces (not $H$-spaces). The map $\eta$, however, does not have a right inverse.
