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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.)

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    $\begingroup$ I cannot answer your question, but consider the folloing example: $X=B\mathbb{Z}/2$, $Y= B\mathbb{Z}/4$ and $f$ the map induced by the non-trivial group homomorphism $\mathbb{Z}/2\to \mathbb{Z}/4$. As $\Omega X \simeq \mathbb{Z}/2$ and $\Omega Y \simeq \mathbb{Z}/4$, the map $\Omega f$ has a left inverse up to homotopy. But there can be no left inverse as an H-map; therefore, $f$ has no left-inverse. This does not answer your question because while $X$ and $Y$ are infinite loop spaces, they are not $Q$ of something. But do you have reason to believe your conjecture to be true? $\endgroup$
    – Lennart Meier
    Sep 14, 2013 at 2:24

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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.

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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.

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