I have asked the below question on MathSE (with a 200 point bounty) but have yet to receive an answer there, and so am trying here. I am happy to remove it if it is nevertheless decided that this question is not appropriate here.

Let $X$ and $Y$ be pointed CW complexes, with $X$ a Co-H-Space with co-multiplication $\mu$.

I believe that the following identity holds.

$$X \rtimes Y \simeq X \vee (X \wedge Y),$$

where $\rtimes$ denotes the half-smash: $X \rtimes Y := X \times Y/(* \times Y)$.

I am looking for a proof, or a reference for one.

I am aware that this identity holds when $X$ is a suspension. A technique for proving it in that case is to prove that $\Sigma A \rtimes Y \simeq \Sigma(A \rtimes Y)$, by taking the homotopy pushout of the diagram $* \leftarrow X \rightarrow *$, taking the product with $Y$ everywhere and quotienting by $Y$ everywhere, leaving us with a diagram that is still a homotopy pushout diagram.

It doesn't seem that this argument can be applied here, though.

The co-multiplication $\mu$ on $X$ induces a co-multiplication $\bar{\mu}$ on $X \rtimes Y$ (and also on $X \wedge Y$). This gives us an obvious map:

$$\phi:=(p_1 \vee q) \circ \bar{\mu}:X \rtimes Y \rightarrow (X \rtimes Y) \vee (X \rtimes Y) \rightarrow X \vee (X \wedge Y)$$

where $p_1$ and $q$ are the projection $X \rtimes Y \rightarrow X$ and the quotient map $X \rtimes Y \rightarrow X \wedge Y$, respectively.

I cannot see how it could be deduced that this map is a homotopy equivalence, or if it is even the map I want.