I have asked the below question [on MathSE](https://math.stackexchange.com/questions/3144414/x-rtimes-y-simeq-x-vee-x-wedge-y-for-x-a-co-h-space) (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.



$X \rtimes Y \simeq X \vee (X \wedge Y)$ for $X$ a co-H-Space
The proof is not hard, but more tedious than I would have thought. 


There is a canonical identification 
$$
X\rtimes Y = X\wedge(Y\_+)
$$
where $Y\_+$ is the effect of adding a disjoint base point to $Y$ (we have forgotten the original basepoint here). 
Furthermore, $X\wedge (Y\_+)$ is a co-H space when $X$ is (smash the comultiplication map of $X$ with the identity map of $Y\_+$).


We have two quotient maps $p: X \wedge Y\_+ \to X$ and $ q:X \wedge Y\_+ \to X \wedge Y$. The first is induced by smashing $Y\_+\to S^0$ with the identity map of $X$ and the second is given by smashing the map $Y\_+ \to Y$ with the identity map of $X$.


Now co-multiply the two maps:
$\require{AMScd}$
$$
\begin{CD}
X \wedge(Y\_+) @>c>> X \wedge(Y\_+) \vee X \wedge(Y\_+) @> p \vee q >> X \vee (X \wedge Y)
\end{CD}
$$
where $c$ is the comultiplication. I claim that this map is a homotopy equivalence. 


To get a homotopy inverse, we need to produce a pair of maps 
$a: X\to X \wedge(Y\_+)$ and $b:X\wedge Y \to X \wedge(Y\_+)$ which we can wedge together:
$$
\begin{CD}
X \vee (X \wedge Y) @> a \vee b >> X \wedge(Y\_+) \vee X \wedge(Y\_+) @> \text{fold} >> X \wedge(Y\_+)
\end{CD}
$$
The map $a$ is given by smashing the canonical map $S^0 \to Y\_+$ with $X$. The map $b$ is given as follows: As $X$ is a co-H space, it is a retract of a suspension $\Sigma Z$. For $\Sigma Z$, there is a map 
$(\Sigma Z) \wedge Y \to (\Sigma Z) \wedge (Y\_+)$ since canonically
$$
(\Sigma Z) \wedge Y \cong Z \wedge \Sigma Y
$$
and there is an obvious map $\Sigma Y \to \Sigma (Y\_+)$ which we can smash with the identity of $Z$. Now take the composite
$$
X \wedge Y \to (\Sigma Z) \wedge Y \to (\Sigma Z) \wedge (Y\_+) \to X \wedge (Y\_+)
$$ 


I will leave it to you to check that the map $X \vee (X\wedge Y) \to X \wedge (Y\_+)$ produced in this way is a homotopy inverse to the map produced in the previous paragraph.


