$X \rtimes Y \simeq X \vee (X \wedge Y)$ for $X$ a co-H-Space 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.
 A: 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.
