$X \rtimes Y \simeq X \vee (X \wedge Y)$ for $X$ a Co-H-Space? Let $X$ and $Y$ be topological spaces (pointed CW complexes), where $X$ is a Co-H-Space with co-multiplication $\mu$.
I believe that the following identity holds. My first question is: Does it hold?
$$X \rtimes Y \simeq X \vee (X \wedge Y)$$
Here, $\rtimes$ denotes the half-smash product: $X \rtimes Y := X \times Y/(* \times Y)$.
I am aware that this identity holds when $X$ is a suspension (and the proof is not difficult). The technique for proving it in that case cannot 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, however, see how it could be deduced that this map is a homotopy equivalence. I have not used the co-multiplication which $X$ induces on $X \wedge Y$, as it stands, but I am struggling to see where that could be useful. 
I suspect that there is a slick way to do this using the universal properties in question, but I can't see it. 
 A: First observe that for any based space $X$ and any space $Y$ we have $X\rtimes Y\cong X\wedge Y_+$, where $Y_+$ denotes $Y$ with a disjoint basepoint added. If we assume that $Y$ also carries a distinguished basepoint. Then we have a cofibration sequence $S^0\rightarrow Y_+\rightarrow Y$, where the two maps are the obvious ones, and smashing this sequence with $X$ gives us a cofibration sequence
$$X\xrightarrow{i}X\wedge Y_+\xrightarrow{q}X\wedge Y.$$
However we also have a map $Y_+\rightarrow S^0$, and smashing this with $X$ gives us a map
$$p:X\wedge Y_+\rightarrow X$$
which satisfies $p\circ i=id_X$. It follows that the cohomology of $X\wedge Y_+$ splits as the sum
$$H^*(X\wedge Y_+)\cong H^*(X)\oplus H^*(X\wedge Y).$$
Now if $X$ is a co-H-space with comultipication $c$, then so is $X\wedge Y_+$, with the comultiplication being given by the composite
$$X\wedge Y_+\xrightarrow{c\wedge 1}(X\vee X)\wedge Y_+\cong (X\wedge Y_+)\vee (X\wedge Y_+).$$
Using this co-H-structure we add the maps $q$ and $p$ to get a map
$$\Psi=in_1\circ p+in_2\circ q:X\wedge Y_+\rightarrow X\vee (X\wedge Y)$$
and check easily that it induces the previous isomorphism on cohomology. 
We can apply the Seifert-van Kampen Theorem to the previous cofibration to get information on $\pi_1(X\wedge Y_+)$. Assuming that both $X$, $Y$ are connected we have that $X\wedge Y$ is simply connected, so that the map $i_*:\pi_1X\rightarrow \pi_1(X\wedge Y_+)$ is onto. However since $p_*:\pi_1(X\wedge Y_+)\rightarrow \pi_1(X)$ is a left inverse to this map it must be that $i_*$ and hence $p_*$ are isomorphisms. The upshot of this is that the map $\Psi$ induces an isomorphism on $\pi_1$. 
Now we have that $X\wedge Y$ is the homotopy pushout of the maps $\ast\leftarrow X\xrightarrow{i} X\wedge Y_+$, and if we glue the resulting square along the map $q$ we see that the homotopy pushout of 
$$X\wedge Y\xleftarrow{q}X\wedge Y_+\xrightarrow{p}X\qquad (\ast)$$
is the 1-point space, since $p\circ i\simeq id_X$. 
Now the homotopy pushout of $X\leftarrow X\vee (X\wedge Y)\rightarrow X\wedge Y$ is again the 1-point space. Use the map $\Psi$ constructed previously to get a map of homotopy pushout squares from the square $(\ast)$ to this new square. By construction this map of squares is the identity on the two corners and on the pushout space (which is contractible in each case). Hence if we take cofibers of the induced maps we get a homotopy pushout diagram
$$\ast\leftarrow C_\Psi\rightarrow \ast$$
which again has contractible pushout. In particular $\Sigma C_\Psi\simeq\ast$. But since $\Psi$ induces an isomorphism on $\pi_1$, its cofiber $C_\Psi$ is simply connected, so the fact that $\Sigma C_\Psi\simeq \ast$ is sufficient for us to conclude that $C_\Psi\simeq \ast$ and this tells us that $\Psi$ is a homotopy equivalence. 
In fact an easy calculation gives us a slightly stronger statement: we find that the map $\Psi$ is co-H-map. Hence $\Psi$ is in fact an equivalence of co-H-Spaces.
If you are not so familiar with homotopy pushouts, then apply the groupoid version of Seifert-Van Kampen theorem where we applied the standard version before to show that the inclusion $i:X\rightarrow X\wedge Y_+$ induces an equivalence of fundamental groupoids. Hence any local coeffient system on $X$ is pulled back from one on $X\wedge Y_+$, and vice versa. The diagrams of squares subsequently assembled allow for comparison of the Mayer-Vietoris sequences of belonging to each square, and one sees directly that the map $\Psi$ induces an isomorphism on homology with any local coeffiecents. Hence we can appeal to the Whitehead theorem and arrive at the same claim as before, namely that $\Psi$ is a homotopy equivalence.
