# $X \rtimes Y \simeq X \vee (X \wedge Y)$ 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.

## 1 Answer

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.

• John: the `obvious map' you have near the end of this discussion is not quite as obvious as you make it seem, as the canonical based map goes the other way. Indeed, this I think uses that Y has a nondegenerate base point. – Nicholas Kuhn Mar 18 at 2:56
• @NicholasKuhn yes, I am aware of that. (But $Y$ is a based CW complex.) – John Klein Mar 18 at 10:37