Decomposable maps of half-smash products

Question

[Cross-posted from MSE]

For a pointed space $X$ and unpointed space $Y$, recall the half-smash product $X\rtimes Y=X\land Y_+=(X\times Y)/(\ast\times Y)$. For unpointed spaces $X,Y$ and a pointed space $Z$, suppose there is a commutative diagram of pointed maps $$\require{AMScd}\begin{CD} S^n\rtimes X @>{F}>> Z\rtimes Y\\ @VVV @VVV\\ S^n @>{f}>> Z \end{CD}.$$ Question: Is this sufficient to conclude that $F$ decomposes as $f\rtimes\phi$ for some map $\phi\colon X\rightarrow Y$ up to homotopy?

This claim of sufficiency is in fact made in Exercise 4.34. of Jeffrey Strom's Modern Classical Homotopy Theory. It features there as a generalization of Lemma 4.32., which claims the result in the case $Z=S^n$. The argument in the book, however, is incorrect for it purports a homotopy $F\simeq f\rtimes\phi$ that is fiberwise over $S^n$, but such a homotopy would imply that the fiberwise maps $F_x\colon X\rightarrow Y$ (whenever defined, i.e. for $x\in S^n$ s.t. $fx\in Z$ is not the basepoint) are all homotopic to another. This is already contradicted by an example like $$F\colon S^1\rtimes S^1\rightarrow S^1\rtimes S^1,\,z\rtimes w\mapsto\begin{cases}z^2\rtimes w,\,\mathrm{Im}(z)\ge0,\\z^2\rtimes\overline{w},\,\mathrm{Im}(z)\le 0.\end{cases}$$ That said, I believe $F\simeq(z\mapsto z^2)\rtimes(z\mapsto\ast)$, so that this is not a counter-example to the actual question.

Answer 1

No, take the case $X=\ast$. This asserts that any map $S^n \rightarrow Z \rtimes Y$ factors through the inclusion of a fiber $Z \rtimes \{y\}$, up to homotopy. This is certainly false, for instance if $M$ is a compact $n$-manifold with trivial tangent bundle than there is some large $N$ such that there is a map $S^N \rightarrow S^{N-n} \rtimes M$ which is homologically nontrivial in dimension $N$. An explicit construction is given by embedding $M$ into $\mathbb{R}^N$ and taking the one point compactification of the inclusion of an open tubular neighborhood of $M$.

Answer 2

My example in the post was this close to a genuine counter-example to not just the general Question, but also Lemma 4.32. specifically. There is a quotient map $q\colon S^2\twoheadrightarrow S^2/\{\pm N\}=S^1\rtimes S^1$, which induces an isomorphism on $H_2$ (this also features in @tkf's answer). Now, consider the squaring map $f\colon S^1\rightarrow S^1,\,z\mapsto z^2$ and arbitrary maps $\varphi,\psi\colon S^1\rightarrow S^1$ and let $$F_{\varphi,\psi}\colon S^1\rtimes S^1\rightarrow S^1\rtimes S^1,\,z\rtimes w\mapsto\begin{cases} z^2\rtimes \varphi(w),\,\mathrm{Im}(z)\ge0,\\ z^2\rtimes \psi(w),\,\mathrm{Im}(z)\le0.\end{cases}$$ Then, $F$ fits into a commutative diagram $$\require{AMScd} \begin{CD} S^2 @>>> S^2\vee S^2\\ @V{q}VV @V{(q\circ\Sigma\varphi,q\circ\Sigma\psi)}VV\\ S^1\rtimes S^1 @>{F}>> S^1\rtimes S^1, \end{CD}$$ where the top horizontal map is the equatorial collapse. Applying $H_2$, we obtain that $F_{\varphi,\psi}$ induces multiplication by $\deg(\varphi)+\deg(\psi)$. Taking $\varphi=\psi$, we see that $H_2(f\rtimes\varphi)=2\deg(\varphi)$ is even for any $\varphi\colon S^1\rightarrow S^1$, but $H_2(F_{\mathrm{id}_{S^1},\mathrm{const}})=1$ is not even, hence this map is a counter-example.

Answer 3

Let $n=2$, $X=\{*\}$, $(Z,+)=(U(1),-1)$ and $Y=S^1$. We identify $S^2$ and $Y$ with the unit sphere and unit circle in $\mathbb{R}^3$ and $\mathbb{R}^2$ respectively.

Then we can let $f(x,y,z)=e^{\pi i z}$, which is compatible with: $$F((x,y,z),*)=\left( e^{\pi i z},(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}})\right)$$

Then $F$ induces an isomorphism on second homology groups, so cannot be homotpic to any map which factors through the circle $S^1\times \{(u,v)\}$, for fixed $u,v$.