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How to prove $S^n \vee S^1$ is not retract of $S^n \times S^1$?

If $n=1$, it has been solved here. For greater $n$, the fundamental groups seem not to work.

Kirby Lee
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  • Have you tried looking at cohomology? Specifically its ring structure? – Tyrone Dec 24 '17 at 07:10
  • @Tyrone Not yet. Tough job. But I might try that. – Kirby Lee Dec 24 '17 at 07:20
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    If you're following Hatcher's book, then you can read about cup products in section 3.2. You'll need the Kunneth formula on pg 214 to calculate $H^(S^n\times S^1)$, and you should be able to get $H^(S^1\vee S^n)$ using the axioms. You should be able to work out a simple contradiction once you've got that. – Tyrone Dec 24 '17 at 07:32
  • There are more high-tech ways of going about this. Basically $S^n\times S^1$ is $S^n\vee S^1$ with an $(n+1)$-cell attached via a Whitehead product $w:S^n=S^0\ast S^{n-1}\rightarrow S^n\vee S^1$. Now the suspension $\Sigma w\simeq \ast$ (which is why $\Sigma(S^n\vee S^1)$ is a retract of $\Sigma(S^n\times S^1)$), but the cohomological cup product will detect the fact that $w$ is non-trivial before suspension. – Tyrone Dec 24 '17 at 07:37
  • @Tyrone It is quite out of my reach. Thank you so much. – Kirby Lee Dec 24 '17 at 11:30

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If $S^n\vee S^1$ were a retract of $S^n\times S^1$ then all the homotopy groups of $S^n\vee S^1$ would retract off of those of $S^n\times S^1$ also. This, however, cannot occur, and to show this it will suffice to calculate $\pi_n$ of each space.

We calculate $\pi_n(S^1\vee S^n)$ using its universal cover $X$, following Hatcher's construction of universal covers in section 1.3 of "Algebraic Topology". The space $X$ consists of a copy of $\mathbb{R}$ with an n-sphere $S^n$ attached at each integer and is thus homotopy equivalent to the countable wedge sum $\bigvee_\mathbb{Z} S^n$. From this we get that $\pi_n(S^n\vee S^1)\cong\pi_n(X)\cong \pi_n(\bigvee_\mathbb{Z} S^n)\cong \mathbb{Z}[t,t^{-1}]$ is the integral Laurent polynomials in variables $t$, $t^{-1}$.

On the other hand we have $\pi_n(S^n\times S^1)\cong \pi_n(S^1)\oplus\pi_n(S^n)\cong 0\oplus\mathbb{Z}=\mathbb{Z}$.

It follows that $\pi_n(S^n\vee S^1)\cong \mathbb{Z}[t,t^{-1}]$ cannot retract off of $\pi_n(S^n\times S^1)\cong \mathbb{Z}$ and the lack of such an algebraic retraction rules out the possibility of the existence a topological retraction. Therefore $S^n\vee S^1$ is not a retract of $S^n\times S^1$.

Tyrone
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