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How do I prove that $S^1\vee S^2\vee S^3$ and $S^1\times S^2$ are not homotopic using homology and cohomology ring respectively?

They have the same homology groups by Kunneth. There is an exercise in Rotman's algebraic topology to prove that these are not homotopic using homology in the chapter of Kunneth formula of homology groups. After that, this again appears in an exercise in the chapter devoted to cup products.

I'm especially curious how to prove this using only homology, because I have never seen a case that two spaces are not homotopic but having the same homology groups.

Thank you in advance.

Hari Rau-Murthy
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Rubertos
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4 Answers4

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If you're allowed to do cohomology, note that $H^*(S^1 \vee S^2 \vee S^3)$ is isomorphic to the direct sum $H^*(S^1)\oplus H^*(S^2) \oplus H^*(S^3)$, so cup product of the generators of $H^1$ and $H^2$ is trivial. Whereas cup product of the generators of $H^1$ and $H^2$ is evidently nontrivial for $S^1 \times S^2$, by Kunneth formula if you wish.

Doing this with homology is a bit tricky: note that universal cover of $S^1 \vee S^2 \vee S^3$ is the topological space which looks like $\Bbb R$ with a copy of $S^2 \vee S^3$ attached to each integer point in $\Bbb R$. This is homotopy equivalent to an infinite wedge of $S^2 \vee S^3$'s, hence the homology group $H_3$ is nontrivial - in fact isomorphic to $\bigoplus \Bbb Z$

On the other universal cover of $S^1 \times S^2$ is $\Bbb R \times S^2$, which is homotopy equivalent to $S^2$. That has trivial homology on dimension $3$, so is not homotopy equivalent to the universal cover of $S^1 \vee S^2 \vee S^3$. Thus, $S^1 \times S^2$ is not homotopy equivalent to $S^1 \vee S^2 \vee S^3$ either.

Balarka Sen
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  • The best way to get an answer for http://math.stackexchange.com/q/2036595/44669 is to take to heart the comments that user @EricTower left. – Lee Mosher Dec 20 '16 at 22:16
  • @LeeMosher All right, but making an excuse, at that time I thought what he commented was merely restating the hint given in the text and I was totally lost what the hint was talking about... but now I proved it myself and I think hint given in the text is somewhat wrong. It should not be decomposed into two copies of $D^n\setminus D$, but into one $D^n\setminus D$ and one $D^n$. – Rubertos Dec 20 '16 at 23:12
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If they were homotopy equivalent, there would be a map $S^1 \vee S^2 \vee S^3 \to S^1 \times S^2$ inducing an isomorphism on homology. The hard part is clearly $S^3$, so let's focus on that.

Because $S^3$ is simply connected, any map $S^3 \to S^1 \times S^2$ lifts to the universal cover $\Bbb R \times S^2$, which deformation retracts onto $\{0\} \times S^2$. Therefore any map $S^3 \to S^1 \times S^2$ is homotopic to one which factors through $\{1\} \times S^2$. In particular, it induces the trivial map on $H_3$, as $H_3(S^2;A) = 0$ for any coefficients $A$.

Because a map from the wedge sum is just a map from each factor sending the basepoint of each to the same place, this means that any map $S^1 \vee S^2 \vee S^3 \to S^1 \times S^2$ induces the trivial map on $H_3$ with any coefficients. So it cannot be a homology isomorphism, and definitely not a homotopy equivalence.

Jefferey Quist
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it is interesting to note after suspending once they are homotopy equivalent, in effect suspending kills the ring structure DS

Dennis Sullivan
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One can also prove directly that $\pi_2$ of the two spaces are different. $\newcommand{\Z}{\mathbb{Z}}$, $\pi_2(S_1 \times S_2)=\pi_2(S_1) \times \pi_2(S_2)=\Z$

Since there is a retraction of the inclusion $S_1 \vee S_2 \xrightarrow{i} S_1 \vee S_2 \vee S_3$, (map $S_3$ to the basepoint) $i_*$ is injective under $\pi_2$, and $\pi_2(S_1 \vee S_2)= \pi_2(\text{universal cover})=\oplus_{i \in \Z} \Z$, so $\pi_2(S_1 \vee S_2 \vee S_3)=(\oplus_{i \in \Z} \Z) \oplus \text{stuff}$.

Hari Rau-Murthy
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