Trivial cup product in a (cohomology) subring

Question

I was going through a proof that $S^n\times S^n$ and $S^n\vee S^n \vee S^{2n}$ are not homotopy equivalent spaces. Both spaces have the same cohomology groups in degree $n$ and $2n$, namely

$$H^k(\ \cdot\ ;\Bbb{Z}) = \begin{cases} \mathbb{Z}\oplus \Bbb{Z} & k = n \\ \mathbb{Z} & k = 2n \end{cases}$$

Now the proof states the following

$H^*(S^n\vee S^n\vee S^{2n};\mathbb{Z})$ is a subring of $H^*(S^n;\mathbb{Z})\times H^*(S^n;\mathbb{Z}) \times H^*(S^{2n};\mathbb{Z})$. Thus the product of any two degree $n$ elements is trivial.

Could someone please elaborate why the subring property implies the product of any two degree $n$ elements is trivial? It would be really helpfull if someone could demonstrate how the product in this case looks like.

And what precisely is meant by trivial cup product anyway? I was under the impression that the cup product is called trivial whenever the product vanishes i.e. if $a\smile b = 0$, but we are also calling products $ab$ trivial if one factor is the unit, don't we? What is meant in the above statement?

Thanks for any help!

Answer 1

As what Nick L has pointed out, there is a more direct argument, and you can even triangulate the space and use the definition to conclude that cup products are trivial (i.e. vanish).

The subring property is given by Mayer-Vietoris sequence (I think). It induces isomorphisms $$\eta:H^p(S^n\vee S^n\vee S^{2n})\to H^p(S^n)\oplus H^p(S^n)\oplus H^p(S^{2n})$$ for all $p>0$, so the direct sum (the following) corresponds to the positive dimension of $H^*(S^n\vee S^n\vee S^{2n})$. $$H^*(S^n\vee S^n\vee S^{2n})\supset\bigoplus_{p>0}(H^p(S^n)\oplus H^p(S^n)\oplus H^p(S^{2n}))$$ Actually, this is already enough for deducing the triviality of cup products.

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Edit:

As requested by the OP, I'll elaborate more on this statement. Because we've decomposed the positive dimension part of $H^*(S^n\vee S^n\vee S^{2n})$ into $H^*(S^n)\oplus H^*(S^n)\oplus H^*(S^{2n})$ (the direct sum of three graded polynomial rings), it should be noted that in the definition of graded ring, we have $$\operatorname{im}(H^k(X)\times H^l(X)\overset{\smile}{\to}H^{k+l}(X))\subset H^{k+l}(X)$$ Now it becomes immediate that the cup product of any two degree $n$ elements vanishes, since $H^{2n}(S^n)\cong 0$ and $H^{2n}(S^n\vee S^n)\cong 0$.

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But the idea of subring comes in dimension $0$, the MV sequence for wedge products gives $$0\to H^0(X\vee Y)\overset{a}{\to}H^0(X)\oplus H^0(Y)\overset{i^*-j^*}{\to}\Bbb{Z}\to\ldots$$ (Apply this twice to your space) So, $H^*(S^n\vee S^n\vee S^{2n})$ contains triads $(\alpha,\beta,\gamma)\in H^0(S^n)\oplus H^0(S^n)\oplus H^0(S^{2n})$ s.t. their restrictions to the $0$-th cocycle are equal, i.e., their images under the pullback of the inclusion $\{*\}\hookrightarrow S^n$ (resp. $S^{2n}$) are equal. And I think this is probably what the proof mean by "subring".

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Supplement (Something that I recall from Evan Chen's Napkin project), :)

It can also be concluded from "the cohomology pseudo-ring" consisting of the direct sum of reduced cohomology groups. Now these two graded pseudo-rings $$\tilde{H}^*(X\vee Y)\cong \tilde{H}^*(X)\oplus \tilde{H}^*(Y)$$ are indeed isomoprhic. Since we also know that the reduced cohomology groups are isomorphic to absolute ones in positive dimensions, this can be passed to the argument of cohomology rings to conclude the triviality of cup products. These two argument actually utilize the same idea to express (part of) the cohomology ring in terms of what we're familiar with.