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Should the multiplication $(\sum_i \alpha_i) (\sum_j \beta_j) = \sum_{i,j} \alpha_i\beta_j$ actually be $(\sum_i \alpha_i) \smile (\sum_j \beta_j) = \sum_{i,j} (\alpha_i \smile \beta_j)$?

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    $\begingroup$ He does mean cup product he is just omitting the symbol. $\endgroup$ – William Apr 11 at 18:41
  • $\begingroup$ If would have been clearer for Hatcher to write $\left(\sum_i \alpha_i\right)\left(\sum_j \beta_j\right) = \sum_{i, j} (\alpha_i \smile \beta_j)$, with the product on the left the multiplication in the ring $H^*(X, R)$ and the products on the right the usual cup product. Still, the point is just that $\alpha \beta$ is shorthand for the mutliplication of $\alpha$ and $\beta$ in the ring $H^*(X, R)$, which is defined in terms of the cup product. (There isn't really another convenient multiplication-like operation on the graded ring $H^*(X, R)$, so it's not confusing in practice. $\endgroup$ – anomaly Apr 11 at 18:48
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It's not so much an error as a silent change of notation. If you take a look at the next 10 pages or more, you'll see that Hatcher is omitting the "cup" symbol more and more and more. It kind of makes sense to do this, as long as you keep context in mind, i.e. as long as you can remember that when you write two cohomology classes right next to each other the intention is that you should take their cup product. It's kind of like the ordinary product in the real numbers: after a while, we stop writing $x \times y$, and just write $xy$.

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  • $\begingroup$ Makes sense. It's just a bit confusing because on the next few pages, he introduces the cross product and then a multiplication in the tensor product. It becomes a bit jumbled up if one is seeing this for the first time. $\endgroup$ – Al Jebr Apr 11 at 18:48

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