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In this example, I feel confused why we can just need to consider the cohomology ring? Of course, $H^*(S^2,Z)\bigoplus H^*(S^4,Z)$ is not isomorphic to $H^*(\mathbb{CP}^2,Z)$. But how is that related to the reduced cohomology ring? I think we still need to prove the reduced case is not isomorphic.


1 Answer 1


This is a typical use of reduced (co)homology. For the wedge sum of spaces, one would like to write formulas like $$H^\bullet (X\vee Y) \cong H^\bullet (X) \oplus H^\bullet (Y).$$ But this is false for trivial reasons, because of the degree $0$ components: if $X$ and $Y$ are connected, both of them will have a copy of $\mathbb{Z}$ in degree $0$ cohomology, and $X\vee Y$ is connected as well, so it will have one copy of $\mathbb{Z}$ in degree $0$. To correct the issue with degree $0$, one passes to the reduced cohomology, where $\tilde{H}^0 = 0$ for connected spaces, and the right formula reads $$\tilde{H}^\bullet (X\vee Y) \cong \tilde{H}^\bullet (X) \oplus \tilde{H}^\bullet (Y).$$

In your problem, you consider a wedge sum, so you really need reduced cohomology. But then keep in mind that $$\tilde{H}^n (X) \cong H^n (X) \quad \text{for }n>0,$$ so the only thing you do by drawing "~" is correcting the degree $0$ components. Nothing happens in higher degrees, where everything will be the same and the cup-products will be the same as well. In your example, $\alpha,\beta,\gamma$ live in higher degrees, so you can think of them as of elements of the reduced cohomology rings.

Of course, $H^\bullet (S^2)\oplus H^\bullet(S^4)$ is not isomorphic to $H^\bullet (\mathbb{CP}^2)$

The text you quote actually shows that $\tilde{H}^\bullet (S^2)\oplus \tilde{H}^\bullet(S^4)$ is not isomorphic to $\tilde{H}^\bullet (\mathbb{CP}^2)$. Without "~" they are not isomorphic for trivial reasons!

You may ask why unreduced rings appear in that text. It is just because formulas like $H^\bullet (\mathbb{CP}^2) = \mathbb{Z} [\beta] / (\beta^3)$ summarize the multiplication table in a very concise way. While reduced rings are rings without multiplicative identity (because we kill the degree zero part).


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