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In an exam recently, I was asked to find the minimal number of contractible sets covering $\mathbb{CP}^3$ by considering the cup-product on relative cohomology. Is there nice a way of doing this, either using the proposed approach or some other?

Note: I am aware of (the existence of) the Lusternik–Schnirelmann category, but since this was not part of the curriculum, I doubt that we were supposed to use it.

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LS category only gives you an upper bound, anyways, and you are asking for the minimal number. –  Mariano Suárez-Alvarez Jun 8 '11 at 18:53
@Mariano: Could you elaborate? If I am not misreading the Wikipedia article, the LS category is exactly the number I am looking for. –  Raeder Jun 8 '11 at 23:33
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1 Answer

up vote 1 down vote accepted

Indeed, there is a solution that uses the fact that cup-length is an obvious lower bound for the L-S category. Namely, if $X$ is covered by $n$ contractible sets $U_i$, multiplication $\tilde H(X)^{\otimes n}\to\tilde H(X)$ factors through $\tilde H(X)^{\otimes n}=\otimes H(X,U_i)\to H(X,\bigcup U_i)=H(X,X)=0$. (Perhaps, it's what was expected.)

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What happens at your first equality sign? Assuming we have this, then the lower bound is the cup-length, in this case $3$, and we have the obvious four $\{z_i\neq 0\}$ contractible sets as an open cover - how do we go from $4$ to $3$ or prove that this isn't possible? –  Raeder Jun 8 '11 at 23:33
@Raeder 1) Since $U_i$ is contractible, $H(X,U_i)=H(X,pt)=\tilde H(X)$. 2) Lower bond we get is that $CP^3$ can't be covered by 3 contractible sets (since there is a generator in $u\in H^1(CP^3)$ s.t. $u^3\ne0$), so it solves the problem. –  Grigory M Jun 9 '11 at 6:28
Thanks for the answer and the explanation! –  Raeder Jun 9 '11 at 7:48
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