Suppose \(X\) is a space with a cover by contractible open sets \(U_1, \ldots, U_n\). Show that the cup product of any \(n\) elements of \(H^{>0}(X)\) is zero, and hence that projective \(n\)-spaces cannot be covered by fewer than \(n + 1\) contractible open sets.
Solution
1. Let \(a_i\in H^{d_i}(X;R)\) with \(d_i>0\), for \(i=1,\ldots,n\). Since \(U_i\) is contractible, the restriction of \(a_i\) to \(U_i\) is zero.
A contractible space has no positive-degree reduced cohomology, so \(H^{d_i}(U_i;R)=0\) for \(d_i>0\).
2. Each \(a_i\) therefore lifts to a relative class\[ \bar a_i\in H^{d_i}(X,U_i;R). \]
The long exact sequence of the pair \((X,U_i)\) contains \[ H^{d_i}(X,U_i)\longrightarrow H^{d_i}(X)\longrightarrow H^{d_i}(U_i). \] The last term receives \(a_i\) as zero by 1, so exactness gives a relative lift.
3. The relative cup product satisfies\[ \bar a_1\smile\cdots\smile\bar a_n \in H^{d_1+\cdots+d_n}(X,U_1\cup\cdots\cup U_n)=H^*(X,X)=0. \]
Iterating the relative cup product \[ H^p(X,A)\otimes H^q(X,B)\to H^{p+q}(X,A\cup B) \] gives the displayed class. Since the \(U_i\) cover \(X\), their union is \(X\).
4. Hence every product of \(n\) positive-degree cohomology classes on \(X\) is zero.
The forgetful map from relative to absolute cohomology sends the relative product in 3 to \(a_1\smile\cdots\smile a_n\). Since the relative product is zero, so is the absolute product.
5. Consequently \(\mathbb CP^n\) cannot be covered by fewer than \(n+1\) contractible open sets.
In \[ H^*(\mathbb CP^n;\mathbb Z)=\mathbb Z[\alpha]/(\alpha^{n+1}),\qquad |\alpha|=2, \] the \(n\)-fold product \(\alpha^n\) is nonzero. If \(\mathbb CP^n\) were covered by \(n\) contractible open sets, 4 would force \(\alpha^n=0\).
6. Likewise \(\mathbb RP^n\) cannot be covered by fewer than \(n+1\) contractible open sets.
Use mod-\(2\) cohomology \[ H^*(\mathbb RP^n;\mathbb F_2)=\mathbb F_2[\beta]/(\beta^{n+1}),\qquad |\beta|=1, \] for which \(\beta^n\ne0\).