Let \(X\) be \(\mathbb{CP}^2\) with a cell \(e^3\) attached by a map \(S^2 \to \mathbb{CP}^1 \subset \mathbb{CP}^2\) of degree \(p\), and let \(Y = M(\mathbb{Z}_p, 2) \vee S^4\). Thus \(X\) and \(Y\) have the same 3-skeleton but differ in the way their 4-cells are attached. Show that \(X\) and \(Y\) have isomorphic cohomology rings with \(\mathbb{Z}\) coefficients but not with \(\mathbb{Z}_p\) coefficients.
Solution
Give \(X\) the CW structure with one cell in dimensions \(0,2,3,4\). The cellular boundary \[ C_3(X;\mathbb Z)\longrightarrow C_2(X;\mathbb Z) \] is multiplication by \(p\), while the other positive-dimensional cellular boundaries are zero. Hence \[ H_2(X;\mathbb Z)\cong\mathbb Z_p,\qquad H_4(X;\mathbb Z)\cong\mathbb Z, \] and all other reduced integral homology groups vanish. The same is true for \[ Y=M(\mathbb Z_p,2)\vee S^4. \] By the universal coefficient theorem, \[ \widetilde H^3(X;\mathbb Z)\cong\mathbb Z_p,\qquad H^4(X;\mathbb Z)\cong\mathbb Z, \] with all other positive-degree integral cohomology zero, and identically for \(Y\). Degree reasons force every product of positive-degree integral classes to vanish, so the integral cohomology rings are isomorphic.
Now use \(\mathbb Z_p\) coefficients. Since multiplication by \(p\) becomes zero, both spaces have one-dimensional cohomology in degrees \(2,3,4\). Let \[ x\in H^2(X;\mathbb Z_p) \] be the class restricting to the standard generator of \(H^2(\mathbb{CP}^2;\mathbb Z_p)\). The attached \(3\)-cell does not alter the \(4\)-cell cup-product calculation, and by naturality with respect to the inclusion of the \(\mathbb{CP}^2\) subcomplex one has \[ x^2\ne0\in H^4(X;\mathbb Z_p); \] in fact it is the generator.
For \(Y\), the unique nonzero degree-two class \(y\) is supported on the Moore-space wedge summand. Products of positive-dimensional classes from different wedge summands vanish, and the Moore space has no \(4\)-dimensional cohomology. Therefore \[ y^2=0. \] Thus \[ H^*(X;\mathbb Z_p)\not\cong H^*(Y;\mathbb Z_p) \] as graded rings, although the integral cohomology rings are isomorphic.