Show that the cross product map \(H^*(X; \mathbb{Z}) \otimes H^*(Y; \mathbb{Z}) \to H^*(X \times Y; \mathbb{Z})\) is not an isomorphism if \(X\) and \(Y\) are infinite discrete sets. [This shows the necessity of the hypothesis of finite generation in Theorem 3.15.]
Solution
For a discrete space \(X\), singular cohomology is concentrated in degree \(0\) and \[ H^0(X;\mathbb Z)=\operatorname{Map}(X,\mathbb Z)=\prod_{x\in X}\mathbb Z. \] Thus for infinite discrete sets \(X,Y\), the cross product in degree zero is \[ \operatorname{Map}(X,\mathbb Z)\otimes \operatorname{Map}(Y,\mathbb Z) \longrightarrow \operatorname{Map}(X\times Y,\mathbb Z), \] sending \(f\otimes g\) to the function \((x,y)\mapsto f(x)g(y)\).
Take countably infinite subsets \(\{x_i\}\subset X\) and \(\{y_j\}\subset Y\) and define \(h:X\times Y\to\mathbb Z\) by \[ h(x_i,y_j)=\delta_{ij}, \] and \(h=0\) elsewhere. If \(h\) were in the image, then on this countable rectangle its matrix \((\delta_{ij})\) would be a finite sum \[ \delta_{ij}=\sum_{r=1}^N a_r(i)b_r(j). \] Every finite submatrix of such a matrix has rank at most \(N\), whereas the \((N+1)\times(N+1)\) identity submatrix of \((\delta_{ij})\) has rank \(N+1\). Contradiction.
Hence the cross product is not surjective, so it is not an isomorphism.