Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 / 1

Let $x,y\in H^n(S^n\times S^n;\mathbb{Z})$ be the pullbacks of the two orientation classes. The cohomology ring of a product of two spheres hasand $xy$ generates $H^{2n}\cong \mathbb{Z}$. WriteThe matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ lies in $G{L}_2(\mathbb{Z})$ because a homeomorphism induces a cohomology ring automorphism.

If $n$ is even, graded commutativity givesThus $ab=cd=0$. Invertibility forces $A$ to be diagonal or anti-diagonal, and its two nonzero entries must each be $\pm 1$. Hence there are exactly eight possible actions: the signed permutation matrices.

Evenness is necessary. For $n=1$, the product is the torus, and every $A\in G{L}_2(\mathbb{Z})$ is induced by an integral linear automorphism of the torus. For example, the infinitely many matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ give distinct actions on $H^1$.

The cellular chain complex for either space has one cell in dimensions $0,2,3,4$, with the only nonzero differential equal to multiplication by $p$ from degree three to degree two. The universal coefficient theorem for cohomology therefore gives, for both $X$ and $Y$,Every product of positive-degree classes vanishes for dimensional reasons, so

Their modulo-$p$ cohomology rings distinguish them. Let $u\in H^2(X;{\mathbb{F}}_p)$ be the class restricting to the standard generator on ${\mathbb{CP}}^2$. The attaching map has degree $p$, so its cellular coboundary vanishes modulo $p$; the classes in degrees two and four restrict isomorphically to those of ${\mathbb{CP}}^2$. Hence $u^2\ne 0$ in $H^4(X;{\mathbb{F}}_p)$. In $Y=M(\mathbb{Z}/p,2)\vee S^4$, the degree-two class comes from the three-dimensional Moore space, so its square is zero; products between distinct wedge summands also vanish. The mod-p cup-square obstruction to a homotopy equivalence now proves

The standard cellular chain complex of the Klein bottle givesThe universal coefficient theorem for cohomology therefore yields $H^0=\mathbb{Z}$, $H^1=\mathbb{Z}$, and $H^2=\mathbb{Z}/2$. The degree-one generator is pulled back from the base circle in the circle-bundle description of $K$, so its square is zero. Thus every positive-degree product vanishes. The same calculation for ${\mathbb{RP}}^2\vee S^1$ gives the integral cohomology ring of the wedge of the real projective plane and a circle, and hence

There is no map $Y\to K$ inducing this isomorphism. The Klein bottle is an aspherical space, and its fundamental group is torsion-free. Any map ${\mathbb{RP}}^2\to K$ therefore induces the trivial homomorphism $\mathbb{Z}/2\to {\pi }_1(K)$ and is null-homotopic. Its pullback on $H^2$ is zero, while the circle summand has no degree-two cohomology. Consequently every map $Y\to K$ is zero on $H^2(K;\mathbb{Z})$ and cannot be a cohomology isomorphism.