Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 127 / 5

Let $i:F\hookrightarrow E$. A pairis transgressive pair when, in the long exact sequence of the pair $(E,F)$,where $H^{r+1}(B,b_0)$ is identified with reduced cohomology. In the Serre spectral sequence, this says that $x$ survives to the transgression andunder the edge identifications, modulo the usual earlier-differential indeterminacy.

The Kudo transgression theorem says that if $(x,y)$ is transgressive and $0\le j\le r$, thenis transgressive. To prove it, use relative Steenrod squares. Naturality givesand stability, equivalently compatibility with the suspension isomorphism, makes squares commute with the connecting map:Applying ${\mathrm{Sq}}^j$ to $\delta x={\pi }^{\ast }y$ proves the theorem. The properties used are naturality, stability, additivity, and the instability conditions ${\mathrm{Sq}}^{j}z=0$ for $j>|z|$ and ${\mathrm{Sq}}^{|z|}z=z^2$.

Let $x\in H^1({\mathbb{RP}}^{\infty };{\mathbb{F}}_2)$ be the generator. Instability givesThe Cartan formula says the total square is multiplicative, soComparing components yields the complete formulawith the binomial coefficient reduced modulo two.

Finally, ${\mathrm{Sq}}^1$ is the Bockstein homomorphism associated withIf a mod-two cocycle representing $y$ is lifted to an integral cochain $A$, write $dA=2B$. Then $B$ modulo two represents ${\mathrm{Sq}}^{1}y$. But $dB=0$, because integral cochains are torsion-free and $2dB=d^{2}A=0$. Thus $B$ itself is a cocycle lift, so its Bockstein vanishes. Thereforefor every space $Y$ and every $y\in H^{\ast }(Y;{\mathbb{F}}_2)$.