Serre spectral sequence of the complex Hopf fibration

Let $U_j=\{[z]:z_j\mathrm{\ne }0\}$. Every line in $U_j$ has the unique unit representative $s_j([z])$ whose $j$th coordinate is positive real. Explicitly, if $z$ is already unit, then $s_j([z])=z\mathrm{\mid }{z}_j\mathrm{\mid }\mathrm{/}{z}_j$. Hence $p^{-1}(U_j)⟶U_j\times S^1,z⟼([z],z_j\mathrm{/}\mathrm{\mid }{z}_j\mathrm{\mid })$ is a bundle chart, with inverse $([z],\lambda )\mapsto \lambda s_j([z])$. For $a=1\mathrm{/}(2(n+1))$, set $r_j([z])=\max \{\mathrm{\mid }{z}_j{\mathrm{\mid }}^2-a,0\},{\rho }_j=r_j\mathrm{/}{\sum }_i{r}_i\mathrm{.}$ Some $\mathrm{\mid }{z}_j{\mathrm{\mid }}^2\ge 1\mathrm{/}(n+1)>a$, so the denominator is positive; moreover $\mathrm{supp}({\rho }_j)\subseteq \{\mathrm{\mid }{z}_j{\mathrm{\mid }}^2\ge a\}\subset U_j$. Thus these finitely many charts are support-subordinate numerating data, and [F1] makes $p$ a Serre fibration.

Put $P^r={\mathbb{C}\mathbb{P}}^r$. The characteristic map $D^{2r}⟶P^r,w⟼[w_0:\cdots :w_{r-1}:\sqrt{1-\mathrm{\parallel }w{\mathrm{\parallel }}^2}]$ sends the boundary into $P^{r-1}$ and its interior homeomorphically onto $P^r\setminus P^{r-1}$: there the last homogeneous coordinate has a unique positive-real unit representative. Compact-to-Hausdorff quotient descent gives the attachment homeomorphism. Induction produces one cell in dimensions $0,2,\dots ,2n$ and none elsewhere. Adjacent cellular chain groups never both occur, so every cellular differential is zero. By [F2, F3], $H^k(P^n;\mathbb{Z})=\{\begin{array}{ll}\mathbb{Z},& k=0,2,\dots ,2n,\\ 0,& \text{otherwise}\mathrm{.}\end{array}$

The base is simply connected: its CW structure has one vertex and no one-cells. Thus the fiber system is constant. Since all base and fiber groups are free, [F4] has $E_2^{p,q}=H^p(P^n;\mathbb{Z})\otimes H^q(S^1;\mathbb{Z}),$ with only rows $q=0,1$ and columns $p=0,2,\dots ,2n$. The only possible differential is $d_2:E_2^{p,1}\to E_2^{p+2,0}$. The total sphere has cohomology only in total degrees $0$ and $2n+1$ by [F3]. Therefore every displayed $d_2$ for $0\le p<2n$ is an isomorphism between infinite cyclic groups, while the class in $(2n,1)$ survives as the total sphere's top class.

Choose $u$ as in the statement and put $x=d_2(u)$. The $p=0$ isomorphism in step 2.1 makes $x$ a primitive generator of $H^2(P^n;\mathbb{Z})$. Since $d_2$ vanishes on the bottom row, the derivation rule gives $d_2(u{x}^k)=x^{k+1}(0\le k<n)\mathrm{.}$ Inductively the isomorphisms in step 2.1 make $x^k$ a generator of $H^{2k}(P^n;\mathbb{Z})$ for every $0\le k\le n$. The CW dimension gives $x^{n+1}=0$. Hence evaluation induces a surjection $\mathbb{Z}[X]\mathrm{/}(X^{n+1})\to H^{\ast }(P^n;\mathbb{Z})$; it is injective because in each allowed degree the image $x^k$ has infinite order. This proves both directions of the ring identification.

There is no extension ambiguity: every total degree on the base ring has one group, and the actual cup powers were identified before passage to the stable page. The top element $u{x}^n$ survives because its target column is $2n+2$, outside the base, exactly accounting for $H^{2n+1}(S^{2n+1})$. For $n=1$ the sole differential is $d_2(u)=x$ and the survivor is $ux$; the zero groups, unit, first and last columns, both rows, and both differential endpoints are therefore included. Reversing $u$ reverses $x$. AC is used only through [F1], [F3], and [F4]; all charts, the finite partition, and the cellular calculation are choice-free. No converse or splitting is asserted.