Path-loop Serre computation of CP infinity

By [F1], the finite-stage quotients $S^{2N+1}\mathrm{/}{S}^1={\mathbb{C}\mathbb{P}}^N$ assemble to the Milnor quotient $B{S}^1={\mathbb{C}\mathbb{P}}^{\mathrm{\infty }}$. The weak equivalence in [F2], together with ${\pi }_1(S^1)=\mathbb{Z}$ and vanishing higher homotopy, gives ${\pi }_i({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }})=\{\begin{array}{ll}\mathbb{Z},& i=2,\\ 0,& i\mathrm{\ne }2,\end{array}$ for $i\ge 1$. Thus the displayed CW colimit is a marked $K(\mathbb{Z},2)$, so [F3] applies and makes the displayed strict path-loop fibration legitimate, with contractible total space and fiber ring $\mathrm{\Lambda }(u)$.

Put $A=H^{\ast }({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }};\mathbb{Z})$. The base is simply connected, hence the fiber cohomology system is constant. Because its two nonzero stalks are the free rank-one group $\mathbb{Z}$, the constant-system comparison is literal, and [F5] gives $E_2^{p,q}=A^p\otimes H^q(S^1;\mathbb{Z}),$ with nonzero rows only at $q=0,1$. Consequently the only possibly nonzero differential is $d_2:E_2^{p,1}=A^{p}u⟶E_2^{p+2,0}=A^{p+2}\mathrm{.}$ By [F6], every positive-total-degree stable term is zero.

Define $c=d_2(u)\in A^2$. It is the transgression of $u$ by [F5]. The upper row has no incoming differential, the bottom row has no outgoing differential, and there are no differentials after $d_2$. Therefore the vanishing stable page says simultaneously that $A^p⟶A^{p+2},a⟼ac,$ is injective (to kill every $au$) and surjective (to hit every positive-degree bottom class), for every $p\ge 0$. Here the Serre Leibniz sign is $(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}$; the induction below makes $A^{\mathrm{o}\mathrm{d}\mathrm{d}}=0$, so this is the displayed multiplication map.

Since $A^0=\mathbb{Z}$, step 2.1 gives $A^2=\mathbb{Z}c$ and shows that $c$ is primitive, not a nonunit multiple of a generator. Since $A^1$ has no possible incoming differential and must vanish at infinity, $A^1=0$. Inductively, multiplication by $c$ identifies $A^{2k}$ with $\mathbb{Z}{c}^{k+1}$ and identifies $A^{2k+1}=0$ with $A^{2k+3}=0$. Hence the homomorphism $\mathbb{Z}[c]\to A$ sending the formal generator to the transgression is an isomorphism in every degree, both injectively and surjectively. There is no additive or multiplicative extension ambiguity: each associated-graded total degree has at most the single bottom-row term.

The base, fiber, and path space are nonempty. The zero and degree-one groups, the unit, $c^0$, the first differential, both rows, both endpoints of every possible differential, and both directions of the ring isomorphism were checked above. Changing $u$ to $-u$ changes $c=d_2(u)$ to $-c$. AC is used through [F2]–[F6]; the finite-stage identification and the integer induction add no choice. This is a computation, not a converse characterization.