Rational cohomology of Eilenberg–Mac Lane spaces in one generator

The calculation cited in [F0] gives ${\pi }_1(S^1)\cong \mathbb{Z}$ and ${\pi }_i(S^1)=0$ for $i>1$, so marked uniqueness in [F1] identifies $K_1$ up to homotopy with $S^1$. By [F8], its rational cohomology is $\mathbb{Q}$ in degrees zero and one and zero elsewhere. If $x_1$ is dual to the marked generator, then $x_1^2=0$ for degree reasons, hence $H^{\ast }(K_1;\mathbb{Q})={\mathrm{\Lambda }}_{\mathbb{Q}}(x_1)$. The equivalence preserves products by [F5].

Let $j\ge 2$ and suppose the theorem holds for $K_{j-1}$. The path fibration in [F0] has contractible total space and strict loop fiber marked-homotopy-equivalent to $K_{j-1}$. The induced cohomology map is a graded-ring isomorphism by [F5], so the induction hypothesis computes the actual fiber ring used by the Serre sequence.

Put $A=H^{\ast }(K_j;\mathbb{Q})$. Since $K_j$ is simply connected, the fiber coefficient system is constant. Each nonzero graded fiber group is one-dimensional by induction, so the constant-coefficient comparison is literal scalar multiplication and [F6] gives a bigraded algebra $E_2\cong A{\otimes }_{\mathbb{Q}}{H}^{\ast }(K_{j-1};\mathbb{Q})\mathrm{.}$ By [F3], $A^0=\mathbb{Q}$, $A^p=0$ for $0<p<j$, and $A^j\cong \mathbb{Q}$. By [F6, F7], $E_{\mathrm{\infty }}$ is $\mathbb{Q}$ at $(0,0)$ and zero in every positive total degree.

Suppose $j$ is even. The induction hypothesis gives $H^{\ast }(K_{j-1};\mathbb{Q})=\mathrm{\Lambda }(y)$ with $\mathrm{\mid }y\mathrm{\mid }=j-1$, so only the rows $0$ and $j-1$ occur. The sole possible nonzero differential is $d_j:E_j^{p,j-1}=A^{p}y⟶E_j^{p+j,0}=A^{p+j}\mathrm{.}$ It sends $y$ to a nonzero element $x_j\in A^j$: otherwise $y$ would survive in positive total degree. The upper row has no incoming differential, while the positive-degree bottom row has no outgoing differential and only this possible incoming one. Vanishing of the positive-degree abutment therefore says that $A^p⟶A^{p+j},a⟼(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}a{x}_j,$ is an isomorphism for every $p\ge 0$. Starting with $A^0=\mathbb{Q}$ and the vanishing in degrees $1,\dots ,j-1$, induction over residue classes modulo $j$ gives $A=\mathbb{Q}[x_j]$. Since $j$ is even, all powers have the required graded-commutative sign.

Suppose $j$ is odd. Now the induction hypothesis gives $H^{\ast }(K_{j-1};\mathbb{Q})=\mathbb{Q}[y]$ with $\mathrm{\mid }y\mathrm{\mid }=j-1$ even. Before page $j$ no differential can join two occupied fiber rows. The class $y$ must die, and its only possible first differential is $d_j(y)=x_j\in A^j,$ which is nonzero and hence generates the one-dimensional group supplied by [F3]. The Leibniz rule gives $d_j(y^k)=k{y}^{k-1}{x}_j(k\ge 1)\mathrm{.}$ Because $j$ is odd, graded commutativity and rational coefficients give $x_j^2=-x_j^2$ and hence $x_j^2=0$.

Assume for contradiction that $A^p\mathrm{\ne }0$ for some $p>j$, and choose the least such $p$. A nonzero bottom-row class $a\in A^p$ cannot be hit by $d_j$: every possible source has base degree $p-j<p$, hence is zero by minimality unless $p-j=j$, where it is a multiple of $x_{j}y$ and $d_j(x_{j}y)=-x_j^2=0$. For a later differential $d_r$ to hit $a$, its source fiber degree $r-1$ must equal $k(j-1)$. If its smaller base degree is neither $0$ nor $j$, minimality makes the source zero. In base degree $0$, every $y^k$ has already been killed by the injective map $d_j(y^k)=k{y}^{k-1}{x}_j$; in base degree $j$, every $x_j{y}^k$ is the boundary $d_j(y^{k+1})\mathrm{/}(k+1)$. Thus no later differential hits $a$. No differential leaves the bottom row, so $a$ survives to $E_{\mathrm{\infty }}$, contradicting step 2.1. Therefore $A^p=0$ for $p>j$, and $A={\mathrm{\Lambda }}_{\mathbb{Q}}(x_j)$.

The Hurewicz isomorphism followed by rational evaluation defines $x_j$ as the class dual to the marked generator. A marked homotopy equivalence commutes with Hurewicz and evaluation and preserves cup products, so [F1, F3, F5] give the stated naturality. The only alternative generator of $\mathbb{Z}$ is its negative, which changes the dual class by $-1$. For $j=1$, $j=2$, the zero class, the unit, the first powers $y^0$ and $y^1$, and either parity, the preceding computations remain literal. The spaces and path fibers are nonempty; zero groups occur in the displayed vanishing ranges. All incoming and outgoing differential possibilities, both rows in the even case, every occupied row in the odd case, and both algebra-identification directions were checked. AC is used through [F0], [F1], [F3], [F6], [F7], and [F8]; all Koszul calculations are finite and add no choice. There is no converse assertion.