The first connective complex K-theory Postnikov invariant is integral Sq-three

We first locate the space classes correctly. For a connective Omega-spectrum, [F4] identifies ${\pi }_i(k{u}_m)$ with ${\pi }_{i-m}(ku)$ for $i\ge 1$: each adjoint map is a weak equivalence, so the bonding maps in that stable colimit are isomorphisms. For $m\ge 2$ the space is connected (its components identify with ${\pi }_1(k{u}_{m+1})={\pi }_{-m}(ku)=0$), and it is simply connected by connectivity. Its first groups are ${\pi }_m=\mathbb{Z}$, ${\pi }_{m+1}=0$ and ${\pi }_{m+2}=\mathbb{Z}t$. Consequently the lower space stage $P_{m+1}k{u}_m$ is a $K(\mathbb{Z},m)$ model. Presenting the next stage as the marked fibration of [F1], its class lies in $H^{m+3}(K(\mathbb{Z},m);\mathbb{Z})$. This uses [F1] only for connected simple spaces, with $n=m+2$, never for a spectrum or a degree-zero space stage.

By [F5] the classes of step 1.1 are evaluations of a single stable operation $\kappa$ and its group has just two elements: $0$ and $\theta ={\beta }_{\mathbb{Z}}S{q}^2{\rho }_2$. The target types are integral to mod-two, then mod-two, then integral, respectively. [F3] supplies these local conventions; [F5], rather than [F2], supplies the stable group calculation and compatibility. In particular we are not inferring an operation group for an Eilenberg–Mac Lane spectrum from the theorem for one space $K(A,m)$.

Suppose $\kappa =0$. Its evaluation at the third space is then zero. Here $Y=SU$, and step 1.1 gives $P_{4}Y=K(\mathbb{Z},3)$ up to the fixed stage equivalence. The fibration $P_{5}Y\to P_{4}Y$ has fiber $K(\mathbb{Z},5)$ and zero class. By [F1] it is fiber homotopy equivalent to the homotopy fiber of the constant map $P_{4}Y\to K(\mathbb{Z},6)$, which has a section given by the constant loop. Hence $H^6(P_{4}Y;\mathbb{Z})\to H^6(P_{5}Y;\mathbb{Z})$ is injective, since section pullback is its left inverse.

Choose the CW Postnikov model in [F1] for $Y=SU$. The pair $(P_{5}Y,Y)$ has only relative cells of dimension at least seven. Its cellular cochain group in degree six is zero, so $H^6(P_{5}Y,Y;\mathbb{Z})=0$ by [F6]. The pair sequence therefore makes $H^6(P_{5}Y;\mathbb{Z})\to H^6(Y;\mathbb{Z})$ injective. Composing with step 3.1 would inject the nonzero class $\theta ({\iota }_3)$ of [F5] into $H^6(SU;\mathbb{Z})=0$, a contradiction. Equivalences of stage models transport these maps and do not affect injectivity.

Thus $\kappa \mathrm{\ne }0$, and the two-element alternative of step 2.1 forces $\kappa =\theta$. Its evaluation gives the stated universal class in each space by [F5]. The two identifications of an integral coefficient generator differ by sign; that sign cannot change an element of an order-two group, since $-\theta =\theta$. Hence the fixed Bott identifications preserve this normalization. In low degrees evaluations may be zero: for instance $S{q}^2$ is zero on degree-zero and degree-one classes by [F3]; this illustrates why nonzero stable operation does not mean nonzero on every space. The nonvanishing used in the proof is specifically the degree-three universal class.