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If $u\in {\tilde{K}}^0(S^{2m})$ is a Bott element generator, then

Multiplication by the Bott element $\beta \in {\tilde{K}}^0(S^2)$ gives the Bott isomorphism

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

For a complex vector bundle $E$, define the kth Adams operation by the kth Newton polynomial in its exterior powers:After applying the splitting principle for complex vector bundles and writing $E=L_1\oplus \cdots \oplus L_r$, the Newton identities giveThis formula proves additivity. Applied simultaneously to splittings of $E$ and $F$, it also givesThus ${\psi }^k$ extends to a natural ring endomorphism of $K^0(X)$ and, for a line bundle,

Write $X={\mathbb{CP}}_k^{k+2}$ and let $x=[\overline{\gamma }]-1$. Its reduced K-theory is freely generated by $x^k,x^{k+1},x^{k+2}$, with $x^{k+3}=0$, and restriction to the bottom cell sends $x^k$ to a Bott element $u\in {\tilde{K}}^0(S^{2k})$ and the two higher powers to zero. If $r:X\to S^{2k}$ retracts the bottom-cell inclusion, thenfor some integers $a,b$. Naturality and the Adams operation on a Bott class giveSince ${\psi }^2(x)=2x+x^2$, comparison modulo $x^{k+3}$ first in degree $k+1$ and then in degree $k+2$ givesThe first equality makes $k$ even. Then $3k+5$ is odd, so integrality of $b$ forces $8\mid k$; it is also congruent to $2$ modulo $3$, so $3\mid k$. Consequently

The Bott isomorphism is multiplication by the Bott element $\beta \in {\tilde{K}}^0(S^2)$:Together with the suspension isomorphism andit gives the Complex K-theory of a sphere

Letbe a CW filtration in which each quotient $Y_r/Y_{r-1}$ is a wedge of even-dimensional spheres. The six-term exact sequence in Topological K-theory, the sphere calculation, and induction giveand a short exact sequence whose new summand in $K^0(Y_r)$ is free abelian on the newly attached cells. Every such extension splits as an extension of free abelian groups, so $K^0(Y)$ is free, with one generator for each cell. This proves the Complex K-theory of an even-cell complex result.

The exterior product definesFor a point it is the identity. Attaching one layer of even cells gives corresponding exact sequences on the source and target; the sphere case is the suspension isomorphism, and induction with the Five lemma proves that the product map remains an isomorphism. This is the Künneth theorem for complex K-theory with an even-cell factor.

For a mapping torus $T_f$, the K-theory Wang sequence of a mapping torus containsWhen $K^{-1}(Z)=0$, exactness gives

For $Z={\mathbb{CP}}^2\times {\mathbb{CP}}^2$, the Complex K-theory of complex projective space and the K-theory Künneth isomorphism giveThe factor swap interchanges $x$ and $y$. Its invariant subgroup has the basisIt follows that the K-theory of the mapping torus of the factor swap on two complex projective planes is

The splitting principle for complex vector bundles says that for every complex vector bundle $E\to X$ there is a map $p:F(E)\to X$ such that $p^{\ast }$ is injective on cohomology andsplits into complex line bundles. Write $x_i=c_1(L_i)$ for the formal Chern roots.

Define the Chern character after this injective pullback byEach homogeneous component is a symmetric polynomial in the $x_i$ with rational coefficients, hence a polynomial in the elementary symmetric functions $c_j(E)$. It therefore descends uniquely to $H^{\mathrm{ev}}(X;\mathbb{Q})$ and depends only on $E$. Seton the Grothendieck group $K^0(X)$; additivity under direct sums makes this well defined.

If $E$ has roots $x_i$ and $F$ has roots $y_j$, then $E\otimes F$ has roots $x_i+y_j$. ConsequentlyIt also sends the trivial line to $1$, so it is a unital ring homomorphism.

For $S^{2n}$, a generator of ${\tilde{K}}^0(S^{2n})$ is the $n$-fold exterior product of the degree-two Bott element. The Chern character respects exterior products, and the degree-two Bott element has Chern character equal, up to sign, to the integral generator of ${\tilde{H}}^2(S^2;\mathbb{Z})$. Its $n$-fold product maps to the integral top-dimensional generator. Hence the Chern character on an even-dimensional sphere is integral.

Let the formal Chern roots of $E\to S^{2n}$ be $x_1,\dots ,x_r$ and write $p_n={\sum }_i{x}_i^n$. Sinceall lower Chern classes $c_1(E),\dots ,c_{n-1}(E)$ vanish. The Newton identities then reduce toThe degree-$2n$ term of the Chern character is thereforeIts evaluation on the fundamental class is an integer by integrality of the reduced Chern character. Thusis divisible by $(n-1)!$, proving the Divisibility of the top Chern number on an even-dimensional sphere.