Chern character is a natural ring homomorphism on K-zero

Work componentwise. A finite CW complex has finitely many path components, and cohomology, bundle isomorphism classes, Whitney sums and tensor products all decompose over this finite disjoint union. On each component where a bundle has positive rank, the splitting principle [F2] applies; on a rank-zero component the character is zero by [F1]. For a pullback $f^{\ast }E$, apply this observation on each source component: the flag bundle of the positive-rank restriction is the pullback of the corresponding flag bundle of $E$, and the roots pull back, so $q^{\ast }{f}^{\ast }{\mathrm{ch}}_k(E)=\frac{1}{k!}{\sum }_i(f^{\ast }{t}_i{)}^k$. The rational Leray--Hirsch injectivity recorded in [F2], not a coefficient extension of the integral claim, and [F1] give ${\mathrm{ch}}_k(f^{\ast }E)=f^{\ast }{\mathrm{ch}}_k(E)$ on every component.

Additivity: on each path component, omit any rank-zero summand and use [F2] to choose a common splitting space for the positive-rank restrictions, with $q^{\ast }E={\oplus }_i{L}_i$ and $q^{\ast }F={\oplus }_j{M}_j$. Then $q^{\ast }(E\oplus F)$ has the combined roots, so $q^{\ast }{\mathrm{ch}}_k(E\oplus F)=\frac{1}{k!}({\sum }_i{t}_i^k+{\sum }_j{u}_j^k)=q^{\ast }({\mathrm{ch}}_k(E)+{\mathrm{ch}}_k(F))$; injectivity gives additivity on that component, and hence on $X$.

Multiplicativity: on a component where both bundles have positive rank, use the common splitting of step 2.1. The tensor product splits as ${\oplus }_{i,j}{L}_i\otimes M_j$ by [F4], and its roots are $t_i+u_j$ by [F3], so $q^{\ast }{\mathrm{ch}}_k(E\otimes F)=\frac{1}{k!}{\sum }_{i,j}(t_i+u_j{)}^k={\sum }_{a+b=k}(\frac{1}{a!}{\sum }_i{t}_i^a\left)\right(\frac{1}{b!}{\sum }_j{u}_j^b)=q^{\ast }{\sum }_{a+b=k}{\mathrm{ch}}_a(E){\mathrm{ch}}_b(F)$. The middle identity is the binomial theorem. Injectivity gives multiplicativity. If either rank is zero, both sides vanish by [F1], so the result holds on every component.

Extension to $K^0$. By step 2.1 the map $E\mapsto \mathrm{ch}(E)$ is an additive monoid homomorphism from ${\mathrm{Vect}}_{\mathbb{C}}(X)$ to the additive group $H^{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}(X;\mathbb{Q})$; by universality of the Grothendieck completion [F5] it extends uniquely to a group homomorphism $K^0(X)\to H^{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}(X;\mathbb{Q})$, still written $\mathrm{ch}$, with $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$.

Since $K^0(X)$ is generated as an abelian group by bundle classes, multiplicativity on generators from step 3.1 extends: for representatives $a=[E]-[F]$, $b=[E^{\mathrm{\prime }}]-[F^{\mathrm{\prime }}]$ the product in $K^0(X)$ is $[E\otimes E^{\mathrm{\prime }}\oplus F\otimes F^{\mathrm{\prime }}]-[E\otimes F^{\mathrm{\prime }}\oplus F\otimes E^{\mathrm{\prime }}]$ by [F5], and applying additivity (step 2.1) and multiplicativity (step 3.1) to the four summands gives $\mathrm{ch}(ab)=\mathrm{ch}(a)\mathrm{ch}(b)$. The unit is $[{\epsilon }^1]$ and $\mathrm{ch}({\epsilon }^1)=1$ by [F1], so $\mathrm{ch}$ is a unital ring homomorphism.

External products. For finite CW complexes $X,Y$ and bundle classes $a=[E]\in K^0(X)$, $b=[F]\in K^0(Y)$, the external product is $a\times b={\mathrm{pr}}_X^{\ast }a\cdot {\mathrm{pr}}_Y^{\ast }b$ in $K^0(X\times Y)$; by steps 1.1 and 4.1, $\mathrm{ch}(a\times b)={\mathrm{pr}}_X^{\ast }\mathrm{ch}(a)\cdot {\mathrm{pr}}_Y^{\ast }\mathrm{ch}(b)=\mathrm{ch}(a)\times \mathrm{ch}(b)$, which is the stated external formula.

Boundary cases. For the trivial bundle ${\epsilon }^n$ one has $\mathrm{ch}({\epsilon }^n)=n$ in degree zero, matching the rank; for the zero bundle $\mathrm{ch}(0)=0$. The trivial group $K^0(\mathrm{\varnothing })$ is allowed and the homomorphism is the zero map. The coefficient field $\mathbb{Q}$ is nonzero and contains $1\mathrm{/}k!$ for every $k$, which is why the rational coefficients are required; over $\mathbb{Z}$ the character is not defined in general. AC enters only through [A1] in the splitting and K-theory suppliers.