Doubt about splitting principle

Question

I have a doubt about splitting principle.

We know that if $E\to M$ is a complex vector bundle of rank $k$, then there exist a manifold $T=T(E)$ and a proper smooth map $f:T\to M$ such that

1. $f^*:H^*(M)\to H^*(T)$ is injective

2. $f^* E\cong L_1\oplus\dots\oplus L_k$, where $L_j\to M$ is a line bundle, $j=1,\dots,k.$

We can use the splitting principle to show, for example, that $ch(E\otimes F)=ch(E) ch(F)$, given two complex vector bundles. Let suppose that it exists a map $\tilde f$ that splits both $E$ and $F$ and whose pullback is an injection:

$$\tilde f^*ch(E\otimes F)=ch(\tilde f^*(E\otimes F))=ch(\tilde f^*E\otimes f^*F)=ch(\tilde f^*E)ch(\tilde f^*F)=\tilde f^*(ch(E)ch(F))$$

and since $\tilde f$ is an injection we have proved the identity. But how can I find the map $\tilde f$.

I have the idea we have two distinct maps for $E$ and for $F$.

Thank you!

Answer 1

You are correct in noting that if $f : T(E) \to M$ is the map for which $f^*E$ splits, there is no reason to expect that $f^*F$ splits. For example, if $E$ is a line bundle then $T(E) = M$ and $f = \operatorname{id}_M$, but $\operatorname{id}_M^*F = F$ which need not be a sum of line bundles. However, there is a simple way to resolve this.

Note that $f^*F \to T(E)$ is a complex vector bundle, so there is a map $g : T(f^*F) \to T(E)$ such that $g^*(f^*F)$ splits and $g^*$ is injective on cohomology. Letting $h = f\circ g : T(f^*E) \to M$, we see that $h^*E$ and $h^*F$ split as a sum of line bundles, and since $f^*$ and $g^*$ are injective of cohomology, so is $h^* = g^*f^*$. Now you can show that $h^*\operatorname{ch}(E\otimes F) = h^*(\operatorname{ch}(E)\operatorname{ch}(F))$ and hence $\operatorname{ch}(E\otimes F) = \operatorname{ch}(E)\operatorname{ch}(F)$.