Suppose we have definied chern classes $c_i$ for line bundles for all $i\geq 0$.

Let $E\to B$ be a rank $r$ vector bundle. By the splitting principle there exists a map $f:Y\to B$ such that $f^*E$ is a sum of line bundles $L_1\oplus\ldots\oplus L_r$ and $f^*:H^*(B)\to H^*(Y)$ is injective.

Does the splitting principle define chern classes for vector bundles if they are known for line bundles?

The splitting principle says only, that if the chern classes are defined in two ways (natural with respect to continuous maps), one can check on line bundles if the two ways end up to be the same. I don't see how the splitting principle could be used to define the classes because I fear that the decomposition into line bundles (or, the space $Y$ such that $f^*E$ splits) is not unique.

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    $\begingroup$ You can define the "Chern classes with respect to a splitting" this way and then prove that it doesn't depend on the choice of splitting. This is the same as any mathematical situation where you define a function on a set of equivalence classes by defining it on elements and then proving that it respects the equivalence relation. $\endgroup$ – Qiaochu Yuan May 22 '13 at 19:01
  • $\begingroup$ Ok, so you are saying this actually is possible but the splitting principle doesn't help and I need other arguments, right? What are those other arguments in this case? $\endgroup$ – Ronald Bernard May 22 '13 at 19:03

To answer the question in the title: yes. You are really asking if the Chern classes defined via a splitting map are independent of the choice of splitting map $f$. First note that there exists a splitting map $f$ making the cohomology $H^*(Y)$ a free module over the cohomology of the base $H^*(B)$---I believe this follows from the usual construction of a splitting map via flags of subbundles.

Given another splitting map $f': Y' \rightarrow B$, form the fiber product $Y' \times_B Y$, with the canonical map to $B$ (induced by $f'$ or $f$). The map to $B$ from this fiber product corresponds to the map $R \rightarrow S' \otimes_R S$ on cohomology, where $R=H^*(B)$, $S'=H^*(Y')$ and $S=H^*(Y)$, and is hence still injective (here we use the Eilenberg-Moore spectral sequence, which actually degenerates immediately by the freeness of $H^*(Y)$ over $H^*(B)$). Inside $H^*(Y' \times_B Y)=S' \otimes_R S$ the Chern classes become equal, proving independence.


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