Splitting principle confusion

I'm confused about something related to the splitting principle. It seems to imply that the Chern classes of any endomorphism bundle are zero, which is not true.

Let $M$ be a manifold of dimension $n$, and $E \to M$ be a smooth complex vector bundle of rank $r$. There exists some space $M'$ and a map $f : M' \to M$ such that $f^*E$ splits as a direct sum of complex line bundles $L_j$, and such that $f^* : H^*(M, \mathbb R) \to H^*(M', \mathbb R)$ is injective.

If $E' = \bigoplus_{j} L_j$, then $\operatorname{End} E'$ is flat: Equip each $L_j$ with some hermitian metric $h_j$, and denote its curvature form by $\omega_j$. The curvature form $\Theta$ of the induced metric on the direct sum is then a diagonal matrix with entries $\omega_j$. But because $\Theta$ is diagonal, it is equal to its transpose, so $$\Theta_{\operatorname{End} E'} = \operatorname{id}_{(E')^*} \otimes\, \Theta - \Theta^{t} \otimes \operatorname{id}_{E'} = 0.$$

Thus all the Chern roots of $\operatorname{End} f^* E \cong \operatorname{End} E'$ are zero, so all its Chern classes are zero. But as the Chern classes are functorial and $f^*$ injective, all the Chern classes of $\operatorname{End} E$ are then zero as well.

Where have I gone wrong here?

• Isn't $f^*E = E'$ instead of $\text{End} f^*E = E'$? – user99914 May 18 '17 at 6:47
• Yes, sorry for the typo. – Gunnar Þór Magnússon May 18 '17 at 7:00
• This may be a dumb comment, but do those tensor products actually commute? – user347489 May 18 '17 at 7:56
• @user347489 NO! Man, I feel silly. – Gunnar Þór Magnússon May 18 '17 at 8:13
• @GunnarÞórMagnússon it happens to the best of us :P – user347489 May 19 '17 at 8:04