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Assume we have a fiber bundle $F\to E\stackrel \pi\to B$.

In the wikipedia article it is stated that $E$ the vertical bundle $V=\ker d\pi$ consisting of vectors along the fibers is canonically defined while the horizontal bundle of vectors along the base is not.
Specifying a horizintal subspace of $TE$ is then called an Ehresmann connection on $E$.

But isn't $\pi^* TB$ a canonical subbundle of $TE$ which consists of vectors along $B$?

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No! There is no reason why $\pi^*TB$ is a subbundle of $TE$ -- this is precisely why you need a connexion.

Recall $\pi^*TB$ is actually constructed as $E\times_B TB$: $$ \pi^*TB=\{(e,(p,v))\in E\times TB\mid \pi(e)=\operatorname{proj}_{TB\to B}(p,v)=p\} $$ where $\operatorname{proj}_{TB\to B}$ is the usual projection $TB\to B$.

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  • $\begingroup$ Then why is it claimed differently here? map.mpim-bonn.mpg.de/Tangent_bundles_of_bundles_(Ex) $\endgroup$ – Peter Jun 6 at 8:59
  • $\begingroup$ A connexion $\nabla$ is precisely a choice of splitting $TE=H\oplus V$ with $H\cong \pi^*TB$. You don't have a canonical choice of $H$. $\endgroup$ – user10354138 Jun 6 at 9:03
  • $\begingroup$ So independently of the connection, there always is an isomorphism $H\to \pi^* TB$? Specifying such an isomorphism does then define a unique connection? $\endgroup$ – Peter Jun 6 at 9:36
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By definition of the pullback, $\pi^*B=\{(x,y,v):y\in\pi^{-1}(x), v\in T_xB\}$ therefore it is not a subbundle of $TE$.

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