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Feb 3, 2014 at 8:39 comment added Ben McKay Sorry, I should have indicated more clearly: if $\sigma(p)=0$ we can define $d\sigma(p) \in T^*_p M \otimes E_p$, rather than in $T^*_p M \otimes T_{\sigma(p)} E$.
Feb 3, 2014 at 7:43 comment added user45859 When $\sigma\in\Gamma(E)$ and $\sigma(p)\neq 0$, why can't we define $d\sigma_p$? Since $\sigma$ is a smooth map from $M$ to $E$, $d\sigma_p$ should be a linear map from $T_pM$ to $T_{\sigma(p)}E$.
Feb 3, 2014 at 7:42 comment added abx I don't understand the last comment: the differential of $\sigma:M\rightarrow E$ is well-defined as $d\sigma (p): T_p(M)\rightarrow T_{\sigma (p)}(E)$.
Feb 3, 2014 at 7:32 history answered Ben McKay CC BY-SA 3.0