This article says that for an Ehresmann connection $H \subseteq TE$ of a vector bundle $E \to M$ to be linear (and so to define a connection in the usual sense as a linear covariant derivative operator) all you need is

$$D(S_\lambda)_e(H_e) = H_{\lambda e}$$ for $\lambda \in \mathbb{R}$ and $e \in E$, where $S_\lambda : E \to E$ is the multiplication by $\lambda$ map. But to me it seems like we also need $$D\sigma_{(e,e')} (H_e \oplus H_e') = H_{e + e'}$$ where $\sigma : E \oplus E \to E$ is the addition map and $e,e'$ lie on the same fiber. Can this somehow be deduced from the previous property, or is the claim in the article mistaken?

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    $\begingroup$ It sure looks like you're correct. We certainly need the sum of parallel sections to be parallel. $\endgroup$ Mar 9 '21 at 17:35
  • $\begingroup$ Actually it seems that the second property can indeed be deduced from the first, by the argument in this answer to what was basically the same question here math.stackexchange.com/a/2659492/9921 $\endgroup$
    – Pedro
    Mar 10 '21 at 3:24
  • $\begingroup$ More specifically, the first property is enough to show (via the usual Leibniz rule trick) that the covariant derivative of a section at a point only depends on the 1-jet of that section, and that this map $(J^1 E)_p \to E_p$ given by covariant derivative along a fixed $X$ is homogeneous; by the argument in that answer it is linear, which is equivalent to the second property. $\endgroup$
    – Pedro
    Mar 10 '21 at 3:27

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