Artificial intelligent assistant

Covariant derivative as a connection on a vector bundle In the Wikipedia article Connexion (vector bundle), such a connection is defined as a function $\Gamma(E) \to \Gamma(E\otimes T^*M)$ . Then the definition of a covariant derivative is given as a function $\Gamma(E) \to \Gamma(E)$. But the article proceeds by saying that a covariant derivative is also a connection. * How could a covariant derivative be a connection since the codomains of those two kinds of objects appear to be different ? * Are those two different definitions of a connection ? Some authors would use the $\Gamma(E) \to \Gamma(E\otimes T^*M)$ version, while others would use $\Gamma(E) \to \Gamma(E)$ ?

You differentiate in the direction of some vector (field). Feeding in the vector field is going to remove the $\otimes T^*M$ :)

**EDIT** : To be concrete, given the map $D\colon\Gamma(E)\to\Gamma(E\otimes T^*M)$, a section $s$, and a vector field $X$ on $M$, we have $Ds \in\Gamma(E\otimes T^*M)$ with $(Ds)(X) = \
abla_X s$. This is the "classic" covariant derivative in the direction of $X$. In coordinates, if $s_1,\dots, s_r$ give a (local) basis for sections of $E$ and $\omega^1,\dots,\omega^n$ give a (local) basis for sections of $T^*M$, $D$ is defined by $$D(s_i) = \sum_{\substack{1\le j\le r\\\ 1\le k\le n}} a_{ik}^j s_j\otimes\omega^k \quad\text{for some functions } a_{ik}^j.$$ Then the covariant derivative viewpoint is given by $$\
abla_X(s_i) = \sum_{j,k} a_{ik}^j \omega^k(X)s_j.$$

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