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You need to write $\omega X$ as $\omega \otimes X$.

If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of ${\Omega} ^2(M,End(TM))$. (That is even more elementary result - see any introductory book on differential geometry.)

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.

You need to write $\omega X$ as $\omega \otimes X$.

If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of ${\Omega} ^2(M,End(TM))$, because then the value of $R(X,Y)S$ at a point $m\in M$ depends only on $X_m,Y_m$ and $S_m$ and not on their derivatives (even though we have defined $R$ via differentiation). The desired identification then follows from linear algebra $Hom(\Lambda^2 V \otimes E,E) \simeq Hom(\Lambda^2 V, \otimes E^* \otimes E)$ & $End(E) \simeq E^*\otimes E$.

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.

You need to write $\omega X$ as $\omega \otimes X$.

If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of $\Omega^2(M,End(TM))$. (That is even more elementary result - see any introductory book on differential geometry.)

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.

You need to write $\omega X$ as $\omega \otimes X$.

If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of ${\Omega} ^2(M,End(TM))$. (That is even more elementary result - see any introductory book on differential geometry.)

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.

You need to write $d\omega X$ as $d\omega \otimes X$, where $X$ is a section of the vector bundle $E$. For the curvature - you need to prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial.

You need to write $\omega X$ as $\omega \otimes X$.

If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of $\Omega^2(M,End(TM))$. (That is even more elementary result - see any introductory book on differential geometry.)

The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.