I am trying to understand a proof from "Lectures on Kähler Geometry" by Andrei Moroianu. I still have lots of trouble working with covariant derivatives. Lemma 5.2 states the following:
Let $\pi: E \longrightarrow M$ be a vector bundle.
For every $X_p \in T_pM$ and $\sigma_0 \in E_p$ there exists a local section $\sigma$ of $E$ which is parallel in the direction of $X$ at $p$ and satisfies $\sigma(p) = \sigma_0$.
Proof:
Let $(x_i)$ be local coordinates on some neighbourhood $U$ of $p$.
To prove this he takes some arbitrary local section $\tau$ of $E$ around $p$ such that $\tau(p) = \sigma_0$. Then he writes $X_p = \sum a_i \frac{\partial}{\partial x^i}|_p$ for some real numbers $a_i$ and $(\nabla_X\tau)(p) = \sum b_j \sigma_j(p)$ for some real numbers $b_j$ and frame $(\sigma_j)$ of E.
He then claims that $\sigma := \tau - \frac{x_1}{a_1}\sum b_j \sigma_j$ (assuming that $a_1 \neq 0$) satisfies $\sigma(p)=\sigma_0$ and $(\nabla_X \sigma)(p) = 0$ which concludes the proof.
I am unable to show that $\sigma(p)=\sigma_0$ and $(\nabla_X \sigma)(p) = 0$.
I tried this:
$\sigma(p) = \tau(p) - \frac{x_1(p)}{a_1}\sum b_j \sigma_j(p)=\sigma_0 - \frac{x_1(p)}{a_1}\sum b_j \sigma_j(p)$. Because $\sigma_j(p)$ are linearly independent and $b_j \neq 0$ at least for one $j$, $x_1(p)$ must be zero? But I don't see how this is the case as we just assume that $(x_i)$ is some arbitrary system of coordinates around $p$.
For the second claim I am completely lost. I tried working with the Leibniz-Rule to calculate the covariant derivative of $\sigma$ but to no end.
We need to choose coordinates to suffice $(x_1, ..., x_n)(p) = 0$.
Then it follows that:
$\sigma(p) = \tau(p) - \frac{x_1(p)}{a_1} \sum b_j\sigma_j(p) = \tau(p)$
$(\nabla_X \sigma)(p) = (\nabla_X \tau)(p) - \partial_{X_p} (\frac{x_1}{a_1})(p) \sum b_j \sigma_j(p) - \frac{x_1(p)}{a_1} (\nabla_X (\sum b_j \sigma_j))(p) = (\nabla_X \tau)(p)(1 - \frac{1}{a_1} \sum a_i \frac{\partial x_1}{\partial x^i}|_p) = 0$