Understanding about definition of connection

Question

I'm trying to understand about definition of connection. Before give definition, following are some notation:

Let $\pi : E \rightarrow M$ be a vector bundle ($E, M$ are smooth manifolds). Set $\mathfrak{X}(M)$ is set of vector fields on $M$, $\Gamma(E)$: set of sections on $E$, $\Omega^k(E)$: $k$-forms on E.

Definition: A connection on $E$ is an $\mathbb{R}$-bilinear map

$$\nabla :\mathfrak{X}(M) \times \Gamma(E) \rightarrow \Gamma(E) \hspace{1cm} (X, s) \mapsto \nabla_Xs$$

such that for any $f \in C^{\infty}(M)$, we have

i) $\nabla_{fX}s=f\nabla_Xs$

ii) $\nabla_X(fs) = (Xf)s + f\nabla_Xs$.

Remark: Equivalently, $\nabla: \Omega^0(E)=\Gamma(E) \rightarrow \Omega^1(E)$ with $s \mapsto \nabla s$ such that

$$\nabla(fs)=df\otimes s + f\nabla s.$$

My questions:

1. Why we have equivalence above?

2. Look at the condition ii) in Definition, I confus that if $fs \in \Gamma(E)$?

Thank you in advance!

Answer 1

1. The equivalence is just like this: If you have a map $\nabla:\mathfrak X(M)\times \Gamma(E)\to \Gamma(E)$ you can get a map $\nabla':\Gamma(E)\to \Omega^1(E)$ given by declaring $\nabla'(s)(X)=\nabla_Xs.$ The same formula allows you to go from $\nabla'$ to $\nabla.$ With this equality, the conditions i and ii are just restatements of the conditions that $\nabla'(fs)=df\otimes s+f\nabla' s$ and the fact that differential forms in $\Omega^1(E)$ are linear over the $C^\infty$ functions on $M.$
2. $fs\in \Gamma(E)$ since at each point $x\in M$ we have $(fs)(x)=f(x)s(x)$ and since $f(x)\in \mathbb R,$ $s(x)\in E_x,$ and $E_x$ is a vector space, the product $f(x)s(x)$ makes sense and still lies in $E_x.$