It suffices to check the skew-symmetry of a point-operator locally

Question

Let $M$ be a smooth manifold $E,F$ be $\mathcal{C}^{\infty}$ vector-bundles over M.

An $\mathbb{R}$-linear map $\alpha:\Gamma{(E)} \to \Gamma{(F)}$ is defined to be a pointwise operator iff whenever for some $s \in \Gamma{(E)}$ and $p \in M$ $s(p) = 0$ then $\alpha(s)(p) = 0$.

This is the Definition given in the book "Differential Geometry" by Loring W.Tu on page $53$.

Then in Prop.12.5 the author introduces a function $\Phi:\mathfrak{X}(M) \times \mathfrak{X}(M) \times \mathfrak{X}(M) \times \mathfrak{X}(M) \to \mathcal{C}^{\infty}$ , $(X,Y,Z,W) \mapsto \langle R(X,Y)Z, W \rangle$.

where $R$ denotes the curvature operator. And claims that this function is skew-symmetric in $Z$ and $W$.

Now my question is: Why does it suffice to show this only locally, i.e. on some open set $U \subset M$?

Answer 1

You are misunderstanding what it means to check something locally. When someone states that e.g. continuity is a local property, it means that a function $f:X\to Y$ is continuous if and only if it is continuous on an open neighbourhood $U$ around every point $x\in X$. The same is true here. It does not mean that you can check the corresponding property on a single open subset.