A bundle connection is local and restricts to open sets

Suppose $s$ vanishes on a neighbourhood $W$ of $p$. Take a smooth cutoff $\chi$ supported in $W$ with $\chi (p)=1$, using the construction in the proof of [F1]. Since $\chi s=0$, the Leibniz rule at any $v\in T_{p}M$ gives $0={\mathrm{\nabla }}_v(\chi s)=d{\chi }_p(v)s(p)+\chi (p){\mathrm{\nabla }}_{v}s={\mathrm{\nabla }}_{v}s$. Apply this to the difference of two sections. The direction depends only on its value by [F2]; no assertion that $d{\chi }_p=0$ was needed in this vanishing argument.

For a local section $s$ on $U$, take a cutoff supported in $U$ and equal to one on a neighbourhood $V$ of $p\in U$. Extend $\chi s$ by zero outside $U$ to a global smooth section $\tilde{s}$: outside its closed support it is identically zero, so the definitions paste smoothly. Define $({\mathrm{\nabla }}^{U}s{)}_p=(\mathrm{\nabla }\tilde{s}{)}_p$. Different extensions agree near $p$, so step 1.1 proves independence. On $V$, one fixed extension works at every point, making ${\mathrm{\nabla }}^{U}s$ smooth. Local functions can be extended by the same cutoff; the global real-linearity and Leibniz laws then imply those laws on $U$. Thus this is a connection.

Any other restriction connection is local by step 1.1 applied on $U$, and agrees on the global extension in step 2.1; hence its value on $s$ at $p$ is forced. This proves uniqueness and, by applying uniqueness twice, composition of restrictions. The empty open set has a unique zero operator. Zero sections and rank-zero bundles give zero derivatives in steps 1.1 and 2.1; rank one and a zero-dimensional base require no modification. A single point is handled with a single cutoff, and the uniquely specified values assemble without a choice of cutoffs for all points.