For a vector bundle $E\to X$ with a given connection $\nabla$. We say that a section $s$ of $E$ is parallel to a vector space $V$ if $\nabla_V s=0$. If $\gamma:[0,1]\to X$ is a smooth path, we say that $s$ is a parallel transport of $v$ along $\gamma$ provided that $s(\gamma(0))=v$ and $\nabla_{\dot{\gamma}}s=0$ (to be precise, one considers an extension of $\dot{\gamma}$ to a local vector field).
When $E$ has a metric and $\nabla$ is compatible with this metric, then the metric compatibility definition implies that parallel transport defines an isometry (abusing notation a bit, metric compatibility gives $d/dt\langle s_1,s_2 \rangle=\langle \nabla_{t}s_1,s_2\rangle + \langle s_1, \nabla_{t}s_2\rangle = 0$ along $\gamma$ for parallel sections).
But if we do not have a metric. Can we prove that parallel transport is an injective map?
