 Solution

Use the homotopy lifting theorem for a covering map: a homotopy of paths has a unique lift once its initial path is lifted. Also use uniqueness of lifts to a covering space: a lift of a constant path with a specified starting point is constant.

Let $\tilde{\gamma }$ be a loop based at ${\tilde{x}}_0$ whose image under $p_{\ast }$ is trivial. There is a homotopy $F:I\times I\to X$ from $p\tilde{\gamma }$ to the constant loop at $x_0$, fixing both endpoints. Lift this homotopy with initial path $\tilde{\gamma }$. Both vertical boundary paths are constant lifts at ${\tilde{x}}_0$. Its final path is a lift of the constant loop and hence is also constant at ${\tilde{x}}_0$. Thus the lifted homotopy contracts $\tilde{\gamma }$ relative to its endpoints. The kernel of the induced group homomorphism is trivial, provingNo universal-cover assumption is needed for this injection; it holds for any actual covering map.