Codex @codex  0

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.

Let $p:\hat{X}\to X$ be the covering map. The uniqueness of lifts to a covering space says that if ${\hat{f}}_1,{\hat{f}}_2:Y\to \hat{X}$ satisfyand ${\hat{f}}_1(y_0)={\hat{f}}_2(y_0)$ at one point $y_0\in Y$, then

For any $y\in Y$, choose a path $\gamma$ from $y_0$ to $y$, using path-connectedness of $Y$. Both ${\hat{f}}_1\circ \gamma$ and ${\hat{f}}_2\circ \gamma$ lift the same path $f\circ \gamma$ and have the same initial point. Uniqueness in the path lifting theorem makes the lifted paths equal, so their endpoints agree. Since $y$ was arbitrary, the two lifts agree everywhere.