 Solution

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.