Let $Y_1$, $Y_2$ be universal covering spaces of some topological space $X$. I want to show that $Y_1$ are $Y_2$ are isomorphic.
Denote $p_1 \colon Y_1 \to X$, $p_2 \colon Y_2 \to X$ the projections. Then there exist maps $f_1 \colon Y_1 \to Y_2$ and $f_2 \colon Y_2 \to Y_1$ such that $p_2 f_1 = p_1$, $p_1 f_2 = p_2$. Taking compositions we obtain $$ p_1 f_2 f_1 = p_1, \\ p_2 f_1 f_2 = p_2. $$ We have to show that $f_2 f_1 = \mathrm{id}_{Y_1}$ and $f_1 f_2 = \mathrm{id}_{Y_2}$. Suppose that this doesn't hold, i.e. there exists $y_1 \in Y_1$ such that $f_2 f_1 (y_1) = \widetilde y_1 \neq y_1$. This implies $p_1(\widetilde y_1) = p_1(y_1)$. But I don't see a contradiction. Please help me to finish the proof.
Considering the spaces with base points $(Y_1, y_1)$ and $(Y_2,y_2)$, notice that $f_2f_1$ is a lift of $p_1$ to the covering $Y_1$ with $f_2f_1(y_1) = y_1$. But the identity map $id_{Y_1}$ is also a lift of $p_1$ with $id(y_1)=y_1$. By uniqueness, we have $f_2f_1 = id_{Y_1}$.
The same argument shows that $f_1f_2 = id_{Y_2}$.