I am currently studying Hatcher's book on Algebraic Topology.
I would like to understand the proof of Proposition 1.34 on page 62, concerning the uniqueness of a lift of a map $f: Y\to X$, given a covering space $p:\tilde X\to X$. The Author is trying to show that some set is at the same time open and closed (and since $Y$ is connected this implies that this set is the whole Y), but I cannot follow his reasoning.
I understand the setting of the proof, but the following is not clear to me: let $\tilde U_1$ and $\tilde U_2$ be the sheets containing $\tilde f_1(y)$ and $\tilde f_2(y)$ respectively, then
If $\tilde f_1(y)\neq \tilde f_2(y)$ then $\tilde U_1\neq \tilde U_2$, hence $\tilde U_1$ and $\tilde U_2$ are disjoint and $f_1\neq f_2$ throughout the neighborhood $N$. On the other hand, if $\tilde f_1(y)=\tilde f_2(y)$ then $\tilde U_1=\tilde U_2$ so $\tilde f_1 = \tilde f_2$ on $N$ since $p \tilde f_1 = p \tilde f_2$ and $p$ is injective on $\tilde U_1=\tilde U_2$. Thus the set of points where $\tilde f_1$ and $\tilde f_2$ agree is both open and closed in Y.
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Where is the Author using that the two maps agree at a point?
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Why $\tilde f_1(y)\neq \tilde f_2(y)\implies \tilde U_1\neq \tilde U_2$ and $\tilde f_1(y)=\tilde f_2(y)\implies \tilde U_1=\tilde U_2$?
I am aware of the fact that other questions on this particular proof have been asked, but they are still obscure to me.
Best Answer
In the proof Hatcher shows that the set $A$ of points of $Y$ where $\tilde f_1$ and $\tilde f_2$ agree is is both open and closed in $Y$. Thus, if $Y$ is connected, then either $A = \emptyset$ or $A = Y$. This is true without any assumption on the lifts.
This implies (but Hatcher does not explicitly mention it) that if the lifts agree at one point of $Y$, then $A = Y$.
Concerning your second question: $p^{-1}(U)$ is the disjoint union of open sets $U_\alpha$ each mapped homeomorphically to $U$ by $p$. Let $p_\alpha : U_\alpha \to U$ denote the homeomorphism obtained by restricting $p$. We have $\tilde f_i(y) \in U_{\alpha_i} = U_i$. Thus, if $U_1 = U_2$, i.e. $\alpha_1 = \alpha_2$, then $\tilde f_1(y) = p^{-1}_{\alpha_1}(y) = p^{-1}_{\alpha_2}(y) = \tilde f_2(y)$. Conversely, if $U_1 \ne U_2$, then trivially $\tilde f_1(y) \ne \tilde f_2(y)$ because the $U_\alpha$ are pairwise disjoint. Thus $U_1 = U_2$ if and only if $\tilde f_1(y) = \tilde f_2(y)$.