Unique Lifting Property - Algebraic Topology

Question

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.

• Where is the Author using that the two maps agree at a point?

• 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.

Answer 1

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 $\tilde f_1(y) = \tilde f_2(y)$.

Answer 2

Suppose $\varphi,\psi:Y\to \tilde X$ are continuous lifts of $f$ which agree at some point of $Y$. Let $S=\{y\in Y:\varphi(y)=\psi(y)\}$. The set $S$ is non-empty.

Suppose $y_0\in S$. Let $\varphi(y_0)=\psi(y_0)=\tilde x_0$ and $p(\tilde x_0)=x_0$. Let $U$ be an evenly covered open set containing $x_0$ and $\tilde U$ be an open subset of $\tilde X$ containing $\tilde x_0$ such that the restriction of $p$ gives a homeomorphism from $\tilde U$ to $U$. As both $\varphi$ and $\psi$ are continuous and the image of $y_0$ under $\varphi$ and $\psi$ is in $\tilde U$, there is an open set $V$ containing $y_0$ such that $\varphi(V)\subseteq \tilde U$ and $\psi(V)\subseteq \tilde U$. Let $y\in V$. As $\varphi$ and $\psi$ are lifts of $f$, we see that $p(\varphi(y))=f(y)=p(\psi(y))$. As $\varphi(y),\psi(y)\in \tilde U$ and the restriction of $p$ to $\tilde U$ is injective, we see that $\varphi(y)=\psi(y)$. So we have found an open set $V$ containing $y_0$ such that $V\subseteq S$. Hence, it follows that $S$ is open.

Suppose $y_0\notin S$. Let $f(y_0)=x_0$, $\varphi(y_0)=\tilde x_1$ and $\psi(y_0)=\tilde x_2$. As $p\circ\varphi=f=p\circ\psi$, it follows that $p(\tilde x_1)=p(\tilde x_2)=x_0$. Let $U$ be an evenly covered open set containing $x_0$. As $\tilde x_1\neq \tilde x_2$, there exist disjoint open sets $\tilde U_1$ and $\tilde U_2$ in $\tilde X$ such that $\tilde x_1\in \tilde U_1$, $\tilde x_2\in \tilde U_2$ and the restrictions of $p$ give homeomorphisms $\tilde U_1\to U$ and $\tilde U_2\to U$. As $\varphi$ and $\psi$ are continuous and the images of $y_0$ under $\varphi$ and $\psi$ are in $\tilde U_1$ and $\tilde U_2$, we can find an open set $V$ containing $y_0$ such that $\varphi(V)\subseteq \tilde U_1$ and $\psi(V)\subseteq \tilde U_2$. As $\tilde U_1$ and $\tilde U_2$ are disjoint, for each $y\in V$ we see that $\varphi(y)\neq \psi(y)$. So we have found an open set $V$ containing $y_0$ such that $V\cap S=\emptyset$. Hence, it follows that $S$ is closed.

Thus, as $Y$ is connected we see that $S=Y$ and so it follows that $\varphi=\psi$.