Alternate Proof of Unique Lifting Property of Covering Spaces

Question

I proved one of Hatcher's propositions on my own and my proof is quite a bit different than his.

The Unique Lifting Property says: Given a covering space $p:\tilde{X} \rightarrow X$ and a map $f: Y \rightarrow X$, if $\tilde{f}_{1}, \tilde{f}_{2}: Y \rightarrow \tilde{X}$ are lifts of $f$ which agree at one point of $Y$ and $Y$ is path connected, then $\tilde{f}_{1}$ and $\tilde{f}_{2}$ agree at all points of $Y$.

My proof is as follows:

Let $A=\{y \in Y : \tilde{f}_{1}(y)=\tilde{f}_{2}(y)\}$ and $B=\{y \in Y : \tilde{f}_{1}(y)\neq \tilde{f}_{2}(y)\}$; then $Y=A \cup B$. Note that $A \neq \emptyset$ by hypothesis. Suppose $y \in A$ and consider $f(y) \in X$. Let $U$ be a neighborhood of $f(y)$ which is evenly covered. Let $\tilde{V}$ be the sheet above $U$ containing $\tilde{f}_{1}(y)=\tilde{f}_{2}(y)$. Let $G=\tilde{f}^{-1}_{1}(\tilde{V}) \cap \tilde{f}^{-1}_{2}(\tilde{V})$; suppose $t \in G$ then $p(\tilde{f}_{1}(t))=p(\tilde{f}_{2}(t))=f(t)$. Since $\tilde{f}_{1}(t),\tilde{f}_{2}(t)\in V$ and $p$ is injective on $V$, we have $\tilde{f}_{1}(t)=\tilde{f}_{2}(t)$. Thus $G$ is an open subset of $A$ containing $y$. Suppose now that $y \in B$ and let $U$ be a neighborhood of $f(y)$ which is evenly covered. Let $\tilde{V}_{1}$ and $\tilde{V}_{2}$ be the sheets above $U$ containing $\tilde{f}_{1}(y)$ and $\tilde{f}_{2}(y)$, respectively. Note that $\tilde{V}_{1}\neq \tilde{V}_{2}$ as this would imply that $\tilde{f}_{1}(y)=\tilde{f}_{2}(y)$. Let $G=\tilde{f}^{-1}_{1}(\tilde{V}_{1}) \cap \tilde{f}^{-1}_{2}(\tilde{V}_{2})$. If $t \in G$ then $\tilde{f}_{1}(t) \in \tilde{V}_{1}$ and $\tilde{f}_{2}(t) \in \tilde{V}_{2}$ so they are not equal. Thus $G$ is an open subset of $B$ containing $y$. It follows that $B = \emptyset$ or this would be a separation of $Y$, contradicting that $Y$ is connected.

Does this proof seem correct?

Answer 1

[I'm not allowed to comment due to missing reputation, but my post might be considered a comment.]

Your proof seems similar to the one given by Edwin H. Spanier in his book "Algebraic Topology". I'm including a screenshot of his proof for reference.

Answer 2

This is a sufficiently old post , but I want to provide a complete proof for reference of others as well as for myself.

Let $p:\bar{X}\to X$ be a covering . and $f,g:Y\to \bar{X}$ be two continuous maps such that $pf=pg$ . Then the set of points on which $f$ and $g$ agree is both open and closed.

Let $A=\{y\in Y: f(y)=g(y)\}$ . If $A=\phi$ then nothing to prove.

If $A\neq \phi$ then say $z\in A$ .

Let $U$ be an evenly covered nbd of $pf(z)=pg(z)\in X$ . then $p^{-1}(U)=\amalg_{\alpha\in A}V_{\alpha}$ . Let $f(z)=g(z)$ belong to $V_{\alpha_{0}}$

Then choose a nbd $W$ of $f(z)=g(z)$ such that $W\subset V_{\alpha_{0}}$ . Then due to continuity , there exists $W_{1}$ and $W_{2}$ nbd of $z$ such that $f(W_{1}),g(W_{2})\subset W$ .

Then $z\in W_{1}\cap W_{2}$ and if $y\in W_{1}\cap W_{2}$ then $f(y),g(y)\in W$ .

As $pf=pg$ , then as $p|_{V_{\alpha_{0}}}$ is a homeomorphism , it means that $f(y)=g(y)$ as $pf(y)=pg(y)$. So as any point of $A$ is an interior point and hence $A$ is open.

Similarly if $f(z)\neq g(z)$ , then $f(z)\in V_{\alpha_{1}}$ and $g(z)\in V_{\alpha_{2}}$ where $\alpha_{1}\neq \alpha_{2}$. This is beacuse $p$ restricted to any $\alpha_{i}$ is a homeomorphism and if they are equal then $f(z)=g(z)$ as $pf(z)=pg(z)$.

So by continuity we pick open nbds of $U_{1},U_{2}$ of $z\in Y$ such that $f(U_{1})\subset V_{\alpha_{1}}$ and $g(U_{2})\subset V_{\alpha_{2}}$ .

Then $U_{1}\cap U_{2}$ is an open nbd of $z$ such that for any $y$ in $U_{1}\cap U_{2}$, $f(y)\in V_{\alpha_{1}}$ and $g(y)\in V_{\alpha_{2}}$ . So by disjointness of the $V_{\alpha_{i}}$'s , $f(y)\neq g(y)$. Thus $Y\setminus A$ is also open.

Hence $A$ is open and closed.

Note that this proves the uniqueness of path lifting when $Y=[0,1]$ is connected. This is because in that case, $A=[0,1]$ .