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Let Y be a topological space and $\pi: X\rightarrow Y$ a covering map. Take two lifts of the covering map $\tilde {\pi_1}$, $\tilde {\pi_2} : X\rightarrow X$ such that they agree on $x_0 \in X$.

Under what hypothesis can I conclude that the two lifts coincide?

I have always seen proofs of existence and uniqueness together but I can't exactly single out what is needed for uniqueness only. I know the condition is connectedness, but I am not sure if I have to suppose it for $X$ or $Y$.

If you could also give a sketch of the proof, I would very much appreciate it.

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More generally, let $f:Z\to Y$ be a map, and $g,h : Z\to X$ two lifts of $f$ that agree on $z_0\in Z$.

Then $\{x\in Z \mid g(x)=h(x)\}$ is nonempty (it contains $z_0$). Moreover, it is open : if $h(x) = g(x)$, then consider a basic open set $U$, neighbourhood of $f(x)$ such that $\pi^{-1}(U) =\displaystyle\coprod_{i\in I}U_i$ with each $U_i$ mapped homeomorphically onto $U$.

Then $\pi(h(x)) \in U$ so $h(x) \in U_i$ for some $i$. Let $\pi_i$ be the restriction/corestriction of $\pi$ to $U_i\to U$. Then, restricted to $h^{-1}(U_i)$, $\pi\circ h = f$, so $\pi_i\circ h_{\mid h^{-1}(U_i)} = f_{\mid h^{-1}(U_i)}$, and so $h= \pi_i^{-1}\circ f$ on $h^{-1}(U_i)$.

Similarly, on $g^{-1}(U_i)$, $g= \pi_i^{-1}\circ f$. So on $h^{-1}(U_i)\cap g^{-1}(U_i)$ (which contains $x$), $g=h$. Therefore our set is open.

It is also closed. This either follows trivially if you're assuming $Y$ to be $T_2$; but even if it's not then if $g(x) \neq h(x)$, then you can perform a similar argument to show that on a small enough neighbourhood of $x$, $g$ lands in $U_i$ while $h$ lands in $U_j, j\neq i$, so $U_i\cap U_j = \emptyset$, so that on this small neighbourhood, $g(y)\neq h(y)$. If this isn't clear you should do this yourself.

Thus we have found a nonempty clopen subset of $Z$: if $Z$ is connected, it follows that it's $Z$.

So we have

If $Z$ is connected and $f:Z\to Y$ has two lifts $g,h : Z\to X$ that agree on a point $z_0\in Z$, then $g=h$.

In your situation (after the edit) it means that if you assume $X$ to be connected, then the uniqueness follows.

If $X$ is not connected, then there are some counterexamples.

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  • $\begingroup$ For very simple counterexamples where $X$ is not connected, you could take $Y$ to be a point and $X$ to be any discrete space, so any map $X\to X$ at all is a lift. $\endgroup$ – Eric Wofsey Jan 1 at 19:49
  • $\begingroup$ Thank you both! $\endgroup$ – Angelo Brillante Romeo Jan 1 at 19:59

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