# Uniqueness of lift in covering space theory

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.

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.

• 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. – Eric Wofsey Jan 1 at 19:49
• Thank you both! – Angelo Brillante Romeo Jan 1 at 19:59