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I've just come across this proof of the following theorem that I can't convince myself is true. Any ideas whether it's correct?

Suppose $\gamma$ and $\lambda$ are homotopic paths starting at $x$ in a Riemann surface $X$ with the same endpoint. Let $Y$ be a covering space for $X$, with $Y$ a Riemann surface, $f$ the holomorphic covering map. Pick $y \in f^{-1}(x)$. Let $\tilde{\gamma},\tilde{\lambda}$ be the unique lifts of $\gamma,\lambda$ to $Y$. Then $\tilde{\gamma}(1)=\tilde{\lambda}(1)$.

The proof simply states that by the homotopy lifting theorem $\tilde{\gamma}$ is homotopic to $\tilde{\lambda}$ by a homotopy starting at $y$, so $\tilde{\gamma}(1)=\tilde{\lambda}(1)$. I don't see why this must be the case though. Surely we could have $y_1\neq y_2 \in Y$ with $f(y_1)=f(y_2)=\gamma(1)=\lambda(1)$ and $\tilde{\gamma}(1)=y_1$, $\tilde{\lambda}(1)=y_2$. What prevents this from happening? I assume that the proof is correct since it's a trick that the lecturer on the course used several times!

Many thanks!

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Are you sure your homotopy lifting theorem says nothing about this? I learned (under the same name) a theorem that guarantees that both curves have the same endpoint. The proof used the local homeomorphism property of $f$ (which is used to lift all curves "between" $\gamma$ and $\lambda$ to a homotopy in $Y$) and the discreteness of $f^{-1}(x)$ to show that all endpoints are identical. – Gregor Bruns May 30 '12 at 11:13
Maybe the discreteness of $f^{-1}(x)$ guarantees that the curves have the same endpoint. Since all curves 'between' $\gamma$ and $\lambda$ lift to curves ending at some point of $f^{-1}(x)$ and the lift of the homotopy is continuous, we must have that they all end at the same point of $f^{-1}(x)$. Does this hold in general when $f^{-1}(x)$ is not necessarily discrete though? I expect it does, possibly using the uniqueness of lifts of constant paths at the endpoint? – Edward Hughes May 30 '12 at 12:40
If $f^{-1}(x)$ is not discrete, then $f$ is not a local homeomorphism and therefore you have to give up the assumption that $f$ is a covering map. This is probably not what you want. – Gregor Bruns May 30 '12 at 15:41
Ah right - so every covering map has $f^{-1}(x)$ discrete then? In that case I see why it's trivially true. – Edward Hughes May 30 '12 at 15:47

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