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I am having difficulties thinking about how an argument for the following exercise should proceed:

Let $p: Y \rightarrow X$ and $q: Z \rightarrow X$ be $G$-coverings (i.e., covering maps such that $X = Y /G = Z/G$ (quotient spaces)), with $X$ connected and locally path connected. Let $\phi: Y \rightarrow Z$ and $\psi: Y \rightarrow Z$ be maps of $G$-coverings (i.e., a covering homomorphism such that $\phi(g \cdot y) = g \cdot \phi(y)$ for all $g \in G$ and $y \in Y$, same with $\psi$). Assume that $\phi(y) = \psi(y)$ for some $y \in Y$. Show that $\phi(y) = \psi(y)$ for every point $y \in Y$.

If $Y$ is connected, this should be an immediate consequence of the unique lifting property for maps, but in the general case I am lost, and I am curious to know if anyone visiting would know how to proceed.

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(Sorry I can't comment)

Show that each connected component of $Y$ contains a preimage of every $z\in Z$. Fix a connected component $Y_0$ of $Y$. Take a $y\in Y$. Using the transitivity of $G$, translate $y$ to $Y_0$. Then use the fact that you know the statement is true for $Y_0$.

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In your first sentence and second sentence, which set contains the connected component (i.e., $Y$)? What do you mean by a lift of a point $z \in \mathbb{Z}$? I get lifting a continuous map, like a path in a space. – Vulcan Feb 7 '12 at 2:55
Sorry I meant fix a connected component of $Y$. By lift of $z$ I just meant a preimage of $z$ in $Y$. – user16544 Feb 7 '12 at 3:28
Also when I said "each connected component" in the first sentence I meant "each connected component of Y". – user16544 Feb 7 '12 at 3:38

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