1
$\begingroup$

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.

1 Answers 1

0

(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 $Y_0$.

  • 0
    Also when I said "each connected component" in the first sentence I meant "each connected component of Y".2012-02-07