A covering map $p: X \to Y$ is said to be regular if for a point $x \in p^{-1}(y)$ we have $p_*(\pi_1(X,x)) \lhd \pi_1(Y,y)$.
Now my notes say that this condition is independent of the choice of the point $x \in p^{-1}(y)$ but I don't understand exactly why. Could you please explain this to me?
Thank you.
[I assume $X$ is path-connected, which is needed for this statement to be true.]
The subgroups $p_*(\pi_1(X,x'))$ for different choices of $x'\in p^{-1}(y)$ are exactly the conjugates of $p_*(\pi_1(X,x))$. Indeed, if you pick a path $\gamma$ from $x$ to $x'$, then you get an isomorphism from $\pi_1(X,x)$ to $\pi_1(X,x')$ by taking a loop $\alpha$ based at $x$ to the loop $\gamma^{-1}*\alpha*\gamma$ based at $x'$. Now when you apply $p_*$ to all of this, $p\circ\gamma$ is a loop, and $p_*(\pi_1(X,x'))=[p\circ\gamma]^{-1}p_*(\pi_1(X,x))[p\circ\gamma]$ as subgroups of $\pi_1(Y,y)$. (Conversely, given any $\delta\in \pi_1(Y,y)$, you can lift it to a path in $X$ from $x$ to some point $x'$ and then $\delta^{-1}p_*(\pi_1(X,x))\delta=p_*(\pi_1(X,x')$.)
In particular, if $p_*(\pi_1(X,x))$ is normal, all of its conjugates are the same, and so $p_*(\pi_1(X,x'))$ is also normal for any other $x'\in p^{-1}(y)$.