Let's say we are looking at a covering space $X$ of $Y$.
Let $K \subset \pi_1(Y,y)$ be the subgroup of elements $$K:= \{ [\gamma] \in \pi_1(Y,y) : x * [ \gamma] = x\}$$
$K$ is the group of paths in $\pi_1(Y,y)$ such that when we lift $\gamma$ to a path $\tilde \gamma$ is starting at $x$ in X, it also ends at $x$.
My prof. writes that, there is a bijection $$ \frac{\pi_1(Y,y)}{K} \cong p^{-1}(y)$$
So there is an isomorphism between these cosets and all points laying above $y$. Can someone give me an intuitive explanation of why this is the case?
Let $\gamma$ be a loop whose unique lift starting at $x$ ends at $x*\gamma$. The statement that $\pi_1(Y,y)$'s action on $p^{-1}(y)$ is transitive means that, given an $x$, for any $x'$ you pick there is a loop $\delta$ such that $x*\delta=x'$.
Obviously picking an element $x'$ arbitrarily is the same as picking an arbitrary element from $p^{-1}(y)$.
But also, all loops $\delta$ whose lift ends at $x'$ are in the same coset of K, so $\pi_1(Y,y)/K$ is in bijection with $p^{-1}(y)$, as we have just described a bijection.
Let us see that loops with lift ending at the same point are in the same coset of K. Let $\delta, \epsilon$ have lifts starting at $x$, ending at $x'$. We wish to show that $\delta \simeq k \epsilon$ for some $k \in K$.But simply take $k=\delta \epsilon^{-1}$ and we are done, as $\bar{\delta} \bar{\epsilon}^{-1}$ starts and ends at $x$.