I'm trying to prove the classification theorem for covering spaces. I've got to the stage where I need to show the following:
If $H$ a subgroup of $\Pi_1(X,x_0)$ then $\exists Y$ covering space of $X$ such that $\Pi_1(Y,y_0) = H$.
Now I've shown that if $\tilde{X}$ the universal cover of X, then we may take $Y$ to be the group of $H$-orbits of $\tilde{X}$, where I'm identifying $H$ with the group of covering transformations relative to the covering map $r:\tilde{X}\rightarrow X$. Indeed let $q:\tilde{X}\rightarrow Y$ be the quotient map, then the natural map $p:Y\rightarrow X$ is a covering map (hard to prove, but shown here).
I then know that $\Pi_1(Y,y_0) = \Gamma(q)$, the group of covering transformations relative to $q$ on $\tilde{X}$. Clearly $H \leq \Gamma(q)$, but how do I prove the other inclusion?
In particular how do I know that $\nexists g \in \Pi_1(X,x_0)$ s.t. $g \notin H$ and $q\circ g = q$?
Many thanks in advance, and do tell me if I'm not being clear!
Edit - maybe this argument works...
Suppose $g$ satisfies the 'bad' properties I gave above. Let $x \in \tilde{X}$. Then in particular $q \circ g = q$ means that $g(x) = h(x)$ for some $h \in H$. But then $g$ is the unique covering transformation sending $x$ to $h(x)$. But indeed $h$ is also a covering transformation sending $x$ to $h(x)$. So $g = h \in H$. But this is a contradiction! Hence $\nexists$ such a $g$, as required.
Could someone possibly verify that this is correct?