# Why is the group of covering transformations relative to the quotient map isomorphic to a subgroup of the Fundamental Group?

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?

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aren't you trying to prove the same thing as theorem 3.1 ? when you say $\Pi_1(Y,y_0) = H$, are you suggesting that loops in $Y$ are also loops in $X$ ? or do you want to prove that $p'_\star$ (as in thereom 3.1) is injective so that it is not a set equality but rather an isomorphism of groups ? –  mercio Apr 4 '12 at 20:49
Indeed I am trying to prove the same thing as in Step 3 of theorem 3.1. However I reckon that the method above is more slick that his method. Do you reckon that the reasoning I put in my 'edit' is valid? –  Edward Hughes Apr 4 '12 at 20:53
ok so I missed the "I'm trying to prove the theorem" part in your first line. And you're viewing all the groups as groups of covering transformations of $\tilde{X}$. –  mercio Apr 4 '12 at 21:10
Yup that's right. Could have used a different notation to make that clear, but that just makes it messier! –  Edward Hughes Apr 4 '12 at 21:29

By the end of step 2, you know that $(\tilde{X},q)$ is the universal covering of $Y$, so there is a map $\Psi : \Pi_1(Y,y_0) \to CoveringMaps(\tilde{X},q)$ defined just like $\Phi$ :
For any loop $\gamma : [0;1] \to Y$, $\Psi(\tilde{\gamma})(x_0)$ is the endpoint of the lift of $\gamma$ starting from $x_0$.
Furthermore, $\Psi$ is a group isomorphism between $\Pi_1(Y,y_0)$ and the set of covering transformations of $\tilde{X}$ relative to (Y,q).
Then, your argument shows that the set of covering transformations of $\tilde{X}$ relative to $(Y,q)$ is exactly $H'$, the subgroup of the covering transformations of $\tilde{X}$ relative to $(X,p)$ corresponding to $H$.
So $\Psi(\Pi_1(Y,y_0)) = \Phi(H)$ (which is proved by the end of step 2).
But what the theorem wants to show is not that $\Pi_1(Y,y_0)$ is isomorphic to $H$ in some abstract way, but that $p'_*$ itself is an isomorphism between those two groups, which leads to step 3, where it is showed that $p'_* = \Phi^{-1} \circ \Psi$
Ah I see. To be honest I'm happy just knowing that they are isomorphic in an abstract way, and appreciating (but not learning) the proof that $p'_*$ is the manifest isomorphism! Thanks for clearing that up for me. –  Edward Hughes Apr 4 '12 at 23:26