another proof that the inverse image of a lie group is a lie group Given two lie groups $G_1$ and $G_2$ and a lie group homomorphism $f: G_1 \to G_2$.  I am trying to show that if $H$ is a lie subgroup of $G_2$ that the inverse image is also a lie group.
I know how to do this problem using the exponential map:  $(df)^{-1} \mathfrak{h}$ is a lie subalgebra $\mathfrak{h_1} \subset \mathfrak{G_1}$.  Then there is a commutative diagram that shows you that the exponentiation of this is the same as $f^{-1} H$.  That is not what I want to do.  (EDIT: The proof using the fact that closed subgroups of a lie group are lie groups is basically the same; you take guys of $\mathfrak{G_2}$ that exponentiate to elements in $H$, show that this is a lie subalgebra and exponentiate it to get $H$.  Therefore I am not interested in this proof)
By using a transversal slice (as in Onishchik and Vinberg) to $H$, we can show that there is a smooth structure on $G_2/H$.  Is there a way of showing that $[H] \in G_2/H$ is a regular value of the composition $G_1 \to G_2 \to G_2/H$?

$H$ is a closed subset of $G_2$ since it is a Lie subgroup, thus the inverse image of $H$ is closed. It is thus a Lie subgroup of $G_1$ since a closed subgroup of a Lie group is a Lie subgroup.
