@Ted None of the above. What's being done is putting an orientation on $f^{-1}(Z)$ by first putting an orientation on the vector space normal to an appropriate tangent space. But he immediately points out tha one could do this by specifying their favorite subspace complementary to the appropriate tangent space, so it doesn't matter.
I know that A is normal and so its inverse image is a subgroup and in this case in not too difficult to show that is also normal, since is en G is either the trivial subgroup or the entire group, because G is normal
The idea is to show that it has to have the entire grooup. So by contradiction if not, so $f^{-1}={1}$ but from here i don't know how to get the contradiction
someone please?
any idea of how to eliminate the possibility of $f^{-1}={1}$
