@TedShifrin: For some fixed fundamental groups, there is a complete topological classification of 4-manifolds with that given fundamental group. Cyclic is one case.
Surely it would. It's not obvious to me why, but it seems absurd to think that the projection map $M \to N$ of a fiber bundle of closed manifolds could ever be null-homotopic.
Looks like the OP is satisfied by my answer, even though I'm not. I don't have time to work on this anymore, though.
If $D > 0$ then the quadratic has two unequal real roots If $D = 0$ then the roots are real and equal If $D < 0$ then the roots are complex and unequal
A lot of them seem to have this inferiority complex about them. The topic of who can make 6 figures comes up quite frequently. I remember reading about one person who said that anyone who didn't make six figures was "pond scum."
Firstly, I think I have fixed my notation and repaired Part A. Secondly, question: so I thought I did state $F$. I said F was $\nabla L$ but clearly that is wrong.
@anon I don't think so, because note that an element $f(x) \in \Bbb Q[[x]]$ is invertible iff the constant term is nonzero, so you can have different series, $f(x),g(x)$, that have the same constant term, then $f(x)-g(x)$ is not invertible, so it would have to be mapped to zero.