@MikeMiller It seems like a Morse theory argument, at least the homeomorphic fibers part. If $p : M \to X$ is your submersion, look at $F_1 = p^{-1}(x_1)$ and F_2 = $p^{-1}(x_2)$. Take a path $\gamma$ from $x_0$ to $x_1$ and look at $p^{-1}(\gamma)$ - that's a cobordism $W$ between $F_1$ and $F_2$. Projecting to the path back again we get a height function on $W$ which has no critical points - sounds like it shouldn't change the topology going from $F_1$ to $F_2$ then.
If this is/is not the right direction, give me a "Yes/No", but please don't reveal further :)
Thinking of "upwards" as "having the $S^1$ coordinate pointing clockwise", can I pick a point in $S^1\times S^2$ such that the projection onto the $S^2$ coordinate is the vector field that flows from the point to its opposite and that at the point is pointing straight upwards and that at the opposite point is pointing straight downwards @TedShifrin
@Balarka The normal bundle to a fiber is trivial. Flowing along each vector field in a trivialization gives you a map F x R^n -> M. This is rather clearly a submersion along F x 0.
@TedShifrin Right, but there's a vector field on $S^2$ that is zero at a point $p$ and its opposite $-p$ and is pointing away from $p$ and towards $-p$ everywhere else
Oh, ignore the "upwards/downwards" bit from before
@MikeMiller I am confused what you're answer to by that comment. Are you telling me the proof of the local trivialization part (after which homeomorphic fibers is just playing the game chartwise)?
is the "graph of a function" the unique set of tupels that belong to the cartesian product? Or can in some contextes the graph mean, well the picture of the function on some arbitary coordinate system?
DogAteMy: If by the time you're on the other side of the world you want a different problem, you can let me know :) Meanwhile, with a bit of pencil and paper, you should figure this one out.
@DHMO I think from here you need to substitute in $x\mapsto\ln x$ instead of solving algebraically? I mean, the original equation is true for all x, so we can replace x with whatever other value we want
@DHMO Oh, also, it makes more sense to substitute in $\ln x$ rather than taking the log of the inequality because it would take a few more steps to show log is increasing
It's not a bad trick. That's how you show that every finite dimensional CW complex with finitely many cells is weak homotopy equivalent to a finite topological space, I think.
@Ted What's an acceptable way to denote someone isn't the first of their name (e.g. their father had the same name) if its unclear if they use a suffix and generally only write in a language (and alphabet) I cannot read?
@Balarka: You could read section 3.3 of the geom notes. What's involved here is essentially using forms and pullback rather than doing a jacobian computation (as is suggested in the exercise). The depends on using two different moving frames (and that is discussed in 3.3).
