Simple Topology Question

jimisrv

Hi,

I have a question that I'm not sure about.
If f:A->C is continuous and B is a subset of C that is simply connected, is f(^-1)(B) necessarily connected or simply connected for that matter? Since the spaces are not necessarily homeomorphic I cannot consider it a topological invariant.

Thanks

quasar987

Consider the old canonical mapping from [0,2pi) to the circle S^1 in the complex plane (f(t)=e^it). Then for r small enough, f^-1(B(1,r)) is clearly not connected (and hence not simply connected either) since it consist of a little neighborhoods of 0 and a little neighborhoods of 2pi.

Doodle Bob

Actually, we can be even simpler than that: let f:R->R be given by f(x)=x^2, and let B={1}. Then f^-1(B)={1,-1}.

jimisrv

Thanks for the help! The last one is a very simple counterexample.

WWGD

Maybe for a more extreme counterexample re connectedness, consider the case of

IR as a covering space of S^1 . Maybe if you had 1-1 -ness. (tho not in this case,

since continuous bijection bet. compact and hausdorff is a homeo., which is sufficient,

tho I don't know if it is necessary).


For a trivial counterexample re simple-connectedness, consider a constant map

defined on an annulus.

mathwonk

trivial counterexamples do exist for zero dimensional B, but try higher dimensional B's.

and try it for algebraic maps.

i.e. if you map an algebraic variety X to an algebraic variety Y, what does the inverse image of an irreducible curve in Y look like?

more specifically, project a surface onto P^2, and ask what the inverse image of a general line looks like?

see the fulton - hansen connectedness theorem, and various versions of bertini's thoerem.

mathwonk

here is a nice theorem of fulton and collaborators:

if L is a linear subspace of a projective space P, having codimension d,

and if X-->P is any morphism from a projective variety X,

having image in P of dimension larger than d,

then the inverse image in X of L is connected.