Simple Topology Question

  • Thread starter jimisrv
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  • #1
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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
 

Answers and Replies

  • #2
quasar987
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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.
 
  • #3
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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}.
 
  • #4
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Thanks for the help! The last one is a very simple counterexample.
 
  • #5
WWGD
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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.
 
  • #6
mathwonk
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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.
 
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  • #7
mathwonk
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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.
 

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