A The fundamental group of preimage of covering map

[1591238460]

i: B to Y is an inclusion, p: X to Y is a covering map. Define $D=p^{-1}(B)$, we assume here B and Y are locally path-connected and semi-locally simply connected. The question 1: if B,Y, X are path-connected in what case D is path-connected (dependent on the fundamental groups)? 2 What's the fundamental group of D at some point?

 

[lavinia]

One can always find a path connected subset whose inverse image is not path connected if the covering is non-trivial.

Take any open set whose inverse image is a collection of disjoint open sets. Such a set always exists around any point in Y. Choose a path in this set. Its inverse image will not be path connected.

 

[zinq]

My intuition is telling me that, under your assumptions, if the inclusion

i: B → Y​

is an isomorphism on the level of fundamental groups:

i_#: π₁(B,b) →_≈ π₁(Y,b),​

then the inverse image of B under the covering map

p: X → Y,​

namely p^-1(B), will be path-connected. I don't think the proof is difficult.

 

[zinq]

The above (#3) may be true, but it is definitely not the whole story. Stay tuned.

 

[1591238460]

One can always find a path connected subset whose inverse image is not path connected if the covering is non-trivial.

Take any open set whose inverse image is a collection of disjoint open sets. Such a set always exists around any point in Y. Choose a path in this set. Its inverse image will not be path connected.

Thanks

 

[1591238460]

My intuition is telling me that, under your assumptions, if the inclusion

i: B → Y​

is an isomorphism on the level of fundamental groups:

i_#: π₁(B,b) →_≈ π₁(Y,b),​

then the inverse image of B under the covering map

p: X → Y,​

namely p^-1(B), will be path-connected. I don't think the proof is difficult.

Thanks