I am reading "Mirror Symmetry" by Hori et al, and have a question on Chap.17 (Chiral rings and geometry of the vacuum bundle). On p.425 the authors say

> Consider the path-integral on the hemisphere. The boundary of the hemisphere is a circle on which our Hilbert space is based. The path-integral will give us a number, and so defines a functional from boundary filed configurations to numbers, equivalently, a state in the Hilbert space...

Then they say 

>To obtain a ground state at the boundary we consider the "neck" of the hemisphere to be infinitely stretched. In other words, we imagine connecting the hemisphere to a semi-infinite flat tube. Noe that on the flat tube the twisted and untwisted theories are equivalent.

They continue 

> Similarly, if we consider the topological path-integral together with the insertion of the corresponding chiral fields, we obtain a correspondence between chiral fields and the ground state....

**Question 1**
Why are the twisted and untwisted theories equivalent on the flat tube?


**Question 2**
What does it mean by inserting chiral field? I don't think this is explained anywhere in the book. Does the insertion mean that the operator acts on the field after some time corresponding to the position of the insertion?

Should I think of this hemisphere as a Riemann surface? 
I am confused because this also looks like the operator formalism and manifolds with boundary. I am kind of lost because they suddenly introduce the hemisphere and identify the states with the Hilbert space on the boundary.  

I think I lack of firm understanding of the subject, so I would appreciate it if someone could kindly explain things from the very basic.