Problem . (a) We have and be open set in hence be open set in , we have also that be open in because and be compact set in .

Assume that are open sets in , if or then and is an open set in therefore is open set in . If and then , here and are compact sets in . We have is compact set in and therefore be open set in .

Assume that is a family of open sets in and for some (maybe empty). If then be open in , and so open in . If then is closed and bound in , also it is compact set in , therefore is open set in .

Thanks to all conditions above we have is a topological space.

Assume are elements in . If then , note that because is Hausdorff space we can find open sets in such that and (Note: and are also open sets in ). If , assume that , denote and . Then are open sets in , and .

Compactness of is a immediately follows from (b).

(b)We have be a surjective because and , where . But it is easy to see that is an injective(checking then ), therefore is a bijective. By above formular, we have is a homeomorphism.( in topology of )

Problem . We have is an open set in and therefore is holomorphic on .

If then we definition by setting .

If then we definition by setting and .

In both cases we have is continuous over and is a bijective, it is easy to find formular of and check and are holomorphic.

Problem .

Problem . The problem is wrong, an example and and . Then and . In my opinion, must is belong to .

Problem . (a)Assume that is open in and is original map. We have and is open in therefore is open in , so $f^{-1}(U)$ is open in , or is continuous.

If and are complex charts of and , respectively, such that . We have therefore is holomorphic.

If is biholomorphic then must is bijection and . In fact, if then but , contradiction!

If then is bijective and therefore by above, and are holomorphic.

(b)We have and use (a).

(c)Use (a) again!