- #1
camel_jockey
- 38
- 0
Hi guys!
Excuse the spam, but I would like to ask something which I read in Armstrongs Basic Topology which I am just not 100% sure about.
He says we wish to define homeomorphism such that a circle cannot be homeomorphic to an interval such as [0,1). A continuous function f : X \mapsto Y is one whose every inverse f^-1(N) (N neighbourhood of a mapped point f(x)) is a neighbourhood in X.
He presents a one-to-one and onto function from [0,1) to the circle
x \mapsto exp ( 2∏i x )
This maps the interval into all of the circle. It has an inverse. Exactly why can one not get a continuous inverse??
I suspect this has something to do with the multi-valued property that the complex LOGARITHM suffers from. But somehow I feel the author does not presume the reader to know this much complex analysis - and that there is a more "primitive" reasons dealing with open sets, neighbourhoods, inverses etc. Is there one? Or is the solution merely that there exists no inverse which is not the logarithm?
Excuse the spam, but I would like to ask something which I read in Armstrongs Basic Topology which I am just not 100% sure about.
He says we wish to define homeomorphism such that a circle cannot be homeomorphic to an interval such as [0,1). A continuous function f : X \mapsto Y is one whose every inverse f^-1(N) (N neighbourhood of a mapped point f(x)) is a neighbourhood in X.
He presents a one-to-one and onto function from [0,1) to the circle
x \mapsto exp ( 2∏i x )
This maps the interval into all of the circle. It has an inverse. Exactly why can one not get a continuous inverse??
I suspect this has something to do with the multi-valued property that the complex LOGARITHM suffers from. But somehow I feel the author does not presume the reader to know this much complex analysis - and that there is a more "primitive" reasons dealing with open sets, neighbourhoods, inverses etc. Is there one? Or is the solution merely that there exists no inverse which is not the logarithm?