Proof biholomorphic map doesn’t exist

[viral lava]

Hi, for following map f: C{0} -> D_1(0) I want to prove that there exists no biholo map. I know this is true because homeomorphisms preserve topological spaces.

My question is, is there another way to show this since we‘re not assumed to have taken any topology class (so at the exam this reasoning might not give any points).

What I thought about is liouvilles theorem as we are only one point away from being able to use it. Furthermore if there exists an analytic continuation then we can basically conclude. Is this the right way to go? Can I assume an analytical continuation?

Thank you!

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[fierce parrot]

I think yes you can if v is a vector in the closure of a set A such that for a continuous function f on A lim z—>a f(x)=l then l is an element of the closure of f(A) it’s quite simple to check if v is in clos(A) then there exists a sequence (xn) that has limit v then f(xn) has limit l so l is in clos(f(A))

so the limit as z approches 0 of f(z) is in clos(D_1(0))

so it’s in the closed disk centered around 0 of radius 1

ie if you let l=limf(z) for z approaching 0 then |l|<=1

The problem is that f isn’t guaranteed to have a limit at 0

But if it does you can extend your function via continuity and then you apply Liouville’s theorem

And via surjectivity of f you have |l|=1

[viral lava]

Hmm, so this is not necessarily generally true? We have that f is analytic on some disc and furthermore f is bounded in a neighborhood of 0 this is equivalent to f having a removable singularity at 0

Thus there exists an analytic cont.? 😳

Therefore constant by liouville and F cannot be biholo. Since a constant function is not bijective