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Anton Petrunin
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Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i \frac{z}{\sqrt{|z|}}$$f(z)=-\tfrac{z}{\sqrt{|z|}}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i \frac{z}{\sqrt{|z|}}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=-\tfrac{z}{\sqrt{|z|}}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

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Pete L. Clark
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Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i\cdot z/\sqrt{z}$$f(z)=i \frac{z}{\sqrt{|z|}}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i\cdot z/\sqrt{z}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i \frac{z}{\sqrt{|z|}}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

added 136 characters in body
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Anton Petrunin
  • 45k
  • 14
  • 135
  • 299

Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i\cdot z/\sqrt{z}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i\cdot z/\sqrt{z}$.

Q1. Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=i\cdot z/\sqrt{z}$.

It seems to answer the other questions.

You may take a spiral in X which is $f$-invariant. This produce a simply connected example.

Source Link
Anton Petrunin
  • 45k
  • 14
  • 135
  • 299
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