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Piotr Hajlasz
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As Alexandre Eremenko pointed out, in general the answer is in the negative. However, in a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

In fact if, $V$ is simply connected and $U$The following result is connected, thenin a positive direction:

If, $V$ is simply connected and $U$ is connected, and $f$ is proper (i.e. preimages of compact sets are compact), then then $f$ is a bijection.

Under the given assumptions $f$ is a bijection. This is well known.covering map, see https://math.stackexchange.com/q/860351/798404

A covering space is a universal covering space if it is simply connected. Assuming that $V$ is simply connected, it is its universal cover and $\operatorname{id}:V\to V$ is a covering map. Now we use the fact:

If the mapping $p: D \to X$ is a universal cover of the space $X$ and the mapping $f : C \to X$ is any cover of the space $X$ where the covering space $C$ is connected, then there exists a covering map $g : D \to C$ such that $f ∘ g = p$.

In our situation $D=X=V$, $p=\operatorname{id}$, $C=U$ and $f$ is $f$. Therefore there is $g:D\to C$ i.e. $g:V\to U$ such that $f\circ g=p$ i.e., $f\circ g=\operatorname{id}$. That proves that $f$ is a bijection (and hence a diffeomorphism).

You can find basic statements about universal cover and covering maps in https://en.wikipedia.org/wiki/Covering_space#Lifting_properties

If you want to learn more, take almost any book in topology. For example:

M. A. Armstrong, Basic Topology.

As Alexandre Eremenko pointed out, in general the answer is in the negative. However, in a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

In fact if, $V$ is simply connected and $U$ is connected, then $f$ is a bijection. This is well known.

A covering space is a universal covering space if it is simply connected. Assuming that $V$ is simply connected, it is its universal cover and $\operatorname{id}:V\to V$ is a covering map. Now we use the fact:

If the mapping $p: D \to X$ is a universal cover of the space $X$ and the mapping $f : C \to X$ is any cover of the space $X$ where the covering space $C$ is connected, then there exists a covering map $g : D \to C$ such that $f ∘ g = p$.

In our situation $D=X=V$, $p=\operatorname{id}$, $C=U$ and $f$ is $f$. Therefore there is $g:D\to C$ i.e. $g:V\to U$ such that $f\circ g=p$ i.e., $f\circ g=\operatorname{id}$. That proves that $f$ is a bijection (and hence a diffeomorphism).

You can find basic statements about universal cover and covering maps in https://en.wikipedia.org/wiki/Covering_space#Lifting_properties

If you want to learn more, take almost any book in topology. For example:

M. A. Armstrong, Basic Topology.

As Alexandre Eremenko pointed out, in general the answer is in the negative. However, in a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

The following result is in a positive direction:

If, $V$ is simply connected and $U$ is connected, and $f$ is proper (i.e. preimages of compact sets are compact), then then $f$ is a bijection.

Under the given assumptions $f$ is a covering map, see https://math.stackexchange.com/q/860351/798404

A covering space is a universal covering space if it is simply connected. Assuming that $V$ is simply connected, it is its universal cover and $\operatorname{id}:V\to V$ is a covering map. Now we use the fact:

If the mapping $p: D \to X$ is a universal cover of the space $X$ and the mapping $f : C \to X$ is any cover of the space $X$ where the covering space $C$ is connected, then there exists a covering map $g : D \to C$ such that $f ∘ g = p$.

In our situation $D=X=V$, $p=\operatorname{id}$, $C=U$ and $f$ is $f$. Therefore there is $g:D\to C$ i.e. $g:V\to U$ such that $f\circ g=p$ i.e., $f\circ g=\operatorname{id}$. That proves that $f$ is a bijection (and hence a diffeomorphism).

You can find basic statements about universal cover and covering maps in https://en.wikipedia.org/wiki/Covering_space#Lifting_properties

If you want to learn more, take almost any book in topology. For example:

M. A. Armstrong, Basic Topology.

Post Deleted by Piotr Hajlasz
added 977 characters in body
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Piotr Hajlasz
  • 28k
  • 5
  • 85
  • 184

InAs Alexandre Eremenko pointed out, in general the answer is in the negative. However, in a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

In fact if, $V$ is simply connected and $U$ is connected, then $f$ is a bijection. This is well known.

A covering space is a universal covering space if it is simply connected. Assuming that $V$ is simply connected, it is its universal cover and $\operatorname{id}:V\to V$ is a covering map. Now we use the fact:

If the mapping $p: D \to X$ is a universal cover of the space $X$ and the mapping $f : C \to X$ is any cover of the space $X$ where the covering space $C$ is connected, then there exists a covering map $g : D \to C$ such that $f ∘ g = p$.

In our situation $D=X=V$, $p=\operatorname{id}$, $C=U$ and $f$ is $f$. Therefore there is $g:D\to C$ i.e. $g:V\to U$ such that $f\circ g=p$ i.e., $f\circ g=\operatorname{id}$. That proves that $f$ is a bijection (and hence a diffeomorphism).

You can find basic statements about universal cover and covering maps in https://en.wikipedia.org/wiki/Covering_space#Lifting_properties

If you want to learn more, take almost any book in topology. For example:

M. A. Armstrong, Basic Topology.

In a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

In fact if, $V$ is simply connected and $U$ is connected, then $f$ is a bijection.

As Alexandre Eremenko pointed out, in general the answer is in the negative. However, in a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

In fact if, $V$ is simply connected and $U$ is connected, then $f$ is a bijection. This is well known.

A covering space is a universal covering space if it is simply connected. Assuming that $V$ is simply connected, it is its universal cover and $\operatorname{id}:V\to V$ is a covering map. Now we use the fact:

If the mapping $p: D \to X$ is a universal cover of the space $X$ and the mapping $f : C \to X$ is any cover of the space $X$ where the covering space $C$ is connected, then there exists a covering map $g : D \to C$ such that $f ∘ g = p$.

In our situation $D=X=V$, $p=\operatorname{id}$, $C=U$ and $f$ is $f$. Therefore there is $g:D\to C$ i.e. $g:V\to U$ such that $f\circ g=p$ i.e., $f\circ g=\operatorname{id}$. That proves that $f$ is a bijection (and hence a diffeomorphism).

You can find basic statements about universal cover and covering maps in https://en.wikipedia.org/wiki/Covering_space#Lifting_properties

If you want to learn more, take almost any book in topology. For example:

M. A. Armstrong, Basic Topology.

Source Link
Piotr Hajlasz
  • 28k
  • 5
  • 85
  • 184

In a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

In fact if, $V$ is simply connected and $U$ is connected, then $f$ is a bijection.