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.
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.