Is $f(W)$ closed if $W$ is closed?

Question

Suppose a function $f : R^n \mapsto R^n$ is $C^1$ and the $Jacobian(f) \not= 0$ $\forall x $ in $R^n$

If $W$ is a closed set of $R^n$, is $f(W)$ closed?

I am not sure how to use the Jacobian to comment about the closure?

Do I use the inverse function theorem to talk about its invertibility? But that would only talk about local invertibility.

Can I think in the direction of continuity? But closed set does not necessarily map to a closed set in case of a continuous function.

How do I relate Jacobian to the notion of closed sets?

Please suggest a detailed explanation. Especially on the Jacobian part and similarly, what other insights I can gain from the Jacobian of a function if its non zero at all points in a domain.

Answer 1

$f(x)=exp(x)$ defined on $\mathbb{R}$, its image is $(0,+\infty)$, $\mathbb{R}$ is closed, but not its image.

Answer 2

No. Consider $f(x)=\arctan(x)$ and $W=\mathbb{R}$.

Answer 3

A simple counterexample in case $n=1$ and $W=\mathbb R$ is $\arctan$.