If $f$ is continuous and $I$ compact, is $f^{-1}(I)$ compact?

Question

Let $f:E\to F$ where $E$ and $F$ are metric space. We suppose $f$ continuous. I know that if $I\subset E$ is compact, then $f(I)$ is also compact. But if $J\subset F$ is compact, do we also have that $f^{-1}(J)$ is compact ?

If yes and if $E$ and $F$ are not necessarily compact, it still works ?

Answer 1

Not necessarily; consider $f:(-2\pi,2\pi)\to[-1,1]$ given by $f(x)=\sin(x)$.

Answer 2

Clayton's answer based on non-injectivity is very good. However, we can also base a counterexample on non-surjectivity.

Consider $f:\mathbb{R}\mapsto[-1,1]$ given by $$ f(x)=\frac{x}{\sqrt{x^2+1}} $$ Then $$ f^{-1}([-1,1])=\mathbb{R} $$