Continuous function from complex matrices to complex numbers

Question

Given $A(t) \in M_n$, where $t \in \mathbb{R}$.

$A(t)$ is smooth and continuous.

Given $f(A(t)) \in \mathbb{C}$

$f(A(t))$ is smooth and continuous. f may contain inverses, transposes, determinants or traces. f is not one-to-one.

let $G$ be an arbitrary closed connected subset of {$f(A(t)), t \in \mathbb{R}$}

Is there always some closed interval of $\mathbb{R}$, say $S = [a..b]$ such that $f(A(S)) = G$ ?

I'm confused as to how to prove this. So I'm taking a subset of the image of a function and asking if there's a connected component in the preimage that maps to it. Thanks.

Answer 1

For example, $f(A(t))$ may be periodic and go around a figure $8$ like this, and $G$ could be the right half of the image (shown in blue).

EDIT: If you want to make $G$ smooth, you could "squeeze it" near the crossing, something like this: