The preimage of continuous function on a closed set is closed. My proof is very different from my reference, hence I am wondering is I got this right?
Apparently, $F$ is continuous, and the identity matrix is closed. Now we want to show that the preimage of continuous function on closed set is closed.
Let $D$ be a closed set, Consider a sequence $x_n \to x_0$ in which $x_n \in f^{-1}(D)$, and we will show that $x_0 \in f^{-1}(D)$.
Since $f$ is continuous, we have a convergent sequence
$$\lim_{n\to \infty} f(x_n) = f(x_0) = y.$$
But we know $y$ is in the range, hence, $x_0$ is in the domain. So the preimage is also closed since it contains all the limit points.
Thank you.

Yes, it looks right.   Alternatively, given a continuous map $f: X \to Y$, if $D \subseteq Y$ is closed, then $X \setminus f^{-1}(D) = f^{-1}(Y \setminus D)$ is open, so $f^{-1}(D)$ is closed.
