Compact Domain and Inverse Image I am trying to show that given $f:M \rightarrow N$, where $M$ is compact, $f$ is continuous and onto, then given $A \subset N$:
$$
f^{-1}(A) \text{ closed} \implies A\text{ closed} 
$$
I am dealing here with any metric space, although I feel that the approach is identical to the $\mathbb{R}$eal case (am I right?). So my attempt is the following.
My attempt: $f^{-1}(A)$ closed and $f^{-1}(A) \subset M \implies f^{-1}(A)$ bounded and hence, compact. Hence, I can consider $\{x_n\} \rightarrow x, x \in f^{-1}(A)$. Using continuity, now we have a $\{[f(x)]_n\} \subset N$ and $\{[f(x)]_n\} \rightarrow f(x)$. Since $x \in f^{-1}(A)$, $f(x)=y \in A$. But how do I know that all limit points of $A$ are generated by convergent subsequences in $f^{-1}(A)$?
Thanks.
 A: My solution assumes $N$ is Hausdorff space (metric spaces would have this property, for example).
As $f$ is onto, we have $f(f^{-1}(A)) = A.$ Now $f$ is continuous and $f^{-1}(A)$ is compact (closed set in a compact space), then $f(f^{-1}(A)) = A$ must be compact, hence closed (compact set in a Hausdorff space). 
(Not relevant for this solution, but note also that $N$ is a compact space, as $f$ is continuous and onto, and $M$ is compact.)
A: Let $\{y_n\}_{n\in\mathbb N}\subset A$, with $y_n\to y\in N$. We need to show that $y\in A$.
Since $f$ is onto, there exist a sequence $\{x_n\}_{n\in\mathbb N}\subset f^{-1}(A)\subset M$,
with $f(x_n)=y_n$. 
But as $M$ is compact, there exists a converging subsequence $x_{n_k}\to x\in M$. By the hypothesis, $f^{-1}(A)$ is closed, and thus $x\in f^{-1}(A)$.
We know that $y_{n_k}=f(x_{n_k})\to f(x)$, and as the limit of  $\{y_n\}_{n\in\mathbb N}$ is unique, we have that $y=f(x)$.
However, $x\in f^{-1}(A)$ implies that $y=f(x)\in f\big(f^{-1}(A)\big)=A$.
The last one, i.e., $f\big(f^{-1}(A)\big)=A$, is a property of the surjections. (Onto maps.)
Thus $A$ is indeed closed.
