Misunderstanding about continuous functions over open and closed sets

Question

How do these two propositions sit together?

1. Let $f\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$ be a continuous function and let $A\subseteq\mathbb{R}^{m}$ be an open set, so $f^{-1}(A)$ is an open set.
2. Every open set $\mathbb{A\subseteq R}^{m}$ is a continuous image of a closed set.

I guess i'm just missing something here, but as i see it, the 2nd states that for every open set $A\subseteq\mathbb{R}^{m}$ exists a closed set $C\subseteq\mathbb{R}^{n}$ and a continuous function $f\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$ such that $f(C)=A$.

But that gives us that $C=f^{-1}(A)$ which contradicts the 1st no?

Answer 1

From $f(C)=A$ you only get $f^{-1}(f(C))=f^{-1}(A)$. It is quite possible that $f^{-1}(f(C))\supsetneq C$.