Obtaining Density From Homeomorphism onto Image

Question

Let $f:X\rightarrow Y$ be a continuous function between metric spaces which is a homeomorphism onto its image and let $K\subseteq X$ be non-empty and $D\subseteq Y$ be dense and satisfy $$ D\cap {f(X)} = f(K). $$ Then is it the case that $K$ is dense in $X$?

Answer 1

No. The assumption that $K$ is nonempty is highly artificial; without it you can simply let $D$ be any dense proper subset of $Y$, let $X=Y\setminus D$ with $f$ the inclusion map, and let $K=\emptyset$. If you insist on having $K$ nonempty, you can tweak the example by letting $X$ contain a single point of $D$, say. For instance, you could have $Y=\mathbb{R}$, $D=\mathbb{R}\setminus\mathbb{Q}$, $X=\mathbb{Q}\cup\{\pi\}$, and $K=\{\pi\}$.