Mapping $U$ homeomorphically onto $f(U)$.

Question

I've got a rather elemental doubt about what "mapping a subspace homeomorphically onto its image" means, but I'd like to be sure what is being implied.

Let $f:X\to Y$ a continuous injective map between topological spaces. Let $U\subseteq X$ be an open subspace. Then we could say in some context that $f$ maps $U$ homeomorphically onto $f(U)$.

My question is

Must $f(U)$ be an open subspace of $Y$?

If we consider $f$ as a map $f:U\to f(U)$ (its restriction to $U$), then $f(U)$ is open in itself, so there is no contradiction with $f$ being a homeomorphism between $U$ and $f(U)$ if $f(U)$ wasn't open in $Y$.

Answer 1

No, it doesn't have to be open in $Y$.

For instance, take $f\colon\{0\}\longrightarrow\mathbb R$, defined by $f(0)=0$. Consider in $\mathbb R$ the usual topology. Then $f$ maps $\{0\}$ homeomorphically onto its image, but $\{0\}$ is not an open subset of $\mathbb R$.

Answer 2

Not always. E.g. $x \to (x,0)$ from $\mathbb{R}$ into $\mathbb{R}^2$.

We consider $f[U]$ as a space in its own right, with the subspace topology and $U$ as well, and $f$ is a homeomorphism between those two spaces.

Answer 3

Not always. Consider $f: \mathbb{R} \to \mathbb{R}^2$ defined by $f(x)=(x,0)$. This is continuous and injective, but does not map open sets to open sets.