Is multiplication by scalars a homeomorphism?

Question

Let V be a complex topological vector space. For any nonzero v in V, consider the map x -> x * v from C to V. This map is continuous, but is it a homeomorphism onto its image? Thank you.

Answer 1

The map $f \colon \mathbb{C} \to V$ given by $x \mapsto xv$ for $v \neq 0$ is only a homeomorphism onto its image if $V$ is assumed to be a Hausdorff topological vector space. Note that $f(\mathbb{C})$ inherits a Hausdorff topology from $V$ and use that there is only one Hausdorff vector space topology on $\mathbb{C}^n$.

Perhaps an easier way to see this is to assume $V$ to be locally convex in addition, and to use the Hahn-Banach theorem to find a continuous linear functional $\varphi \colon V \to \mathbb{C}$ such that $\varphi(f(1)) = 1$, so that $\varphi|_{f(\mathbb{C})}$ is bijective and continuous.

Answer 2

If $x\ne 0$, the map $v\mapsto xv$ is a homeomorphism since its reciprocal is multiplication by $x^{-1}$. It is even an isomorphism of topological vector spaces.

The map $\;\begin{aligned}[t]\mathbf C&\to\langle v\rangle\\[-1ex]x&\mapsto xv\end{aligned}$ is also an isomorphism of topological vector spaces, if $V$ is Hausdorff, because on the finite dimensional vector space $\langle v\rangle$, there is only one topology $\mathcal T$ such that $(\langle v\rangle, \mathcal T)$ is Hausdorff