True or false: continuous image of convex is convex

Question

Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be a continuous function.

Is it true that $f(A)$ is convex given that $A$ is convex?

This claim seems to be quite intuitively true, but maybe I am using my intuition too much for $\mathbb{R} \to \mathbb{R}$ type functions. I cannot find references on MSE, nor can I find a general result elsewhere.

What I was able to find is the following,

Given $f: \mathbb{R}^n \to \mathbb{R}^m$, if $m \leq n$, and $f$ is $C^{1,1}$, the map $f^\prime(a): \mathbb{R}^n \to \mathbb{R}^m$ surjective at some point $a \in \mathbb{R}^n$, then image $K = f(B(a,\epsilon))$ of ball $B(a,\epsilon) = \{x \in \mathbb{R}^n: \|x-a\|\leq \epsilon\}$ of sufficiently small radius $\epsilon$ is convex.

in B. T. Polyak, “Convexity of nonlinear image of a small ball with applications to optimization".

which is not a global result...does it mean that it is not true in general?

Answer 1

The function $f:\Bbb R^2\to\Bbb R^2:\langle x,y\rangle\mapsto \langle x,y+x^2\rangle$ is continuous and sends the $x$-axis, which is convex, to the graph of $y=x^2$, which is not.

Answer 2

No. For example, take $A$ to be the open upper half-plane and $f(z) = z^2$ regarded as a function $(\operatorname{Re}z,\operatorname{Im}z)\mapsto (\operatorname{Re} z^2,\operatorname{Im}z^2)$. Then $f(A)$ is a slit plane (the origin and positive x-axis is removed).