Open homogeneous functions are onto.

Question

Let $f:\mathbb{R}^{n}\to\mathbb{R}^{m}, n,m\in \mathbb{N}$ be a continuous function such that $f(\lambda x)=\lambda^{p}f(x)$(homogeneous), $p\geq0$.

Show that f is an open map, then f is surjective.

I don't have any idea.

Answer 1

Notation : $0_l \in \mathbb R^l$ denotes the $l$ dimensional zero vector.

First of all, note that if $f$ is homogeneous, then $f(0_n) = f(2 \times 0_n) = 2^pf(0_n)$, which implies that $f(0_n) = 0_m$ since $p$ is a positive integer.

Consequently, consider the open set $U = \{x \in \mathbb R^n : \sum_{i=1}^n x_1^2 < 1\}$. This is the open unit ball in $\mathbb R^n$. Since $f$ is open, it follows that $f(U)$ is open. Since $0_n \in U$, we have $0_m \in f(U)$, so $0_m$ is an interior point of $f(U)$ i.e. there exists $\delta > 0$ such that the set $V = \{y \in \mathbb R^m : \sum_{i=1}^m y_i^2 < \delta\}$ is a subset of $f(U)$.

Now, let $z \in \mathbb R^m = (z_1,...,z_m)$. Find $\lambda > 0 $ large enough so that for all $i$, we have $\frac{|z_i|}{\lambda} < \frac{\sqrt \delta}{\sqrt m}$. Now, it follows that the vector $\frac z \lambda$ lies in $V$, since $\sum \frac{z_i^2}{\lambda^2} < \frac{m \delta}{m} < \delta$. Finally, since $V \subset f(U)$, there exists $x$ such that $f(x) = \frac z{\lambda}$. Finally, $f(x \lambda^{\frac 1p}) = \lambda f(x) = z$.

Thus, $f$ is surjective.

Note that we only used the fact that $f$ is open at the point zero. Beyond this, we used the scaling property of open balls with homogeneity to conclude.