I am researching different generalizations of the familiar open mapping theorem from functional analysis. Every "proof" I attempt while simply assuming positive-homogeneity, even in the finite-dim case, has hit a giant brick wall, making me think it is not true, but I cannot seem to write down a counter-example.

Is there an example in which a map $f\colon\mathbb{R}^m\longrightarrow\mathbb{R}^n$ is positively-homogeneous, onto, and not open?

For simplicity, why not take $f$ to be analytic and homogeneous of degree $k\in\mathbb{N}$ and $m\geq n$?

  • $\begingroup$ "Homogeneous" here means $f(cx) = c^k f(x)$? $\endgroup$ – Nate Eldredge Nov 8 '15 at 16:10

Here is an example. In complex notation, $f(x+iy):=(x^2+iy^2)^4$ defines a homogeneous polynomial map from $\mathbb{R}^2$ to $\mathbb{R}^2$, which is clearly surjective, but not open (for instance, the image of the open set $\{(x,y)\in\mathbb{R}^2: |y|<|x|\}$ is the upper open half-plane plus the half-line $\{(x,0)\in\mathbb{R}^2: x>0\}$, not open).

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    $\begingroup$ Interesting, fold the plane up into the first quadrant, then stretch it back around to the full plane. Cool example, thanks! Taking the "regularity" a step further, can you think of one such that $f$ is also multi-linear? $\endgroup$ – charlestoncrabb Nov 9 '15 at 18:07

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