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Suppose $f:\mathbb{R}^2\to\mathbb{R}$ is a nonconstant polynomial function.

If $f$ is an open mapping, then $f$ must be a surjection?

Any help will be appreciated.

David Chan
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  • I think so. If f isn't surjective then Image(f) will be bounded above or below [I think that's easy to show as a polynomial is continuous] and have a max or min (i.e. the sup Image(f) is in Image(f); I think that is easy to show.) The max and min will not be an interior point of Image(f) so Image(f) isn't open, so it isn't an open mapping. I think. – fleablood Oct 23 '15 at 04:51

1 Answers1

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Consider $f(x,y)=x^2+(xy-1)^2$. Then $f$ is always positive, so it is not surjective; I claim $f$ is open. Indeed, it is easy to check using calculus that $f$ has no local minima or maxima, so $f$ is open by my previous answer.

Community
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Eric Wofsey
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