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seen Sep 14 at 4:04

May
30
comment Height and minimal number of generators of an ideal
See here for your last comment.
May
24
revised Are maps locally preserving collinearity homographies?
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May
20
comment Are maps locally preserving collinearity homographies?
By the way, it's unfair that such a good answer didn't accept desirable upvotes.
May
20
comment Are maps locally preserving collinearity homographies?
I think it even works for any domains (connected open regions), since $\mathbb R^2$ is locally path-connected and we can use consecutive discs to cover the path connecting two points. I haven't written up the rigorous proof.
May
18
comment Why is the projection map proper?
Okay, edited. I hope you can check it.
May
18
revised Why is the projection map proper?
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May
18
comment Why is the projection map proper?
Thanks. When the degrees (the sum of multiplicities for each point) are the same, we can choose a neighborhood $W$ of $y_0$ such that the cardinality of $f^{-1}(y)$ is just the degree of $y_0$ for $y\in W\setminus\{y_0\}$, then my EDIT1 should work since the local behavior of $f$ is just $z\mapsto z^n$, right? At a glance, I think you're using Rouché theorem just to show the very fact of the local behavior. I don't like go into sequential compactness even for metric spaces.
May
18
revised Why is the projection map proper?
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May
18
accepted Why is the projection map proper?
May
18
revised Why is the projection map proper?
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May
18
revised Why is the projection map proper?
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May
18
revised Why is the projection map proper?
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May
18
revised Why is the projection map proper?
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May
18
comment Why is the projection map proper?
The proof seems quite involved. I got a plausible generalization and I hope you can help me to check it, and if fortunate, it could be used to simplify the proof and clarify the logic.
May
18
revised Polynomial that is surjective $\mod n$ for all $n$?
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May
17
answered Polynomial that is surjective $\mod n$ for all $n$?
May
17
comment Polynomial that is surjective $\mod n$ for all $n$?
and $x\mapsto\pm x+m$, $m\in\mathbb Z$.
May
17
revised Why is the projection map proper?
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May
17
asked Why is the projection map proper?
May
17
comment Prove that f is differentiable at $0$! Not continuous though, Right!?
It's differentiable at $0$ since $\lim_{x\to0}f(x)/x=0$, for $0\le\lvert f(x)/x\rvert\le\lvert x\rvert$. It's indeed continuous at $0$.