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If $U\subset\mathbb{R}^n$ is an open and contractible subset such that there is a continuous function $f\colon U\to\mathbb{R}$ with only one minimum and the level curves of $f$ are connected by paths, then is $U$ homeomorphic to $\mathbb{R}^n$?

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If the function were a Morse function (in particular, smooth) this would follow from Morse theory—maybe that is enough for your? (Could the level sets not be path connected in your situation?) –  Mariano Suárez-Alvarez Jun 25 '12 at 17:57
    
@Mariano: You can make a spiral construction in $\mathbb{R}^2$ such that that the inverse of a value (say 1) is the unit circle and some curves in the unit disc spiraling out towards the unit circle, but never getting there. You can put the unique minimum at 0 taking the value 0. The inverse image of 1 would be connected but not path connected. –  Thomas Kragh Jun 25 '12 at 18:21
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If $U$ contains the origin, then $f=x_1^2+...+x_n^2$ satisfies the first condition and is a Morse function, but the set is not necessarily homeomorphic to $\mathbb R^n$. ( math.stackexchange.com/questions/55114/… ) Level sets can easily fail to be path-connected, for instance take $x_1^2+...x_{n-1}^2<1$, then '$\{f>1\}$' has two components. –  Will Sawin Jun 25 '12 at 18:22
    
You should assume your function to be proper. Perhaps, in case of not differentiability, please assume also that, away from the minimum point, $f$ is locally homeomorphic to the map $(x_1, \dots, x_n) \mapsto x_1$. This should suffices. –  Daniele Zuddas Jun 25 '12 at 18:27
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What about Whitehead manifold? –  Anton Petrunin Jun 25 '12 at 18:36
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2 Answers

See Hidden convexity (almost, but not quite a duplicate)

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@Mariano: In my situation the function is not a Morse function and the curves of level are connected by paths, I think that this hypothesis about U are enough, but I can't proof this affirmation.

@Anton: I don't know if the Whitehead manifold has a function with this properties.

I want to proof that the compact sets $K_m = U-f^{-1}([0,m])$ (assuming that $0$ is the minimum value of $f$) are simply-connected, this will implies that $U$ is simply-connected to infinite.

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This should really have been a comment below Mariano's answer (well, the comment to Anton would have gone in the comment thread below the answer). There is an extensive wikipedia page on the Whitehead manifold. It is an open 3-manifold which is contractible but not homeomorphic to $\mathbb{R}^3$. You'd still have to find a map from it to $\mathbb{R}$ with only one minimum, but basically Anton was giving evidence that the answer to your question is no. The terminology "simply-connected to infinity" is not standard, but does appear in the SE thread Will Sawin linked to –  David White Jul 24 '12 at 12:44
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