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If $A\subset \mathbb R^n$ is a convex subset and $A$ is homeomorphic to a subset $B\subset R^n$, can I also say $B$ is convex? Any counterexamples?

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    For a question such as this, it is worth realizing a topological space as a rubber sheet (or plasticine, whichever you prefer), understanding mappings (functions) as stretching of the rubber.2017-01-16
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    Affine maps preserve convexity.2017-01-16
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    Take a circle in $\mathbb R^2$ and pinch it.2017-01-16

2 Answers 2

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Define, for $(x,y) \in \mathbb R^2$: $$ \varphi(x,y) := (x-|y|,y) $$ Then $\varphi$ is a homeomorphism $\mathbb R^2 \to \mathbb R^2$ but the image of $\{(x,y)\mid x = 0\}$ under $\varphi$ is not convex.

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    Do you know if a diffeomorphism of class $C^2$ is convex-preserving?2017-01-16
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    @user42912 Take $\varphi$ to be the transformation to polar coordinates and consider the image of a retangle.2017-01-16
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    @user42912 every simply-connected subset of $\mathbb{C}$ is equivalent to the unit disc via a bi-holomorphic map. That is, infinitely differentiable (actually, analytic) conformal, bijective with an inverse that has the same properties.2017-01-16
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My preferred is $\varphi:(x,y)\mapsto (x,y^3)$. It is a homeomorphism ($\varphi^{-1}:(x,y)\mapsto (x,\sqrt[3]{y}$) and the image of the first diagonal is $y=x^3$ which is not convex.