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Assume that $f:\mathbb{R}^N\to\mathbb{R}^N$ is a surjective function and in addition suppose that $f$ is a local diffeomorphism. Take two points in the image of $f$, let's say, $f(x),f(y)$ with $f(x)\neq f(y)$.

Is it possible to find a continuous curve $\alpha :[0,1]\to\mathbb{R}^N$ such that $\alpha(0)=x$, $\alpha(1)=y$ and $f(\alpha (t))=(1-t)f(x)+tf(y)$.

Remark: This question is related to this one

Remark 1: The answer given here by @smiley06, solves this problem, when only one boundary condition is prescribed.

Thank you

Tomás
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2 Answers2

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All the credit goes to the user8268 but I made a little animation.

enter image description here

tom
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For $N=2$; I replace $\mathbb R^2$ with a disk (as it is diffeomorphic); here is a surjective map from a disk to a "disk" (still diffeomorphic to $\mathbb R^2$) and a straight segment that can't be lifted.

really horrible picture

user8268
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