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Is there any real analytic diffeomorphism from two dimensional disk to itself, except to the identity, such that whose restriction to the boundary is identity?

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    @Ryan: you're right. My bad!2011-12-01

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$f(x,y) = (x,y) + (-x,-y)(x^2+y^2)(1-x^2-y^2)$

Doesn't the above map do the job? I'm using the disc in $\mathbb R^2$ given by $x^2+y^2 \leq 1$.

If you want one without a fixed point in the interior,

$f(x,y) = (x,y) + \left(\frac{1}{10},0\right)(1-x^2-y^2)$

The fraction $\frac{1}{10}$ just needs to be a positive number strictly smaller than $1/2$.

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    I'll edit in one.2011-11-30
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$f(z)=ze^{2\pi|z|^2i}$ should do the trick.

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    We perturbed the identity map in different directions.2011-11-30
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Some that spring to mind are $(x,y) \to (x+ c (1 - x^2 - y^2), y)$ where $-1/2 < c < 1/2$, $c \ne 0$.