# Diffeomorphisms, Vector fields and Push Forwards.

Studying for an exam, and trying to get my head around the concepts of push forwards.

The question I'm attempting to answer is:

"Give an example of a continuously differentiable diffeomorphism F and a continuously differentiable vector field X, such that the push forward of X is continuous but not differentiable." The answer given in the back of the book is:

On the real numbers, take

$F(x) =$ $x^2$ if $x \ge 0$; $−x^2$ if $x \gt 0$,

and $X(x) = 1$.

(Apologies for the awful formatting - I'll edit again in future when I work out how to use MathJaX to get piecewise functions!)

I'd really appreciate if someone could explain why this example satisfies the question.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – amWhy Nov 14 '12 at 14:56
Thanks for the advice. I'll edit it to make it clearer! – user49394 Nov 14 '12 at 15:43
Thanks for being so responsive. – amWhy Nov 14 '12 at 16:02

If $F: M \to N$ is a diffeomorphism and $X$ is a vector field on $M$ then the pushforward vector field $dF~X$ is defined by $(dF~X)(y) = dF\vert_{F^{-1}(y)} X_{F^{-1}(Y)}$. In your case $dF\vert_x$ is multiplication by $2|x|$ and the inverse of any $y \in \mathbb R$ is given by $\sqrt{|y|}$ if $y > 0$ and $-\sqrt{|y|}$ if $y < 0$.
Therefore $(dF~X)(y) = 2\sqrt{|y|}$, which is continuous but not differentiable at the origin.