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Why if $\varphi:U\subseteq\mathbb{R}^N\to V\subseteq\mathbb{R}^N$ is bijective and $\varphi, \varphi^{-1}$ are of class $C^1$ then the jacobian:

$\det(D\varphi (x))\neq 0,\ \forall\ x\in U$???

Bogdan
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2 Answers2

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Because both $\varphi$ and $\varphi^{-1}$ are $C^1$ you can apply the chain rule and you get $$D\varphi^{-1}(\varphi(x))\circ D\varphi(x)=D(\varphi^{-1}\circ \varphi)(x)=D(id_U)(x)=Id.$$ Hence $$\det(D\varphi^{-1}(\varphi(x)))\det\left ( D\varphi(x)\right)=\det(Id)=1$$ so $\det\left ( D\varphi(x)\right)\neq 0$.

Adam Chalumeau
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  • 6
    it might be good to emphasize that we can apply the chain rule precisely because both $\varphi$ and $\varphi^{-1}$ are already known to be differentiable – peek-a-boo Jul 03 '19 at 15:32
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Let $x\in U \subseteq \mathbb R^n$

$$\varphi^{-1}(\varphi(x))=x=id(x)$$

Take the Jacobian on both sides* and apply the chain rule on the left-hand side**:

$$D\varphi^{-1}(\varphi(x))\cdot D\varphi(x) = I_n$$

So $D\varphi(x)$ is invertible $\forall x \in U$.

*Can do this since functions are differentiable. **Can do this since they are $C^1$.

Lance
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