Say we have a smooth bijective function $f(z)$ where $f: \mathbb{R}^N \to \mathbb{R}^N$ whose Jacobian matrix can be factored using a singular value decomposition $\frac{df(z)}{dz}=USV^T$.
Can we always find smooth bijective functions $u(.), s(.), \bar{v}(.)$ so that $f=u\circ s \circ \bar{v}$ and the Jacobian matrices of $u,s,\bar{v}$ are $U,S,V^T$ respectively?
My guess is that for any $f$ the answer is no, but I'm not sure how to go about showing this.