# Existence of a particular inverse transformation

Let $$h : \mathbb{R}^D \rightarrow \mathbb{R}^d$$, where $$d < D$$, be a differentiable function. I would like to find minimal conditions under which there exists a differentiable function $$g : \mathbb{R}^{D} \rightarrow \mathbb{R}^{D-d}$$ such that the function $$f : \mathbb{R}^D \rightarrow \mathbb{R}^D$$ defined by $$f(x)=(h(x)^\top, g(x)^\top)^\top$$ is invertible. If possible, I would also like to obtain a construction of this $$g$$ function. I hypothesize that the following conditions might be enough, but I am not sure:

-$$h$$ is surjective.

-$$h$$ cannot have the same value on a set with non-zero measure, that is, for every $$y \in \mathbb{R}^{d}$$, the set $$h^{-1}(\{y\})=\{x\in\mathbb{R}^D : h(x)=y\}$$ has Lebesgue measure 0.

The first condition is clearly necessary, and the reason why I believe that the second condition might be enough is the following:

In order for $$f$$ to be invertible, $$f^{-1}(\{z\})$$ has to be a singleton for every $$z \in \mathbb{R}^D$$. If $$g$$ was such that $$g(x_1)\neq g(x_2)$$ for every $$x_1$$ and $$x_2$$ such that $$h(x_1)=h(x_2)$$, that would ensure that $$f^{-1}(\{z\})$$ is indeed a singleton for every $$z \in \mathbb{R}^D$$. My intuition is that the second condition might ensure that such a $$g$$ function actually exists. Furthermore, if such a $$g$$ exists that also makes $$f$$ surjective, the result would follow.

Any help would be very appreciated, either in proving my above conjecture, disproving it, or providing non trivial assumptions about $$h$$ that would result in the existence of $$g$$.

Thank you very much!

Assume that a required function $$f$$ exists. Since $$f$$ is differentiable, it is continuous, so by the invariance of domain, $$f$$ is a homeomorphism.

Then for each $$t\in\Bbb R^{D-d}$$ a restriction $$f^{-1}|\Bbb R^{d}\times\{t\}$$ is a homeomorphism onto the image $$L_t=f^{-1}(\Bbb R^{d}\times\{t\})$$. But $$f(x)=(h(x)^\top, g(x)^\top)^\top= (h(x)^\top, t^\top)^\top$$ for each $$x\in L_t$$. Thus $$h|L_t$$ is a homeomorphism onto the image. We shall call a subset $$L$$ of $$\Bbb R^D$$ an $$h$$-layer, if $$h|L:L\to\Bbb R^d$$ is a homeomorphism onto the image. Thus we have that $$\Bbb R^D$$ is a disjoint union of $$h$$-layers.

Also for each $$s\in\Bbb R^d$$ a set $$h^{-1}(s)=f^{-1}(\{s\}\times \Bbb R^{D-d})$$ is a homeomorphic image of a space $$\Bbb R^{D-d}$$.

Both conditions can fail for a function satisfying the conjecture. For instance, let $$D=2$$, $$d=1$$, and $$h(x,y)=x^3-x$$ for each $$(x,y)\in\Bbb R^D$$. Then there are no $$h$$ layers. Indeed, let $$L$$ be any connected set such that $$h(L)=\Bbb R^d$$. Then there exists $$y_{-1}$$ and $$y_1$$ in $$\Bbb R$$ such that both points $$p_{-1}=(-1,y_{-1})$$ and $$p_1=(1,y_{1})$$ belongs to $$L$$. But $$h(p_{-1})=h(p_1)$$, so $$L$$ is not an $$h$$-layer. The second condition is violated because a set $$h^{-1}(0)=\{(x,y):x\in\{-1,0,1\},y\in\Bbb R\}$$ is disconnected and hence not homeomorphic to $$\Bbb R^{1}$$.

Moreover, it can be easily checked that both partitions of $$\Bbb R^D$$ into $$h$$-layers and preimages $$h^{-1}(s)$$ are parallel (see [BH] for a definition) with respect to a metric $$d$$ such that $$d(x,y)=\|f(x)-f(y)\|$$ for each $$x,y\in\Bbb R^D$$. Then similarly to the proof of the implication $$(1)\Rightarrow (2)$$ in Theorem 1 from [BH] we can show that each of these partitions $$\mathcal C$$ is lower semicontinuous and compactly upper semicontinuous. The latter means that for each compact subset $$F$$ of $$\Bbb R^D$$ its $$\mathcal C$$-star $$St(F;\mathcal C)$$ is closed in $$\Bbb R^D$$.

I guess that we can strengthen the above necessary conditions, if we define an $$h$$-layer $$L$$ as a submanifold of $$\Bbb R^D$$ such that $$h|L$$ is a diffeomorphism and require that a set $$h^{-1}(s)$$ for $$s\in\Bbb R^d$$ is a submanifold and a diffeomorphic image of a space $$\Bbb R^{D-d}$$. But I didn’t investigate this topic because I am a general topologist, not a differential one.

References

[BH] Taras Banakh, Olena Hryniv, A parallel metrization theorem, European Journal of Mathematics (2019), 1-4.

• Thank you, this does indeed disprove my conjecture. I just wanted to point out two small typos: in the definition of L_t it should be t and not 0. – Vokram8 Apr 3 at 23:23
• @Vokram8 Thanks. I corrected and updated the answer. – Alex Ravsky Apr 4 at 0:42