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Let $h : mathbbR^D rightarrow mathbbR^d$, where $d < D$, be a differentiable function. I would like to find minimal conditions under which there exists a differentiable function $g : mathbbR^D rightarrow mathbbR^D-d$ such that the function $f : mathbbR^D rightarrow mathbbR^D$ 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 $y in mathbbR^d$, the set $h^-1(y)=xinmathbbR^D : h(x)=y$ 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 $z in mathbbR^D$. 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 $z in mathbbR^D$. 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 $tinBbb R^D-d$ a restriction $f^-1|Bbb R^dtimest$ is a homeomorphism onto the image $L_t=f^-1(Bbb R^dtimest)$. But $f(x)=(h(x)^top, g(x)^top)^top= (h(x)^top, t^top)^top$ for each $xin L_t$. Thus $h|L_t$ is a homeomorphism onto the image. We shall call a subset $L$ of $Bbb R^D$ an $h$-layer, if $h|L:LtoBbb R^d$ is a homeomorphism onto the image. Thus we have that $Bbb R^D$ is a disjoint union of $h$-layers.

Also for each $sinBbb R^d$ a set $h^-1(s)=f^-1(stimes Bbb R^D-d)$ 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 $(x,y)inBbb R^D$. 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):xin-1,0,1,yinBbb 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 $x,yinBbb R^D$. 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 $sinBbb R^d$ 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.