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.


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

  • 1
    $\begingroup$ 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. $\endgroup$ – Vokram8 Apr 3 at 23:23
  • $\begingroup$ @Vokram8 Thanks. I corrected and updated the answer. $\endgroup$ – Alex Ravsky Apr 4 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.