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We learnt the Inverse Function Theorem for multi-variable functions, and it only dealt with "local" inverses, not "global" inverses. Is my interpretation of a global inverse just that there exists an inverse around ALL points in the domain?

Here is the question related.

Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $$ f(x,y) = \begin{pmatrix} e^x\cos y \\ e^x\sin y \end{pmatrix}. $$ Show that $f$ has a local $C^1$ inverse at every point in $\mathbb{R}^2$, but it has not any global inverse on $\mathbb{R}^2$.

What I did was basically found the Jacobian matrix of $f$, and showed that its determinant is zero iff $x$ approaches infinity. So for every $(x,y)$ with $x$ finite, then Jacobian of $f$ is invertible and hence there is a local $C^1$ inverse. But with large $x$, there is no inverse, so no global inverse?

user1551
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Twenty-six colours
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2 Answers2

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Before getting started, let us go back on some definitions:

Definition. Let $U$ be an open subset of $\mathbb{R}^n$ and let $f\colon U\rightarrow V\subseteq\mathbb{R}^n$ be a $C^1$-mapping, then $f$ is a $C^1$-diffeomorphism or globally invertible if and only if there exists $g\colon V\rightarrow U$ a $C^1$-mapping such that: $$f\circ g=\textrm{id}_{V}\textrm{ and }g\circ f=\textrm{id}_U.$$

In other words, $f$ is a bijection whose inverse is smooth. This is not to be confused with:

Definition. The mapping $f$ is said to be a local diffeomorphism or locally invertible if and only if when retricted to an open subset of $U$, $f$ is a diffeomorphism onto its image.

Please notice that being globally invertible does not mean being locally invertible around any point.

As a reminder, let us state the inverse function theorem once again:

Theorem. Let $U$ be an open subset of $\mathbb{R}^n$, let $x$ be a point of $U$ and let $f\colon U\rightarrow\mathbb{R}^n$ be a $C^1$-mapping. Assume that $\mathrm{d}_xf\colon\mathbb{R}^n\rightarrow\mathbb{R}^n$ is an invertible linear map, then there exists $V$ an open neighbourhood of $x$ such that $f\colon V\rightarrow f(V)$ is a $C^1$-diffeomorphism.

In our case, for all $(x,y)\in\mathbb{R}^2$, the matrix of $\mathrm{d}_{(x,y)}f$ in the canonical basis of $\mathbb{R}^2$ is: $$\begin{pmatrix}e^x\cos(y)&-e^x\sin(y)\\e^x\sin(y)&e^x\cos(y)\end{pmatrix}.$$ Its determinant is $e^{2x}$ which is nonzero for all $(x,y)$. Theorefore, according to the theorem, for all $(x,y)\in\mathbb{R}^2$, $f$ is a $C^1$-diffeomorphism in a neighborhood of $(x,y)$.

Assume by contradiction that there exists $g\colon\mathbb{R}^2\rightarrow\mathbb{R}^2$ a $C^1$-mapping such that: $$g\circ f=\textrm{id}_{\mathbb{R}^2}.$$ Then, let $(X,Y)\in(\mathbb{R}^2)^2$ such that $f(X)=f(Y)$, then $g(f(X))=g(f(Y))$ i.e. $X=Y$ and $f$ is injective. However, $f$ is non injective since $f(0,0)=(1,0)=f(0,2\pi)$ and $(0,0)\neq (0,2\pi)$, a contradiction. Whence, $f$ is not a $C^1$-diffeomorphism.

Maybe it will give more insight to notice that if one identifies $\mathbb{R}^2$ with $\mathbb{C}$ through $(x,y)\mapsto x+iy$, then the considered mapping is the complex exponential, $z\mapsto e^z$ which is invertible on any horizontal strip of length at most $2\pi$.

C. Falcon
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  • Thanks. Why did you have to show by contradiction? Couldn't you have already shown that $f$ is not injective with that example? – Twenty-six colours May 14 '17 at 04:58
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    The whole point is why $f$ must be injective? That is why is assume that $f$ is globally invertible, then deduced that $f$ is injective, which is a contradiction. – C. Falcon May 14 '17 at 12:31
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    Could you please elaborate on "globally invertible does not mean being locally invertible around any point"? Do you mean that it is not equivalent (as your answer proves) or that there exist globally invertible functions which are not locally invertible around at least one point? – anderstood Aug 30 '17 at 20:52
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    @anderstood If $f$ is a smooth map, $f$ is invertible means that $f$ is a diffeomorphism, namely there exists $g$ a smooth map such that $f\circ g$ and $g\circ f$ are the identity maps. On the other hand, $f$ is locally invertible at a point means that there exists a neighbourhood of this point on which $f$ is a diffeomorphism. My point is that a function can be locally invertible everywhere without being a diffeomorphism. – C. Falcon Aug 30 '17 at 21:51
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    Thanks for the clarification. I was not sure if you meant "mean" as $\Leftrightarrow$ or as $\Rightarrow$... if you see what I mean :) – anderstood Aug 31 '17 at 01:08
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The idea is that the inverse function theorem guarantees, on a neighbourhood of each $(x,y)$, the existence of a local inverse (because the Jacobian is never zero, and the function is $C^1$). However there cannot be a global inverse; note, for example, that $f$ is not injective.

  • I might not be understanding the difference between a local inverse and a global inverse. if there is a neighbourhood of each $(x,y)$ that there is existence of a local inverse, then why can we not say that it is invertible everywhere (since for every point, theres a local inverse)? – Twenty-six colours Apr 28 '17 at 11:29
  • Consider $f(x)=x^2$. The domain of f is $\mathbb{R}$. What is the global inverse of $f$? Does it exist? Now restrict f to the interval $[1,2]$, for example. Is this restriction invertible? –  Apr 28 '17 at 11:35
  • With the question in OP, so it is not globally invertible because it isn't injective, how would we show this? Does this mean that by the Inverse Function Theorem, it is locally injective (and surjective) for each $(x,y)$ for some neighbourhood of $(x,y)$? – Twenty-six colours Apr 28 '17 at 12:14
  • @Twenty-sixcolours: it fails to locally invertible about $x=0$ as $f'(0)=0$, and this is what breaks us from concluding that local invertibility on either side give rise to global invertibility – obataku Dec 27 '17 at 06:34