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The definition of an invertible function that my book (Apostol's Mathematical Analysis) gives is:

A function $f:S \to\mathbf{R}^n$, where $S$ is open in $\mathbf{R}^n$, has a unique inverse if $f$ is $C^1$ on $S$ and $J_f(\mathbf{a})\not = 0$ on $S$ (where $J_f(\mathbf{a})$ is the Jacobian determinant of $f$ at $\mathbf{a}$).

This definition is fine for functions from $\mathbf{R}^n$ to $\mathbf{R}^n$, but what about for $g:\mathbf{R}^n\to\mathbf{R}^m$? In this case the determinant does not exist, so this theorem does not apply! How do I determine whether such a $g$ is invertible?

Also, for a function $f$ between topologies we know that $f$ has an inverse if it is bijective - does that apply here as well? Is it sufficent to show that $g$ is bijective? Is the inverse of $g$ always unique?

Myshkin
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alexvas
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  • for a function f between topologies we know that f has an inverse if it is bijective: I might be misunderstanding what you mean here, but a even a continuous bijection may not have a continuous inverse. Inverse functions are always unique, though. – rschwieb Mar 20 '13 at 17:29
  • Perhaps I am misremembering, but I seem to recall that if $f:A\to B$ is injective then it has a right inverse (not necessarily continuous) and if $f$ is surjective then it has a right inverse (also not necessarily continuous). So if $f$ is bijective then it has a good-old-fashioned inverse $g$ such that $g\circ f(x) = x$ and $f\circ g(y) = y$. – alexvas Mar 20 '13 at 17:35
  • In general, if you are talking about sets, then yes, a set function has an inverse iff it is bijective. When you are talking about topological spaces and continuous maps between them, you would call a continuous map $f$ invertible iff there is a continuous map $g$ such that $fg$ and $gf$ are identities on their respective spaces. (This implies that they are bijections.) But in general, a continuous bijection may not have a continuous inverse, and so it may not be invertible. – rschwieb Mar 20 '13 at 18:00

2 Answers2

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In the case of different dimensions, you can't normally hope for an inverse function, if you are thinking of the problem in an analytic/differential set-up. In fact, it is a simple consequence of Sard's theorem that if $g$ is $C^1$, then the image of $g$ will have Lebesgue measure zero if $m > n$, so it will be rather "small". In particular, $g^{-1}$ won't make sense on any open set.

On the other hand, if the derivative of $g$ (the Jacobian matrix, or the tangent map, if you prefer) has full rank, i.e. $n$ (assuming $m > n$), then the image of $g$ will locally be manifold, and $g$ will locally be invertible. By this I mean that for a point $x \in R^n$, you can find a small neighbourhood, such that $g$ restricted to that neigbourhood maps it to bijectively and smoothly to its image.

Jakub Konieczny
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  • "if the derivative of $g$ has full rank, then [...] g will locally be invertible." That's exactly what I was looking for. Thanks! – alexvas Mar 20 '13 at 17:40
  • One question, though: does the Jacobian having full rank tell me anything about the inverse being $C^1$? In your last sentence, did you mean "such that the inverse of $g$ restricted to that neigbourhood maps bijectively and smoothly to its image."? – alexvas Mar 20 '13 at 18:09
  • @alexvas: What I meant was: for any $x \in R^n$, you can find an open neighbourhood $U$ of $x$, such that $g:\ U \to g(U)$ is smooth and bijective, and $g^{-1}:\ g(U) \to U$ is smooth and bijective (smooth = as smooth as $g$ is assumed to be). Mind that $g(U)$ is a submanifold of $R^m$, with dimension $n$. – Jakub Konieczny Mar 20 '13 at 18:14
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Consider a nice smooth map from $\mathbb{R}^1$ into $\mathbb{R}^2$ whose image is like a loop de loop. You would have a two-to-one point.

Consider a nice smooth map from $\mathbb{R}^2$ into $\mathbb{R}^1$. Locally you will have smashed a two-dimensional vector space into a one-dimensional (or lower) space, so there are many-to-one points.

2'5 9'2
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  • Sure - around the point that breaks injectivity the function would not be invertible (this is consistent with bijective => invertible), but there are plenty of non-bijective maps from $\mathbf{R}^1$ to $\mathbf{R}^2$ that are locally invertible. For instance, any graph is invertible in any ball where it is bijective. I'm trying to find the necessary condition for a function between dimensions to be invertible. – alexvas Mar 20 '13 at 17:45
  • @alexvas You should check out the examples given at http://math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous – rschwieb Mar 20 '13 at 18:02
  • @alexvas I didn't interpret your question to be about local invertibility, so my answer was meant to show that even very nice conditions on the map (smoothness) will not be able to guarantee invertibility if $m\neq n$. – 2'5 9'2 Mar 20 '13 at 22:05