# Why is the image of the implicit function in the implicit function theorem not open?

We have a continuously differentiable function $f$ from $\mathbb{R}^{n+m}$ to $\mathbb{R}^n$, and we find a continuously differentiable function $g$ which maps points from $\mathbb{R}^m$ into $\mathbb{R}^n$ (this function defines $x \in \mathbb{R}^n$ "implicitly" in terms of $y \in \mathbb{R}^m$).

What regularity properties must the image of $g$ have as a direct result of the proof of the implicit function theorem (if any) in order to allow meaningful differentiable geometry (e.g. manifolds) to "happen"?

In the proof of the Inverse (not implicit) function theorem, we need to care about the properties of the image of $f$. Yet for some reason the properties of the image of $g$ in the Implicit (not inverse) function theorem don't matter? Why is that?

What properties, if any, of $g$ in the implicit function theorem, then, are important for the definitions of manifolds and tangent spaces and bundles?

This question seems to address a related issue, but the discussion seems to assume that the image of g is open. However, Rudin does not mention this in his proof, nor does it seem to necessarily follow from his proof (we only show the existence of two open sets in $\mathbb{R}^{n+m}$ and one in $\mathbb{R}^m$, but I'm asking about the image of the latter as a set in $\mathbb{R}^n$).

EDIT: "Isn't this unsatisfactory? How can we use the conclusions of the implicit function theorem if the implicit function is neither a homeomorphism nor a diffeomorphism?" --- I realize now that it can't be a diffeomorphism except for the case where $n=m$, because otherwise we would have a contradiction.

Context:

This is in some sense the sequel to a question I asked here.

In that question I asked why we wanted the image of the continuously differentiable function $f$ to be open.

The answer seemed to come down to having a sufficiently meaningful/useful definition of derivative apply to the inverse function so that we had in particular continuity of the inverse function and a homeomorphism. That having the image be open is necessary for this seems somewhat unclear, but appears to be justifiable using Brouwer's invariance of domain theorem.

However, in Rudin's proof of the implicit function theorem, it does not seem like it is necessary that the image of the "implicit function" $g$ be open.

Why don't we have the same problem for the implicit function theorem?

Wouldn't we need the image to be open for the inverse image of some set under $g$ to be a manifold? (Assuming it is supposed to somehow be a manifold?)

• I don't know how to answer, but I would like to point you to Bruce Blackadar's manuscript: wolfweb.unr.edu/homepage/bruceb (follow the link for the manuscript on Real Analysis). The chapter on inverse/implicit function theorem is very carefully crafted and it treats many subtle details that usually go uncared for in standard textbooks. – Giuseppe Negro Apr 19 '16 at 20:57
• I'll take a look; thank you! – Chill2Macht Apr 19 '16 at 22:16
• The manuscript definitely has a lot of information -- I haven't finished reading the section on the implicit function theorem yet, but I will try to finish it soon, and if I don't find the answer there, I will look in the references which the author mentioned. – Chill2Macht Apr 22 '16 at 19:53
• It's very unclear to me what you're asking for. If $f$ is just the standard coordinate projection, then $g$ is a constant function. Is there something you find wrong with this example? I can't recall ever seeing any application of the implicit function theorem where you care about properties of the image of $g$... – Eric Wofsey Apr 22 '16 at 22:48
• You care about the image of $f$, but why would you care about the image of $g$? The analogue of $f$ from the inverse function is $f$ here, not $g$... – Eric Wofsey Apr 22 '16 at 23:07

## 1 Answer

The function $g$ is much less important than you seem to think it is. You don't actually care about $g$; the function you care about is $f$, and you are just using $g$ to get a simple description of what $f$ looks like.

In particular, you never ask for the inverse image of a set under $g$ to be a manifold; you ask for the inverse image of a set under $f$ to be a manifold. For instance, when you take the inverse image of a single point $p$ under $f$, the implicit function theorem tells you that you get the graph of a differentiable function $g$, and the graph of any differentiable function is always a differentiable submanifold. Indeed, we know the graph is a differentiable submanifold because we can parametrize it by the function $\mathbb{R}^n\to\mathbb{R}^{n+m}$ sending $x$ to $(x,g(x))$. This has nothing to do with any special regularity properties of $g$ besides just being a function and being differentiable.

Note that it is a highly nontrivial statement that there even exists such a function $g$: it says that locally, once you choose the first $n$ coordinates of a point of $\mathbb{R}^{n+m}$, there is always exactly one way to choose the last $m$ coordinates to get a point which $f$ sends to $p$. The content of the implicit function theorem isn't that the function $g$ is super-nice, but merely that it exists at all. It is $f$ which you want to be super-nice, and the mere existence of $g$ is all you need to conclude useful facts about $f$ (e.g., that $f^{-1}(\{p\})$ is a manifold).

• This is really helpful -- I'll come back in 23 hours and award the bounty when the site finally lets me. – Chill2Macht Apr 22 '16 at 23:27
• One last stupid question though -- how does the implicit function theorem show that the inverse image of p under f is the graph of g? I have trouble imagining this visually/spatially, and I'm not sure how to demonstrate it analytically either. But it seems like the crucial point connecting the implicit function theorem with the theory of manifolds, so I would really like to have a strong understanding of it. – Chill2Macht Apr 22 '16 at 23:28
• This depends on exactly what your statement of the implicit function theorem is. For some versions (e.g. Wikipedia's), the fact that the inverse image of $p$ is the graph of $g$ is exactly what the theorem says – Eric Wofsey Apr 22 '16 at 23:50
• I have the statement from Baby Rudin in mind: "Let f be a continuously differentiable mapping of an open set $E \subset R^{n+m}$ into $R^n$ such that f(a,b)=0 for some (a,b) in E. Put A=f'(a,b) and assume that $A_x$ is invertible. Then there exist open sets $U \subset R^{n+m}$ and $W \subset R^m$, with (a,b) in U and b in W, having the following property: to every y in W corresponds a unique x in $R^n$ such that (x,y) is in U and f(x,y)=0. If this x is defined to be g(y), then g is a continuously differentiable mapping of W into $R^n$, g(b)=a, f(g(y),y)=0 for y in W, and g'(b)=$-(A_x)^{-1}A_y$ – Chill2Macht Apr 23 '16 at 0:05
• What I'm calling $p$ is just $0$ in Rudin's formulation (because we are looking at solutions of $f(a,b)=0$). The graph of $g$ is by definition the set of points of the form $(g(y),y)$, and Rudin's formulation tells you that every point $(x,y)$ in $U$ such that $f(x,y)=0$ is of this form. (In my answer, I called the graph the set of points of the form $(x,g(x))$ instead, but obviously this doesn't really matter.) – Eric Wofsey Apr 23 '16 at 0:41