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For which $m$, $n$ (if any) is the following true: if $M$ and $M'$ are smooth manifolds of dimension $n$, and $\Phi$ is a bijection from $M$ to $M'$ such that for any subset $S$ of $M$, $\Phi(S)$ is an embedded submanifold of $M'$ of dimension $m$ iff $S$ is an embedded submanifold of $M$ of dimension $m$, then $\Phi$ is a diffeomorphism?

This is clearly false when $m=0$ and when $m=n$. I thought it looked plausible when, for example, $m=1$ and $n\geq 2$; but I can't see how to prove it.

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I think a similar question (possibly in the topological category?) came up on MO. –  Qiaochu Yuan Mar 14 '11 at 17:31

2 Answers 2

(This is a bit long for a comment, but it's not a full answer)

Here's where I would start when looking at this question.

Theorem: The category of smooth manifolds is a full subcategory of the category of Frölicher spaces.

First, we need to know what a Frölicher space is! It is triple $(X,C,F)$ where $X$ is a set, $C \subseteq \operatorname{Map}(\mathbb{R},X)$ is a family of curves in $X$ and $F \subseteq \operatorname{Map}(X,\mathbb{R})$ is a family of functionals on $X$. Note that these are just set maps. The curves and functionals have to satisfy a compatibility condition:

  1. $\alpha \in C$ if and only if $\phi \circ \alpha \in C^\infty(\mathbb{R},\mathbb{R})$ for all $\phi \in F$, and
  2. $\phi \in F$ if and only if $\phi \circ \alpha \in C^\infty(\mathbb{R},\mathbb{R})$ for all $\alpha in C$.

A morphism of Frölicher spaces is a set map $f \colon X \to Y$ which satisfies the following (equivalent) conditions:

  1. $f \circ \alpha \in C_Y$ for all $\alpha \in C_X$,
  2. $\phi \circ f \in F_X$ for all $\phi \in F_Y$,
  3. $\phi \circ f \circ \alpha \in C^\infty(\mathbb{R},\mathbb{R})$ for all $\alpha \in C_X$ and $\phi \in F_Y$.

The chain rule is enough to show that there is a functor from the category of smooth manifolds to that of Frölicher spaces, where we assign to a smooth manifold $M$ the Frölicher space $(M, C^\infty(\mathbb{R},M), C^\infty(M,\mathbb{R}))$. We want to show that this embeds the category of smooth manifolds as a full subcategory.

Firstly, that it is faithful is obvious since both categories are concrete (that is, equipped with a faithful functor to $\operatorname{Set}$) and the functor preserves this structure.

So we just need to show that it is full. As both categories are concrete (and the functor preserves this structure), it is sufficient to show that if $f \colon M \to N$ is a set map which is not smooth, then it is not a morphism in the Frölicher category. So let $M$ and $N$ be two smooth manifolds and $f \colon M \to N$ a set map which is not smooth. Then there is a chart for $M$ and a chart for $N$ which detects this non-smoothness. That is to say, there are smooth maps $\theta_M \colon U_M \to V_M \subseteq M$ and $\theta_N \colon U_N \to V_N \subseteq N$ such that $g = \theta_N \circ f \circ \theta_M$ both makes sense and is not smooth.

Now $g$ is a map from an open subset of some Euclidean space to an open subset of some Euclidean space. A map in to an open subset of a Euclidean space is smooth if and only if it is smooth when considered as a map into the ambient space, and then a map into a Euclidean space is smooth if and only if all the compositions with the projections are smooth. Thus we can find some projection $\pi_i \colon \mathbb{R}^n \to \mathbb{R}$ such that $\pi_i \circ \iota \circ g \colon U_M \to \mathbb{R}$ is not smooth. Furthermore, as $U_M$ is an open subset of some Euclidean space, we can find an open disc such that the restriction of this map to that disc is not smooth, and then composing with a diffeomorphism of the ambient space to that disc provides us with a non-smooth map $\mathbb{R}^m \to \mathbb{R}$.

Now comes the magic step. We use a result of Jan Boman which says that a map $\mathbb{R}^m \to \mathbb{R}$ is smooth if and only if its compositions with all smooth curves are smooth. That is, $h \colon \mathbb{R}^m \to \mathbb{R}$ is smooth if and only if $h \circ \gamma \colon \mathbb{R} \to \mathbb{R}$ is smooth for all $\gamma \in C^\infty(\mathbb{R},\mathbb{R}^m)$. So as our map was not smooth, there is a smooth curve $\gamma \colon \mathbb{R} \to \mathbb{R}^m$ which detects this non-smoothness. We can now unpick all the constructions to transfer this curve to a smooth curve in $M$, similarly we can transfer our projection to a smooth functional on $N$, and produce a smooth curve $\alpha \in C_M$ and functional $\phi \in F_N$ such that $\phi \circ f \circ \alpha$ is not smooth. Hence $f$ is not a morphism in the category of Frölicher spaces.

This proves the theorem.

Now the question as stated asks about embedded submanifolds, not all maps from, say, $\mathbb{R}$. So the place to look is to see if Boman's result works if one is allowed to only test with curves that are parametrisations of submanifolds. As this is a local result, and immersions are local embeddings, it is enough to test with curves $\alpha \in C^\infty(\mathbb{R},M)$ with non-vanishing first derivative.

My instinct would be to say that the result still holds, but I don't have a good enough grasp of the details of Boman's proof to be sure. I do know that you don't need all smooth curves for it to work. For example, you can leave out all curves that have non-negligible intersection with, say, the $x$-axis.


  • Boman, J. (1967). Differentiability of a function and of its compositions with functions of one variable. Math. Scand., 20, 249–268.

  • Kriegl, A. and M., Peter W. (1997). The convenient setting of global analysis (Vol. 53). Mathematical Surveys and Monographs. Providence, RI: American Mathematical Society.

  • nLab pages on, starting at Frölicher spaces.

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@Andrew, there is a tricky point here, if we consider e.g. the case $M=M'= \mathbb R$ (both with the standard structure) and the map $\Phi(t) = t^3$. This is clearly not a diffeomorphism. However, it does induce a bijection from the set of embedded submanifolds of $\mathbb R$ to itself. This is a special case of the remark that Cian made in the question statement, that it is clearly false when $m=n$. How does this show up in the above? –  Sam Lisi Mar 25 '11 at 11:01
Thanks for this - the Boman theorem certainly does look to be deeply relevant to this question, whatever the answer is. However, I think we need to distinguish my question from the different question whether a homeomorphism that preserves _embeddings of $\mathbb{R}^k$_ has to be a diffeomorphism. $\Phi$ will be a counterexample to my conjecture if it maps each embedded submanifold $S$ of $M$ to an embedded submanifold of $M'$---the smooth parameterisation of $\Phi(S)$ doesn't have to be the result of applying $\Phi$ to a smooth parameterisation of $S$. –  Cian Dorr Mar 25 '11 at 11:18
@Sam: That's why I prefixed this with "this is more of a comment than an answer". I don't know what difference using submanifolds makes as opposed to curves because I haven't thought about it before. Again, I may be missing something, but a submanifold of R is either a discrete set of points or an open interval. So any homeomorphism preserves those, you don't need to work with an "almost diffeomorphism"! Now that I write that, I see that the subtlety is in the fact that the submanifolds before and after need not be diffeomorphic. –  Loop Space Mar 25 '11 at 11:23
That's why the answer to my question is no when $n$ is 1. In a one-dimensional manifold, the embedded one-dimensional subsets are just the open sets; by being a homeomorphism, $\Phi$ can preserve these without having to be a diffeomorphism. So any proof that the answer is "yes" when $n=2$ and $k=1$ would have to turn on some distinctive property which distinguishes 2-dimensional manifolds from 1-dimensional ones. –  Cian Dorr Mar 25 '11 at 11:23
@Cian: That's what I just realised. Fortunately I littered my "answer" with lots of "get out" clauses! My instinct is now veering the other way. –  Loop Space Mar 25 '11 at 11:24

The claim is false as stated, but I think a refinement can fix the problem. (Not sure what the correct refinement is yet.)

My counterexample is essentially a generalization of taking $\Phi: \mathbb R \to \mathbb R, x \mapsto x^3$. For $n=2, m=1$, we will take $M=M' = \mathbb R^2 = \mathbb C$. Using complex coordinates $z = x+iy$, let $\Phi(z) = z |z|^2$. This is a diffeomorphism of $\mathbb C \setminus \{ 0 \}$, and is a homeomorphism of $\mathbb C$. The inverse map is $\Phi^{-1}(z) = z |z|^{-2/3}$ (with $\Phi^{-1}(0) = 0$).

I will show that if $S$ is embedded in $\mathbb C$ then $\Phi(S)$ is embedded. The other direction follows the same argument, but using the inverse function instead. If $S$ avoids $0$, there is nothing to check. If $S$ intersects $0$, we reduce to the local problem of a smooth curve $\gamma: [-1,1] \to \mathbb{C}$, $\gamma(0) = 0$, $\gamma'(t) \ne 0$. We then check that $\gamma( t^{1/3})$ satisfies that $\Phi( \gamma( t^{1/3}))$ is smooth. (The easy way to see this is to note that $\gamma(t) = ut + O(t^2)$ for a non-zero constant $u$, and thus $\Phi( \gamma(t) ) = u|u|^2 t^3 + O(t^4).)

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Thanks, this has been very helpful. Unfortunately I haven't been able to reconstruct the last part of the proof that you suggest. The problem I'm coming up against is specific to the direction where you show that if $\Phi(S)$ is embedded, $S$ is embedded. I see that this reduces to the problem of showing that if $\gamma$ is a smooth curve as specified, $\Phi^{-1}(\gamma(t^3))$ is a smooth function of $t$. And I think I could show that if I could assume that $g(t) = (f(t^3))^{1/3}$ is smooth whenever $f$ is. But I'm not seeing why that has to be true. Can anyone help me here? –  Cian Dorr Mar 21 '11 at 15:04
@Cian, the $g(t)$ will be $C^1$ if $f$ is, using also that $f(0) = 0$. Away from $t=0$, we have that $f(t) = bt + o(t)$, so we have $f(t^3) = bt^3 + o(t^3)$. Then, $$g'(t) = \left ( \frac{t^3}{f(t^3)} \right )^{\frac{2}{3}}.$$ This clearly has a continuous extension across $0$, establishing $C^1$. –  Sam Lisi Mar 24 '11 at 11:55
There is some more work to be done with higher derivatives... actually, it's maybe cleaner than what I wrote above. Consider the case of $f$ real analytic. The smooth case should follow, but with a little more bookkeeping. Then, write $f(t) = a_1 t + a_2 t^2 \dots$, and $b_i = a_i/a_1$. Then, $f(t^3) = a_1 t^3( 1 + b_2 t^3 + b_3 t^6 + \dots)$. Then, $g(t) = (a_1)^{1/3} t (1 + b_2 t^3 + \dots)^{1/3}$. Notice that $(1+x)^{1/3}$ is analytic for $|x| < 1$, so the function $g$ is real analytic too. –  Sam Lisi Mar 24 '11 at 11:57
That looks convincing for real analytic functions. However based on something I learnt from another question on math.stackexchange.com - number 14811 - I believe (or have believed up to now) that it can't be true for all smooth functions. In his answer there, Andrew Stacy proves that if $g^{-1}\circ f \circ g$ is $C^\infty$ for all $C^\infty$ f, then $g$ must not only be smooth (answering my question) but must in fact be a diffeomorphism. Since $g(x)=x^3$ isn't a diffeomorphism, it seems to follow that for some smooth $f$, $f(x^3)^{-1/3}$ is not smooth. Or am I misreading the result? –  Cian Dorr Mar 24 '11 at 14:38
Maybe there is a problem for smooth then. I'll have to think about it further. I suspect that for the original question, it doesn't matter, since we are also allowed to change the differentiable structure on $S$... but it makes the argument complicated enough that it would be worth writing it out in detail to make sure it holds up to scrutiny. –  Sam Lisi Mar 24 '11 at 17:34

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