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Note 1: same question as this, but please don't use sard's theorem or anything similar. preferably, no measure theory.

Note 2: Didier's comment:

I think another proof can use the constant rank theorem but would be a bit messy

Let $N$ be smooth $n$-manifold. Let $M$ be smooth $m$-manifold. Let $F: N \to M$ be smooth map.

What I understand:

  • A. If $F$ is a submersion on $N$, then $n = \dim N \ge \dim M = m$.

  • B. If $F$ is a submersion on $N$, then $F$ is open.

  • C(i). If $F$ is smooth surjective, then $n = \dim N \ge \dim M = m$.

  • C(ii). If $F$ is a submersion at a point $p \in N$ but not necessarily at other points in $N$, then $\dim N \ge \dim M$ still because $n = \dim N = \dim T_pN \ge \dim T_{F(p)}M = \dim M = m$.

Question: If $F$ is open but not necessarily a submersion, then is $n = \dim N \ge \dim M = m$ still, i.e. under (B), we may relax (A) from submersion on $N$ to open?

All I know so far:

  • D. The image $F(N)$ is open in $M$ (I could be wrong, but I think this condition is equivalent to saying that $F$ is open), i.e. $F(N)$ is a smooth embedded $m$-submanifold of $M$.

  • E. The induced map $\tilde F: N \to F(N)$ is smooth and open.

    • E.1. $\tilde F$ is the unique map s.t. $\iota \circ \tilde F = F$, where $\iota: F(N) \to M$ is the inclusion map (which is smooth embedding if and only if $F(N)$ is a smooth embedded $k$-submanifold and so is I guess more than a smooth embedding since $k=m$. Update: I believe it's an open smooth embedding, i.e. an injective local diffeo.)

  • F. In re (C) and (D), i think codimension $0$ is pretty close to surjective at least when we're thinking about dimensions. Like how an open disk/disc in $\mathbb R^2$ is diffeomorphic to the whole of $\mathbb R^2$.

  • G. Constant rank level set theorem: For $q \in M$, the preimage $F^{-1}(q)$ is a regular submanifold of codimension $k$ if '$F$ has constant rank $k$ in some neighbourhood $U$ of $F^{-1}(q)$ in $N$'

    • G.1. where I believe the condition in quotes is equivalent to: 'for $U$ s.t. $F^{-1}(q) \subseteq U \in \mathscr T(N)$, where $\mathscr T(N)$ is the topology of $N$, we have that $F_{*,p}$ has the same rank $k$ for each $p \in F^{-1}(q)$'

    • G.2. Btw, it didn't occur to me before, but I guess there's supposed to be some well-definedness remark here like if in some neighbourhood of the preimage we have constant rank $k$, then for every neighbourhood we will get constant rank $k$ (OR at least some neighbourhoods will give constant rank $k$ and other neighbourhoods will give non-constant rank, but I think we shouldn't have a neighbourhood where we get constant rank that isn't $k$).

    • G.3. I guess either $q \in image(F)=F(N)$ or it's ok for $F^{-1}(q)$ to be empty because...I guess for every or for some (not sure w/c is the appropriate quantifier here) neighbourhood of the empty set, I guess $F$ has...non-constant rank.

    • G.4. I guess we pick $q \in F(N)$, otherwise $F^{-1}(q)$ is just an empty regular level set...then idk

BCLC
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    You should more focus on $2.$ (1. has already been said to be true in the other question) – Didier Apr 21 '21 at 07:55
  • @Didier oh yeah lol. will edit. thanks. – BCLC Apr 21 '21 at 07:56
  • @Didier Done. now for constant rank... – BCLC Apr 21 '21 at 07:58
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    Here is a nice exercise in measure theory: if $f: R^m\to R^n$ is $C^1$ (even locally Lipschitz or differentiable everywhere will suffice) and $m<n$, then the image of $f$ has Lebesgue measure zero. If you cannot solve this, read Rudin. – Moishe Kohan Apr 21 '21 at 11:51
  • @MoisheKohan this is sard's right? the point of this question re-ask is to avoid sard's... – BCLC Apr 21 '21 at 11:52
  • No, this is much less than Sard (note that I assume very little smoothness) and, actually, easier to prove. – Moishe Kohan Apr 21 '21 at 11:54
  • @MoisheKohan ah nitpicks. but it's still the same proof anyway as when you use sard's right? – BCLC Apr 21 '21 at 11:54
  • It depends on what you mean by "the same"; the proof boils down to the fact that the diameter of $f(B(x,r))\le C r$ for some uniform constant $C$ and all $x, r$, and the definition of the outer measure. – Moishe Kohan Apr 21 '21 at 11:56
  • @MoisheKohan i mean, you're still gonna say measure zero implies not open right? that was the thing in the answer for the previous question. anyhoo, i updated question to include 'preferably, no measure theory'. i mean i can't believe we really have to use measure theory. surely there must be something with the tangent spaces that that openness will say s.t. $n \ge m$. I was thinking somehow for some $p$ like the differential $F_{*,p}$ is somehow surjective or something. i mean, ok maybe $F$ is not necessarily a submersion at some $p$ just because $F$ is open, but...idk something. i was hoping – BCLC Apr 21 '21 at 12:00
  • @MoisheKohan that Didier might be on to something with the constant rank theorem. editing post to include constant rank stuff. – BCLC Apr 21 '21 at 12:00
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    I see, you want to avoid measure theory. It's unclear to me why would you want this, but, indeed, I think, there is a messy proof using CRT and stratification of the domain of $f$ into subsets according to which partial derivatives vanish. – Moishe Kohan Apr 21 '21 at 12:23
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    Incidentally, there are continuous, open maps $R^3\to R^4$. – Moishe Kohan Apr 21 '21 at 12:24
  • @MoisheKohan oh holy crap. then i guess 'smooth' is really relevant here. hmmm.... thanks! – BCLC Apr 21 '21 at 12:25
  • @MoisheKohan 'It's unclear to me why would you want this' --> the same reason we can conclude some inequality or equality about $n$ and $m$ from knowing $F$ is an immersion, a submersion, a surjection or an injection? why can we avoid measure theory in all those but when it comes to open we have to use measure theory? well, seem to recall there's no conclusion when $F$ is closed... but anyway whenever there is a conclusion for condition X on F, then i expect we don't have to use measure theory at least if understanding condition X doesn't require measure theory – BCLC Apr 21 '21 at 12:27
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    @BCLC Note that Sard's theorem really is a theorem about smooth functions. It says that indeed, the image of a smooth function $f : M^m \to N^n $ with $m <n$ can be dense, but cannot be too dense, and use the differentiable property in a crucial manner. Basic measure theory (= just zero-measure sets) is the right way to define what too dense means. – Didier Apr 28 '21 at 09:16
  • @Didier thanks for the info re intuition. but besides info, you're justifying why it's natural to use sard's theorem for this? – BCLC Apr 29 '21 at 07:09
  • @Didier What's wrong with this? 1 - $F(N)$ open in $M$ means $F(N)$ is a regular/an embedded smooth $m$-submanifold of $M$. 2 - Now, $\tilde F: N \to F(N)$ is smooth surjective. 3 - Therefore, by (1) and (2), $n = \dim N \stackrel{(2)}{\ge} \dim F(N) \stackrel{(1)}{=} \dim M = m$. i'll post as answer – BCLC Apr 30 '21 at 07:53
  • @MoisheKohan What's wrong with this? 1 - $F(N)$ open in $M$ means $F(N)$ is a regular/an embedded smooth $m$-submanifold of $M$. 2 - Now, $\tilde F: N \to F(N)$ is smooth surjective. 3 - Therefore, by (1) and (2), $n = \dim N \stackrel{(2)}{\ge} \dim F(N) \stackrel{(1)}{=} \dim M = m$. i'll post as answer – BCLC Apr 30 '21 at 07:53
  • Why would $\tilde{F}$ be defined on $N$? What you said is "if $U$ is open in $M$ then there exists $\tilde{F} \colon N \to U$ smooth surjective" and this is false. – Didier Apr 30 '21 at 08:09
  • @Didier What? $\tilde F$ is just $F$ with restricted range, so $\tilde F$ is defined on $N$ because $F$ is defined on $N$. See theorem 11.15 of tu's an introduction to manifolds – BCLC Apr 30 '21 at 08:12
  • @Didier i didn't say for arbitrary $U$. of course not. $U$ has to contain the image of $F$. so $\tilde F: N \to S$ is smooth for all smooth embedded/regular $k$-submanifolds $S$ of $M$ s.t. $F(N) \subseteq S$. in this case open submanifold is $k=m$-submanifold and then image is open because map is open. – BCLC Apr 30 '21 at 08:13
  • While saying "therefore $\dim N \geqslant \dim F(N)$", aren't you using what you want to prove? – Didier Apr 30 '21 at 08:15
  • @Didier yes, I don't think i am using what i am trying to prove. i don't (yet) know about the relationship of $n$ and $m$ in $F: N \to M$. but i know about the relationship between $n$ and $l$ in $\tilde F: N \to F(N)$. anyway, i posted an answer. – BCLC Apr 30 '21 at 08:17
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    Seems fair. It(s still the morning, I think I need some coffee. Btw maybe you should check the proof of that fact you are using. – Didier Apr 30 '21 at 08:24
  • Thank you @Didier ! Still, I do think i'm overlooking something pretty obvious. – BCLC Apr 30 '21 at 08:25
  • @Didier Moishe Kohan pointed out what's the flaw. hehe. the proof is correct, but $C(i)$ afaik relies on measure theory too. So far, i don't yet know of any non-measure theory thing for $C(i)$ – BCLC Apr 30 '21 at 12:27
  • @Didier new question: what do injective and immersion have in common please? Surjective, open and submersion all have in common: image has non-empty interior. What about injective and immersion? – BCLC May 24 '21 at 05:06

2 Answers2

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This is a restructured version of the original answer. I try to make clearer the logical steps, because there are some details to make more precise.

  1. Assumptions: Everywhere here, the metrics considered on open subsets of $\mathbb{R}^d$ is always the $\|\cdot\|_{\infty}$. Its closed unit “ball” is denoted $C_k=[-1,1]^k$, while $C’_k$ is the open unit ball, ie the interior of $C_k$.

I. First, we’ll prove a combinatorial property (which we’ll name CP) that replaces measure theory (with a primitive version of Hausdorff dimension theory, maybe?), which is as follows:

Let $n,m$ be two positive integers. Assume that there exists $A,B>1$ such that, for each integer $t >0$, there is a subset $S_t \subset C_m$ with at most $Bt^n$ elements such that any point of $C_m$ is at distance at most $A/t$ from $S_t$. Then $n \geq m$.

Proof: the contradiction with measure theory is obvious with a large enough $t$, but we have to use less sophisticated tools.

For now, let $t=2^p$, where $p >0$ is an integer. We have a subset $G \subset C_m$ of at most $Bt^n$ points such that any $z \in C_m$ is at distance at most $A/t$ of an element of $G$.

Fix some integer $s=2^q$. $C_m$ can be divided in $(2s)^m$ smaller cubes (of side length $1/s$). The center of any of these smaller cubes is at distance $1/(2s)$ of the exterior of the cube. So if $1/(2s) < A/t$, then every one of these smaller cubes must contain a point of $G$, and in particular $Bt^n \geq |G| \geq (2s)^m$.

In particular, we know that if $2^{p-q-1} > A$, then $2^{B’+pn} \geq 2^{(q+1)m}$ where $B=2^{B’}$, $B’>0$.

In other words, if $A=2^{A’-1}$ (so $A’>2$) then for all positive integers $p,q>0$, $p-q \geq A’ \Rightarrow n(p-q) \geq (m-n)q-B’+m$.

Now, take $q \rightarrow \infty$, $p$ an integer such that $q+A’ \leq p < q+A’+1$, then $n(p-q) \in [nA’,n(A’+1)]$, so $(m-n)q-B’+m \leq n(A’+1)$. This implies $n\geq m$. QED

II. Now, an easy application is the following fact (which we’ll name $A1$):

Let $f_1,\ldots,f_N:C_n \rightarrow C_m$ be Lipschitz continuous such that $\bigcup_{i=1}^N{f_i(C_n)}$ contains some $rC_m$ with $1>r>0$. Then $n \geq m$.

Proof: Let $K$ be a Lipschitz constant for all the $f_i$. Let $\pi: C_m \rightarrow rC_m$ the projection, which is Lipschitz continuous with constant $K’$ (I think $K’=1$ but it doesn’t matter).

Let $t>0$ be an integer. Then $C_n$ can be divided in $(2t)^n$ smaller cubes with side length $1/t$, and their centroids are at distance $1/(2t)$ of each cube. Let $U_t$ be the set of these centroids. Let $S_t=\bigcup_{i=1}^N{f_i(U_t)}$. Then $S_t$ has cardinality at most $N2^nt^n$. Moreover, if $z \in rC_m$, $z=f_i(x)$ for some $i$ and some $x \in C_n$. There is some $y \in U_t$ at distance at most $1/(2t)$ of $x$, and thus the distance of $z$ to $S_t$ is at most $\|z-\pi\circ f(y)\|_{\infty}=\|\pi\circ f_i(x)-\pi \circ f)y)\|_{\infty}\leq KK’/(2t)$.

By CP (I), $n \geq m$. QED

III. Now:

Let $N,M$ be smooth manifolds with dimensions $n,m$ respectively, and let $f:N \rightarrow M$ be smooth. Assume that $f$ is open. Then $n \geq m$.

Proof: let $U$ be a small open subset of $M$ contained in $f(N)$, so that $U$ is diffeomorphic to $2C’_m$, and let $N’=f^{-1}(U)$. Let $V \subset f^{-1}(U)$ be a small open subset diffeomorphic to $2C’_n$. Then the map $g: 2C’_n \equiv V \rightarrow U \rightarrow 2C’_m$ is Lipschitz continuous and open if $U,V$ are chosen small enough (eg their closures are contained in bigger open subsets that themselves are diffeomorphic to open subsets of $\mathbb{R}^n$).

In particular, $g(C_n)$ is a neighborhood of $g(0)$, ie $g(C_n) \supset g(0)+rC_m$ for some $0 <r<1$. Let $h: x \in C_n \longmapsto \frac{g(x)-g(0)}{2} \in C_n$. Then $h$ is Lipschitz continuous and its image contains $r’C_m$, $r’=r/2>0$.

By (A1), $n \geq m$, QED.

Remark: This argument works even if $f$ is only open “at one point”, ie there is $x \in N$ such that if $V$ is a neighborhood of $x$, then $f(V)$ is a neighborhood of $f(x)$.

IV. We can deduce, with a slightly more sophisticated argument, a more general result:

Let $N,M$ be smooth manifolds with respective dimensions $n,m$ and $f:N \rightarrow M$ smooth. Assume that the image $f(N)$ has nonempty interior (eg $f$ is open or surjective). Then $n \geq m$.

Proof: up to taking a small open subset of $M$ contained in the interior of $f(N)$, we can assume that $f$ is surjective and that $M$ is $C’_m$.

There is an increasing sequence $K_l$ of compacts of $N$ with reunion $N$. By the Baire category theorem it follows that some $f(K_l)$ has nonempty interior. Therefore, there are smooth maps $g_1,\ldots,g_r: 2C’_n \rightarrow N$ such that the reunion of the $g_i(C_n)$ contains $K_l$. In particular, the $h_i=(f \circ g_i)_{|C_n}$ are all Lipschitz continuous and the reunion of their images contains some $\alpha+tC_m$, with $\alpha \in C’_m$ and $0<t<1$. Now we apply $A1$ to the $f_i=\frac{h_i-\alpha}{2}$ and we get $n \geq m$, QED.

Remark: This general result shows that, for smooth $f$, any of the 3 conditions 'open', 'surjective and 'submersion' imply $n \geq m$ because they all have in common that they each imply the weaker condition 'The image $f(N)$ has nonempty interior' (which implies $n \geq m$).

BCLC
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Aphelli
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  • ran out of upvotes for the day, but HELL. thank you Mindlack – BCLC Apr 30 '21 at 14:06
  • this is for answering just the open implies $n \ge m$? or also the surjective implies $n \ge m$? – BCLC Apr 30 '21 at 14:06
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    The proof really shows that there cannot be a Lipschitz surjection $C_n \rightarrow C_m$ with $n <m$. By messing around (rescaling, projecting), that implies the claim for open maps $C_n \rightarrow C_m$. I’m still surprised that it reduces to an elementary discrete problem. – Aphelli Apr 30 '21 at 14:10
  • wait so you prove 1 - some claim $X$ that proves claim $Y$ and then 2 - claim $X$ proves the surjective implies $n \ge m$ and then 3 - claim $Y$ proves the open implies $n \ge m$? at this point i actually don't really care about the details of the answer. i'm just gonna give you the upvote, bounty and accept, but what i do want to understand is what is being proven and what is the ordo cognoscendi or whatchamacallit here – BCLC Apr 30 '21 at 14:19
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    Er... so you can break down the argument as follows. The second part of the post proves that the “combinatorial property” (CP for short) implies $n \geq m$. The first part is messier, but here’s what it says: if $f:N \rightarrow M$ is smooth and open, then (taking charts) that implies that there is a smooth map $g:C_n \rightarrow C_m$ such that its image contains an open subset. Projecting on a cube included in this open subset, it implies CP, and thus that $n \geq m$. Surjective is a little trickier though, I’ll update and edit when I’m sure I’m not confused. – Aphelli Apr 30 '21 at 14:27
  • oh wow thanks Mindlack. i reached vote limit again. will have to do later on. also, i was supposed to ask about the surjective thingy in another post. but anyhoo my main takeaway from this answer is the last part: which explains what surjective and open (and submersion on domain) have in common. – BCLC May 01 '21 at 04:30
  • quick follow-up i hope does submersion at a point $p$ also perhaps imply $f(N)$ has non-empty interior? (no need for elaborations or edits in answer re this. but you may if you want) – BCLC May 01 '21 at 04:30
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    Yes. Submersion at a point implies open in a neighborhood of this point by the “normal form of submersions”. – Aphelli May 01 '21 at 08:09
  • Mindlack thanks! – BCLC May 01 '21 at 08:18
  • Wait Mindlack to double check, does $f(N)$ non-empty interior imply $n \ge m$ even if $f(N)$ isn't a smooth $m$-submanifold of $M$ (w/c is smooth $m$-manifold)? like not just not same dimension $m$ but really also that $f(N)$ may be just immersed submanifold or maybe just a topological subspace. – BCLC May 24 '21 at 04:27
  • also Mindlack new question: what do injective and immersion have in common please? Surjective, open and submersion all have in common: image has non-empty interior. What about injective and immersion? – BCLC May 24 '21 at 05:06
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UPDATE: I concede. $C(i)$ (smooth surjective implies $n \ge m$. I got it from here actually, but I believe this needs measure theory, probably for the same reason as the injective thing. I hereby shall make this answer community wiki. I offer to award the bounty to Moishe Kohan, who asked how I would prove $C(i)$ without measure theory.

I'm probably missing something really obvious, but here goes:

  1. $F(N)$ open in $M$ means $F(N)$ is a regular/an embedded smooth $l$-submanifold of $M$, where $l$ turns out $l=m$.

  2. Now that we know that $F(N)$ is a smooth ($l$-)manifold, it makes to talk about whether or not the induced map $\tilde F: N \to F(N)$ is smooth. Actually, ​$\tilde F$ is indeed smooth (and I think open, but I think it's not relevant as to whether or not $\tilde F$ is open). Since $\tilde F$ is surjective, we have by $C(i)$ that $n = \dim N \stackrel{(2)}{\ge} \dim F(N) = l$.

  3. Therefore, by (1) and (2), $n = \dim N \stackrel{(2)}{\ge} \dim F(N) = l \stackrel{(1)}{=} \dim M = m$.

Remark: Sard's Theorem again comes up. I was wondering if actually surjective smooth implies submersion, i.e. $F(N)=M$ implies for each $p \in N$ that $F_{*,p}(T_pN)=T_{F(p)}M$. It really shouldn't be the case, otherwise it would be weird for books to not say this. It's indeed not the case as explained here in this answer, w/c mentions Sard's Theorem.

BCLC
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