Does open map imply dimension of domain is greater than or equal to dimension of range? (please don't use sard's theorem)

Question

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

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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

Answer 1

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.

0. 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 $C_k$.

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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

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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

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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 $f(x)$.

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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 $n \geq m$).

Answer 2

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.

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

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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.