in various papers in symplectic geometry, I have encountered the following argument.

Statement: Suppose $f: M \rightarrow N$ is a smooth map of constant rank. Then its image $f(M)$ can be equipped with a structure of an immersed submanifold of $N$, that is a topology and a smooth structure making the canonical inclusion $f(M) \hookrightarrow N$ into a smooth injective immersion.

Some people even say that with respect to this $f: M \rightarrow f(M)$ becomes a surjective submersion. This I do not believe, as this would mean that every smooth map of constant rank can be written as a composition of a smooth surjective submersion followed by a smooth injective immersion, which is not (people say) true for global manifolds.

!!Edit:!! Maybe it is true that every subimmersion (and thus also a map of constant rank) is actually always a composition of a submersion followed by an immersion. I have found this both in Bourbaki (Book X, section about subimmersions) or in super-detailed exposition in the book:


See Proposition 5.2.7 there. Maybe this somehow does not apply here as they consider a bigger class of smooth manifolds (namely, they do not force them to be Hausdorff and second countable). In particular, they use the manifold of germs of submanifolds, as does Bourbaki.

Edit 2: Bourbaki et al say that subimmersion is a composition of submersion followed by immersion. However, they do not say that submersion is surjective and immersion injective. Hence it is useless here.

I have tried two approaches, both in a sense using the constant rank theorem.

Attempt 1: This uses the rank theorem directly. Let $n = \dim(M)$, $k = \dim(N)$ and $r = rank(f)$. Then, for every $m \in M$, there exist coordinate charts $(U,\varphi)$ and $(V,\psi)$ for $M$ and $N$, respectively, such that

(i) $m \in U$, $\varphi(U) = W \times W'$, where $W \subseteq \mathbb{R}^{r}$, $W' \subseteq \mathbb{R}^{n-r}$ are open sets and $\varphi(m) = (0,0)$.

(ii) $f(U) \subseteq V$, $\psi(V) = W \times W''$, where $W \subseteq \mathbb{R}^{r}$, $W'' \subseteq \mathbb{R}^{k-r}$ are open and $\psi(f(m)) = (0,0)$.

(ii) The local form $\hat{f} = \psi \circ f \circ \varphi^{-1}$ in these coordinates has the form \begin{equation}\hat{f}(x,y) = (x,0), \text{ for all } (x,y) \in W \times W'.\end{equation}

The idea is to directly use these coordinates to define the coordinate atlas on $f(M)$. Namely, consider the sets $U_{0} := f(U)$ together with a map $\psi_{0}: U_{0} \rightarrow W$ which is obtained from $\psi$ by restricting it to $f(U)$. This makes sense as $\psi(f(U)) = W \times \{0\}$.

Suppose there is another such defined chart, say $\psi'_{0}: U'_{0} \rightarrow Z$, which is obtained from some local charts $(U',\varphi')$ and $(V',\psi')$ having the properties (i) - (iii) where we replace the open sets $W,W',W''$ with $Z,Z',Z''$, such that $U_{0} \cap U'_{0} \neq \emptyset$.

In order to say something about the transition maps between these two charts, one would have to prove that $\psi'_{0}(U_{0} \cap U'_{0})$ and $\psi_{0}(U_{0} \cap U'_{0})$ are open subsets of $\mathbb{R}^{r}$, This is where I fail and I even doubt that this is true in general. If one would be able to prove this, it is then easy to see that the transition maps are smooth and the composition $\psi \circ I \circ \psi_{0}^{-1}$ is just inclusion $x \mapsto (x,0)$, hence $I$ is a smooth injective immersion.

Remark 1: This construction in fact should fail, as the composition $\psi_{0} \circ f \circ \varphi^{-1}$ is the projection $(x,y) \mapsto x$, hence $f: M \rightarrow f(M)$ a smooth surjective submersion. This would contradict the above observation that not every smooth map of constant rank can be decomposed as a smooth surjective submersion followed by a smooth injective immersion.

Remark 2: If $f$ is assumed open or closed, the statement is true. In fact, then $f(M)$ is an embedded submanifold of $N$. This is because (say if $f$ is open) the sets $f(U) \cap V$ together with the restriction of $\psi$ work as submanifold charts for $f(M)$.

Attempt 2: Here I will be brief, as I in fact suspect it to be just the first paragraph in disguise. Every map $f$ of constant rank is so called subimmmersion, each point $m \in M$ has a neighborhood $U$, such that there exists a smooth manifold $M$ and a pair of maps $h: U \rightarrow P$ and $g: P \rightarrow N$, such that $f|_{U} = g \circ h$ and

(i) $h$ is a smooth surjective submersion,

(ii) $g$ is a smooth injective immersion.

The idea is now to equip $f(U) = g(P)$ with a smooth structure and topology by declaring it diffeomorphic to $P$. To finish, one would have to prove that on overlaps $f(U) \cap f(U')$, the induced smooth structures would be compatible. To my understanding, one would have to prove that the sets $g^{-1}(f(U'))$ and $g'^{-1}(f(U))$ would have to be shown diffeomorphic. Yet it is not clear that they are even open in $P$ and $P'$, respectively.

Any suggestions? I would be happy to read it in the literature, yet I have not found anything reasonable (usually, only level manifolds for constant rank maps are examined, or they do not like immersed submanifolds at all).

Concluding remarks: Maybe something completely different has to be employed, say some Frobenius theorem (gives immersed submanifolds) or some transversality business?


The statement that the image of a smooth map $f\colon M \to N$ of constant rank can be equipped with the structure of an immersed submanifold of $N$ is not true. For example, the map $f\colon \mathbb{R} \to \mathbb{R}^2$ given by $f(t) = (t^2-1, t^3-t)$ is a smooth map of rank 1, but its image is a nodal cubic, which is not a manifold (see L. Tu, $\mathit{An\ Introduction\ to\ Manifolds}$, 2nd ed., Fig. 11.3, p. 121).

If you assume that the smooth map $f$ is injective and of constant rank, then the statement is true.

  • $\begingroup$ What if I take a function $g$ defined on $(-\infty,-1) \cup (-1,\infty)$ given by the same formula as $f$? Then it has the same image as $f$ does, it is an immersion and it is injective, and thus makes $f(M)$ into an immersed submanifold.... $\endgroup$ – Jan Vysoky May 5 '19 at 6:06
  • $\begingroup$ Yes, you are right. What my example shows is that the image of a smooth map of constant rank need not be an embedded submanifold, but it could be an immersed submanifold. I should delete my example. $\endgroup$ – Loring Tu May 5 '19 at 20:25
  • $\begingroup$ Please do not delete your example - it shows that altough there can be an immersed submanifold structure on $f(M)$, the map $f: M \rightarrow f(M)$ is not even continous. Viewed as a map $f: \mathbb{R} \rightarrow (-\infty,-1) \cup (-1,\infty)$, it looks like identity on $\mathbb{R} - \{-1\}$ and $f(-1) = 1$. $\endgroup$ – Jan Vysoky May 6 '19 at 4:36

Claim 1: A subset $N\subset M$ is an immersed submanifold iff for every $n\in N$ there is an embedding $f_n : U_n \rightarrow M$ with $n\in f_n(U_n) \subset N$.

Proof: "$\Rightarrow$" Let $i: Z\rightarrow M$ be an immersion with $i(Z)=N$. Near any $z\in Z$, there is a compact neighborhood on which $i$ is injective. Hence, it is a homeomorphism onto its image.

"$\Leftarrow$" Set $Z:=(\sqcup_{n\in N} U_n)/ \sim$, where $x \sim x'$ iff $f_n(x)=f_{n'}(x')$, and equip it with the quotient topology. We can assume that $f_n(U_n) = \psi_n^{-1}(\mathbb{R}^k\times\{0\}^{m-k})$ for charts $\psi_n : V_n \rightarrow \mathbb{R}^m$. Using $f_n(U_n\cap U_{n'}) = f_{n'}(U_{n'})\cap f_n(U_n)$, one can check that $\psi_n$ induce an atlas on $Z$ such that the obvious map $i: Z\rightarrow M$ is an immersion with image $N$. Q.E.D.

Claim 2: Let $N$ be the image of a map $f: Z \rightarrow M$ with constant rank. Then it is an immersed submanifold.

Proof: Locally, $f$ is equivalent to $(x,y)\in V\mapsto (x,0)\in W$ for some open $V\subset \mathbb{R}^n$, $W\subset \mathbb{R}^m$. Let $W'\subset \pi(W)$ be a precompact open neighborhood of a point $x_0$, where $\pi: \mathbb{R}^m \rightarrow \mathbb{R}^k$ is the projection to the columns corresponding to $x$. Hence, $x\in W' \mapsto (x,0)\in W$ is an embedding. Claim 1 finishes the proof. Q.E.D.

P.S. For me, an immersed submanifold is just the image of an immersion. E.g., in dimension 1, it is a regular curve which can have self-intersections (failure of injectivity) or, if it is non-compact, it can have infinitesimal contact with one its end (failure of homeomorphism onto the image with subspace topology).

  • $\begingroup$ Thank you. It really seems that the statement indeed works if you allow for non-injective immersions. I have also found an approach using the leaf spaces - take the foliation $\mathcal{F}$ corresponding to the kernel of $T(f)$ and there is a way to make the space of its leaves $Z/\mathcal{F}$ into a (unfortunately possibly non-Hausdorff) smooth manifold, and canonical map $Z/\mathcal{F} \rightarrow M$ becomes a (possibly non-injective) immersion. $\endgroup$ – Jan Vysoky Jun 13 '19 at 6:05
  • $\begingroup$ Can you give an example when two different points in $Z/\mathcal{F}$ can not be separated by open neighborhoods? P.S. we might be using different notations, right? $\endgroup$ – Pavel Jun 13 '19 at 6:18
  • $\begingroup$ I am interested in an example when a given subset $S\subset M$ can not be written as the image of any immersion from any Hausdorff manifold but is the image of an immersion from a non-Hausdorff manifold. Does Hausdorffness fail in my construction? I am also interested in an example of an immersion from a non-Hausdorff manifold into a Hausdorff manifold (this is perhaps just identifying the two zero's in the line wth two origins...) or the other way round (this is perhaps again mapping the line into the line with two origins by choosing an origin...). $\endgroup$ – Pavel Jun 13 '19 at 6:30

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