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In summary: Actually, I think the confusion arises from a distinction between (local diffeomorphism)-onto image and local-(diffeomorphism onto image). See (C1) at the end.


Firstly, I believe this is the definition for local homeomorphism onto image:

Let $M$ and $N$ be topological spaces. Let $F: N \to M$ be a map. We say $F$ is a local homeomorphism onto its image, $F(N)$ (under the subspace topology) if $\tilde F: N \to F(N)$ is a local homeomorphism

Now, any subset $A$ of a topological space $B$ can always be made into a topological space by making $A$ a topological subspace of $B$ with the subspace topology. This does not hold for (smooth) manifolds: If $B$ is now a manifold, then we can't always make $A$ into a manifold too. However apparently, we can discuss whether or not $A$ is "diffeomorphic" to other manifolds or any subset of any manifolds by this.

My issue then is translating the above definition for local homeomorphism onto image to "local diffeomorphism onto image": I'm not sure if $F(N)$ is a manifold, specifically a regular/an embedded submanifold (I guess we need this specifically just as we need subspace topology for local homeomorphism onto image).


I have the seen the term "local diffeomorphism onto image" in 3 separate posts, and I would like to clarify the definition.


I notice the definition from the second post above doesn't seem to specify whether or not "$fX$" is a (regular/an embedded) submanifold or even manifold. I'm not sure all 3 posts have the same definition.

Question: The following is my understanding of what's going on. Is this correct?

There are 2 definitions of local diffeomorphsim onto image here.

Let $M$ and $N$ be smooth manifolds with dimensions. Let $F: N \to M$ be a smooth map. We say $F$ is a local diffeomorphism onto its image, $F(N)$ (under the subspace topology) if

  1. A1. $F(N)$ is a regular/an embedded submanifold of $M$, and $\tilde F: N \to F(N)$ is a local diffeomorphism.

  2. A2. $F(N)$ may or may not be a regular/an embedded submanifold, but $\tilde F: N \to F(N)$ is still a "local diffeomorphism", defined based on An Introduction to Manifolds by Loring W. Tu Definition 22.1 and Remark 22.5 or "Diffeomorphisms of subsets of manifolds" from Wikipedia, where such definition may or may not imply $F(N)$ is a regular/an embedded submanifold.

  3. Other

I think (A1) is equivalent to all of the following

  • A1.1. the definition in the second link above, as I try to prove here.

  • A1.2. a local embedding that is open onto its image. In particular I think this is precisely what the definition in the second link above is.

  • A1.3. an immersion that is open onto its image, since immersions are equivalent to local embeddings.

With (A1), we have for $X$ and $Y$ smooth manifolds with dimensions.

  • Local diffeomorphism:

    A map $f:X\to Y$, is a local diffeomorphism, if for each point x in X, there exists an open set $U$ containing $x$, such that $f(U)$ is a submanifold with dimension of $Y$, $f|_{U}:U\to Y$ is an embedding and $f(U)$ is open in $Y$. (So $f(U)$ is a submanifold of codimension 0.)

  • Local diffeomorphism onto image:

    A map $f:X\to Y$, is a local diffeomorphism onto image, if for each point x in X, there exists an open set $U$ containing $x$, such that $f(U)$ is a submanifold with dimension of $Y$, $f|_{U}:U\to Y$ is an embedding and $f(U)$ is open in $f(X)$. (This says nothing about $f(X)$ explicitly, but it will turn out $f(X)$, like $f(U)$ is a submanifold of $Y$.)

  • Local embedding/Immersion:

    A map $f:X\to Y$, is a local embedding/an immersion, if for each point x in X, there exists an open set $U$ containing $x$, such that $f(U)$ is a submanifold of $Y$ with dimension and $f|_{U}:U\to Y$ is an embedding. (This says nothing about $f(X)$ explicitly, but it will turn out $f(X)$, like $f(U)$ is an immersed submanifold of $Y$. However, $f(X)$, unlike $f(U)$, is not necessarily a regular/an embedded submanifold of $Y$.)

Note: Depending on your definition of embedding, "$f(U)$ is a submanifold with dimension of $Y$" may be redundant in the 3 preceding definitions.

Therefore, (A1) gives us:

  1. B1. $\text{local diffeomorphism} \implies \text{local diffeomorphism onto image} \implies \text{immersion and image is submanifold} \implies \text{immersion} \iff \text{local embedding}$

  2. B2. $\text{surjective local diffeomorphism} \iff \text{surjective local diffeomorphism onto image}$

  3. B3. $\text{local diffeomorphism onto image} \iff \text{immersion and open onto image} \iff \text{immersion and image is submanifold}$

    • (B3) is related to this and this, I guess.
  4. B4. $\text{local diffeomorphism onto image} \nLeftarrow \text{immersion}$

  5. B5. $\text{local diffeomorphism} \iff \text{local diffeomorphism onto image and image is open} \iff \text{local diffeomorphism onto image and open map}$

    • (B5) is related to this and this, I guess.

However, the first and third posts above suggest immersions are "local diffeomorphisms onto images", contrary to(B4). Thus, I think the definition in those is different from the one in the second post unless those immersions have submanifold images, by (B3). Since immersions are equivalent local embeddings and embeddings are equivalent to diffeomorphisms onto submanifold images, we might say immersions are local-(diffeomorphisms onto images).

Therefore, my understanding of what's going on is that there is a distinction between

$$\text{(local diffeomorphism)-onto image} \ \text{and} \ \text{local-(diffeomorphism onto image).} \tag{C1}$$

The first and third posts above describe immersions as local embeddings and so use definition (A2) or of local embedding (I didn't check if (A2) is equivalent to immersion) unless those immersions have submanifold images, by (B3) while the second link above uses definition (A1) which is stronger and not equivalent to immersion by (B4).

  • In the first post, I think the idea is weakening a diffeomorphism $F$ not to an immersion $F$ but to an immersion $F$ with submanifold image, i.e. a local diffeomorphism onto image.

    • Update: I think this is what happened. It's really just immersion, and the comments in the first post, both the one of user10354138 and the one of lEm, are mistaken unless they define "local diffeomorphism onto image" as local-(diffeomorphism onto image), i.e. local embedding, i.e. immersion. However their idea is right. Their idea was still to define the pushforward of vector fields. Since $F$ is not a diffeomorphism, get another diffeomorphism to define pushforward. For each $p \in N$, we hope that there exists a $U$ such that $F(U)$ is a submanifold of $M$ and $\tilde{F|_{U}} : U \to F(U)$ is a diffeomorphism. We have $\tilde{F|_{U}}$ as our required diffeomorphism not only when $F$ is a local diffeomorphism or a local diffeomorphism onto image but also when $F$ is an immersion, i.e. a local embedding (where $F(U)$ is not necessarily open in either $F(N)$ or in $M$ but is still a submanifold of $M$): This is because $F(U)$ is a submanifold of $M$ in all 3 cases!
  • In the third post, there may be an additional assumption besides just immersion. I may have missed something.

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  1. The notion of a local diffeomorphism makes sense only if the domain and the range are smooth manifolds. If the image of a map happens to be a smooth submanifold of the target manifold, one can say " $f$ is a local diffeomorphism onto its image" by restricting the range. Any other use is just made-up (by various MSE users, it seems) and should be avoided (at least until you are very comfortable with the subject). Instead you can simply say:

...The image of a map $f: X\to Y$ is a smooth submanifold and $f: X\to f(X)$ is a local diffeomorphism.

You may also sometimes encounter the following, describing what an immersion is:

A map $f: X\to Y$ of smooth manifolds is an immersion if and only if locally, it is a diffeomorphism to its image, meaning that $\forall x\in X \exists$ a neighborhood $U$ of $x$ such that $f(U)$ is a smooth submanifold of $Y$ and $f: U\to f(U)$ is a diffeomorphism.

But, again, given ambiguity of the language, it is better to avoid using this terminology in the beginning. The ambiguity comes from the word "image": It can either mean the image of the original map or the image of the map with the restricted domain.

  1. Everything that you wrote up to the line "However, the first and third posts..." is correct and proofs are very straightforward.

However: I did not check your guesses on how it may or may not be related to various MSE posts.

One thing, you should not repeat ad nauseum "with dimension". (Every manifold has dimension and, except for the empty set, its dimension as a smooth manifold equals its dimension as a topological space. As for the empty set: For every $n\ge 0$, the empty set is a manifold of dimension $n$. At the same time, from the general topology viewpoint, the empty set has dimension $-1$.)

  1. As for what various MSE users meant in their answers and comments, I prefer not to discuss: Frequently, there is no consistency in their use of mathematical terminology. (Many are only beginners, many have trouble with English, etc.)

Addendum. I am not sure who came up with the idea of allowing manifolds to have variable dimension on different connected components, but I wish this never happened as this just leads to a confusion. I checked several sources in geometry and topology, and the only author allowing manifolds to have variable dimension is Lang.

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  • $\begingroup$ Thanks. About "Every manifold has dimension", what is the dimension of $(0,1) \cup \{2\}$ ? See the links. $\endgroup$ – Selene Auckland Jul 30 at 10:51
  • $\begingroup$ "The ambiguity comes from the word "image"." Please don't make this parenthetical. This is precisely the source of all the confusion I described! $\endgroup$ – Selene Auckland Jul 30 at 11:50
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    $\begingroup$ @SeleneAuckland: This example is not a manifold so one cannot talk of its dimension as a manifold. Its dimension as a topological space is 1. $\endgroup$ – Moishe Kohan Jul 30 at 16:17
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    $\begingroup$ @MoisheKohan Can you tell me what topological definition of "dimension" you are using? Are you referring to the Lebesgue covering dimension? $\endgroup$ – SpiralRain Jul 31 at 1:24
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    $\begingroup$ @SeleneAuckland: I posted my answer to the question in your link "Okay so what..." $\endgroup$ – Moishe Kohan Aug 7 at 20:53

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