Suppose $f: M \to N$ is smooth and an immersion, i.e $df_p : T_p(M) \to T_p(N)$ is one-to-one. Since $f$ is an immersion, we have the following theorem,

$\textbf{Local Immersion Theorem:}$ Suppose that $f: M \to N$ is an immersion at $x$. Let $y=f(x)$. Then there exists local coordinates around $x$ and $y$ such that $$ f(x_1, x_2, \dots, x_k) = (x_1, x_2, \dots, x_k, 0, \dots, 0 )$$

In other words, $f$ is a locally one-to-one, and thus an embedding locally. Does this imply that $f$ is a local diffeomorphism?

I am looking for a answer as to the relationship between the three concepts: local immersion theorem, local embedding, and local diffeomorphism.

I know some similar questions have been asked, but in more specific circumstances


Heavens, no! The differential $df$ maps from a $k$-dimensional vector space to an $n>k$ dimensional vector space. It cannot be an isomorphism.

However, by the local coordinates condition you've imposed, the differential is full-rank, and so $f$ is a local diffeomorphism onto its image.

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    $\begingroup$ Why is an immersion a local diffeo onto its image? For that we'd need an open cover $(U_i)$ of the domain $M$ such that $f|_{U_i}$ are smooth embeddings and such that $fU_i\subset fM$ is open. Why is the latter true? $\endgroup$ – Arrow Dec 29 '17 at 1:13
  • $\begingroup$ Neal, why does immersion imply local diffeo onto image? $\endgroup$ – Selene Auckland Jun 27 at 8:43
  • $\begingroup$ @Arrow I'm not the only one wondering about the exact relationships between local diffeomorphisms, local diffeomorphisms onto image, immersions, embeddings and local embeddings! What a relief! I'm actually about to compose ask about those in a single question and am studying your question before posting. $\endgroup$ – Selene Auckland Jun 27 at 8:59
  • $\begingroup$ @Arrow and Neal, what is a local diffeomorphism onto image exactly please? I know what local diffeomorphisms, local homeomorphisms onto image and local homeomorphisms are. The issue with copying the definition of local homeomorphism onto image to local diffeomorphism onto image has the problem of the image possibly not being a submanifold or manifold while there is no such issue for local homeomorphism onto image since image can always be made into a subspace. $\endgroup$ – Selene Auckland Jul 22 at 5:06
  • $\begingroup$ Neal and @Arrow Oh wait I think I get it now. An immersion is not a local diffeomorphism onto image, but the immersion described in the post is a local diffeomorphism onto image because there are other assumptions besides immersion? Edit: Wait Neal, I think Yuugi does not assume anything besides immersion. Did I miss something? $\endgroup$ – Selene Auckland Jul 25 at 5:24

See these:

What if potential errors in an answer are pointed out in comments but not addressed?

What is/are the definitions of local diffeomorphism onto image?

Neal says here that immersions are "local diffeomorphisms onto images". If we read "local diffeomorphisms onto images" as "local-(diffeomorphisms onto images)" rather than "(local diffeomorphisms)-onto images", then this is correct because diffeomorphisms onto (submanifold) images are equivalent to embeddings and because immersions are equivalent to local embeddings.

However, "(local diffeomorphisms)-onto images" imply images are regular/embedded submanifolds and not just immersed submanifolds. Therefore, Neal is wrong if Neal claims that immersions are "(local diffeomorphisms)-onto images".

Therefore, reading "local diffeomorphisms onto images" as "local-(diffeomorphisms onto images)", we have

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

These are the definitions:

Let $X$ and $Y$ be 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 $Y$.)

The difference in all these 3 is what $f(U)$ is. In all cases, $f(U)$ is a submanifold of $Y$, so indeed you still get a "diffeomorphism" out of an immersion.

Observe that while local diffeomorphism implies immersion but not conversely, local diffeomorphisms are equivalent to open immersions, to immersions whose domain and range have equal dimensions and to immersions that are also submersions (submersions are open maps).

  • $\begingroup$ I am puzzled by this response. If one possible interpretation of someone's statement is reasonable, and the other interpretation is nonsense, I'm not sure why you'd be confused by what they meant. After all, if I said out loud "My cat has grey fur," you wouldn't parse that as "My cat has grey. fur." Also please stop saying "with dimension" and linking to a question that is apparently unrelated. It's very distracting. Either require your manifolds be path connected or work locally, so dimension is well-defined. $\endgroup$ – jgon Jul 29 at 4:40
  • $\begingroup$ Also to that last point, all the links point to the same question. Why do that? $\endgroup$ – jgon Jul 29 at 4:47
  • $\begingroup$ @jgon Re dimensions: Tu considers $(0,1) \cup \{2\}$ to be a manifold. I was just trying to be considerate of other conventions. $\endgroup$ – Selene Auckland Jul 30 at 10:53
  • $\begingroup$ @jgon Re interpretations: I hadn't considered local-(diffeomorphism onto image) because I didn't know immersions are equivalent to local embeddings when I first encountered the term "local diffeomorphism onto image" $\endgroup$ – Selene Auckland Jul 30 at 10:54

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