Embedding, local diffeomorphism, and local immersion theorem.

Question

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

Answer 1

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.

Answer 2

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?

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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).