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By the embedding of a smooth manifold I mean a smooth $f : M \to N$ such that $f(M)$ is a submanifold of $N$. The usual criterion that one sees in books requires $f$ to be an immersion and a homeomorphism onto its image. It seems to me that one needs only $f$ to be an immersion that is open on its image, i.e. the global injectivity of $f$ is useless (the problematic self intersections are taken care of by the openness condition). Am I correct? If I am, why don't the books mention this? The proof is exactly the same.

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    I think a map $\mathbb R\to\mathbb R^2$ which turns the line into a "lowercase letter b" is an injective immersion which is open onto its image but is not an embedding. It fails to be a homeomorphism onto its image, as it's not a closed map. – Alex K May 30 '20 at 16:50
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    @AlexK - Is that map open onto its image? It seems like the vertical part of the "b" is the image of an open set, but it's not open. – Phillip Andreae May 30 '20 at 18:34
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    How about the covering map $\mathbb{R} \to S^1$? – Phillip Andreae May 30 '20 at 18:36
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    @PhillipAndreae Of course, you're right! Your example is also more in line with what the post is asking - openness doesn't prevent the map from wrapping around itself globally. – Alex K May 30 '20 at 20:56
  • @PhillipAndreae I don't understand what you mean: the lowercase b function is indeed not open as you said but the image of the covering $\mathbb{R} \to S^1$ (and indeed of any covering of any manifold since the image of a covering is a union of connected compnents) is a submanifold of $S^1$, being the whole $S^1$. –  May 31 '20 at 09:56
  • @student_du_05 - I assumed that you wanted injectivity to be part of the definition of embedding, which is standard. Can you clarify what your definition of embedding is, and what your question is asking? "smooth $f: M \to N$ such that $f(M)$ is a submanifold of $N$" is not the standard definition (and e.g., any surjective $f$ satisfies it). – Phillip Andreae May 31 '20 at 15:00
  • I phrased my post badly. I want to know whether it is true that any map $f : M \to N$ that is an immersion and is open on its image is such that $f(M)$ is a submanifold of $N$. That is, if you remove the injectivity condition on embeddings, is the image still a submanifold? Of course then the non-injective 'embedding' won't be a diffeomorphism onto its image like real embeddings are. –  May 31 '20 at 16:10
  • This question because the proof of the fact that $f(M)$ is a submanifold when $f$ is a(n actual) embedding doesn't seem to me to use (global) injectivity. –  May 31 '20 at 16:17
  • So you're asking whether this is true, is that right? "If $f: M \to N$ is an immersion and is open onto its image, then $f(M)$ is a submanifold of $N$." I think this answers that question (yes): https://math.stackexchange.com/questions/568160/submanifold-given-by-an-open-immersion – Phillip Andreae Jun 01 '20 at 17:00
  • Yes! Thank you for the link. –  Jun 01 '20 at 18:23

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