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I have seen several definitions of embedding in differential geometry. Let $f:M \to N$ be an injective smooth map between manifolds. Then I have heard that $f$ is an embedding if:

  • $f_*$ is injective, that is, $f$ is an immersion
  • for any $p \in M$ there exists a neighbourhood $U$ of $p$ with $f^{-1}:f(U) \to M$ smooth
  • $f$ is diffeomorphic onto its image
  • $f$ is an immersion which is also homeomorphic onto its image

My question is: are all these definitions equivalent, are some of them slightly incorrect, or are different definitions relevant for different purposes? Thanks.

gj255
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    Part 2 is definitely wrong, you can salvage it with some work though (replace $U$ with $U\cap f(N)$). Parts 3 and 4 are equivalent. This is what one usually mean by an "embedding" in the differentiable category. Part 1 is strictly weaker, but is equivalent to an embedding if $f$ is a proper map, e.g. if $M$ is compact. One can call it an "injective immersion". – Moishe Kohan Apr 21 '16 at 12:53
  • I actually copied it incorrectly. I've modified the definition, does it make sense now? – gj255 Apr 21 '16 at 13:04
  • With this modification, 1 is equivalent to 2. – Moishe Kohan Apr 21 '16 at 13:11
  • OK, thank you for the help. I don't suppose you could point me to a resource which shows 1,2 and 3,4 to be equivalent? – gj255 Apr 21 '16 at 13:14
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    This is a good exercise to work out on your own. Start with 1 and 2: Is there a theorem about local properties of immersions that can help you here? For 3=4, use 1=2 (a smooth homeomorphic map is a diffeomorphism iff its inverse is smooth). – Moishe Kohan Apr 21 '16 at 13:19
  • As studiousus says, $1$ is weaker without some extra hypothesis. Here's an explicit example: Map $\mathbb{R}$ into $\mathbb{R}^2$ with image the letter $P$. Then neighborhoods of the triple point don't look like neghborhoods in $\mathbb{R}$, but the mapping is a smooth injective immersion. – Jason DeVito - on hiatus Apr 21 '16 at 14:33
  • @studiosus I've read that an immersion is locally an embedding, but I don't have much experience with this subject --- what local property were you referring to? – gj255 Apr 21 '16 at 15:04
  • Which textbook are you reading? Guillemin and Pollack? Hirsch? Pretty much any book on differential topology will have this theorem. This is really basic. – Moishe Kohan Apr 21 '16 at 15:06
  • I'm actually studying general relativity, and courses on GR usually skip over some of the more basic theory of manifolds and jump straight into Riemannian geometry. I've picked up what I know from various places. – gj255 Apr 21 '16 at 15:08
  • I see. Then read chapter 1 of Guillemin and Pollack. It will help a lot. – Moishe Kohan Apr 21 '16 at 15:11
  • OK, there's one thing I still can't show, which is how we get from 4 to 3: I need to show that $f^{-1}$ is smooth, but the fact that $f$ is an immersion only tells me that $f^{-1}$ is locally smooth, and I don't know how to employ the fact that $f$ is homeomorphic to its image here... – gj255 Apr 21 '16 at 15:46
  • @gj255: Smoothness is a local property, you verify it at each point of the domain. – Moishe Kohan Apr 23 '16 at 02:44

1 Answers1

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The definition that I am familiar with is the last definition. For an embedding, the two properties give different ideas:

  • A bijective immersion means that there is some topology that can be placed on the image so that the image is homeomorphic to the original manifold.

  • The condition of being homeomorphic with its image using the subspace topology says that the topology in the bullet above is actually the subspace topology (not some other weird one that differs from $N$'s topology).

Now, some of what you write may be equivalent (certainly not the first one), and the diffeomorphism in the third one needs to be from the subspace topology. In general, an embedding means that the image is a regular submanifold of the codomain, so $f(M)$ looks like $M$ and also looks like a submanifold of $N$.

Michael Burr
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  • Thanks for the answer. When I say that $f$ is diffeomorphic onto its image, presumably it's almost always implied that that I give the image the natural topological and differentiable structure that one can give a subset of a manifold? One other question: how do we know that $\mathrm{im} f$ is actually a manifold, such that it makes sense to call $f$ a diffemorphism? – gj255 Apr 21 '16 at 14:57
  • By unsaid implication it's a topological space. And given the map $f$ you should be able to construct an Atlas - thus it's a manifold. It can only be diffeomorphic if you either specify charts here and there or use functions/curves and the push forward to define differentiation. With either of those notions of differentiation you can establish smootheness and tadaa.. it's diffeomorphism time. – mike Apr 21 '16 at 15:13
  • So if $f$ is an injective immersion then $\mathrm{im} f$ is not necessarily a manifold, right? So something about $f$ also being homeomorphic onto its image somehow ensures $\mathrm{im} f$ is indeed a manifold... – gj255 Apr 21 '16 at 15:25
  • An embedding is equivalent to a regular submanifold, so the image is always a manifold. In general, any bijective map will give a manifold provided that you give the image the induced topology. – Michael Burr Apr 21 '16 at 15:33
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    The standard example of an immersion whose image is not a submanifold is the image of $\mathbb{R}$ in $\mathbb{R}^2$ where you curve one infinite tail of $\mathbb{R}$ so that it bends back and accumulates on a point (think of a candy-cane whose curved part comes back and almost touches the straight part). – Michael Burr Apr 21 '16 at 15:35