The map between smooth manifolds $f:M \to N$ is called an immersion if the tangent map $Tf$ is injective for each $x \in M$. There is a corresponding notion of submersion. About embeddings I met two definitions: the first states that a map $f:M \to N$ is called an embedding if it is an immersion and is one-to-one: the second definition requires that the map is the embedding in the topological sense.
1. Are these definitions equivalent? If not then:
2. Which definition is correct?
3. If the first definition doesn't implies the second why it is called embedding while it would not be an embedding in the topological sense? If however these definitions are equivalent, why do we assume that the map in an immersion?
(I wonder whether is there some clever way to declare which maps $f:M \to N$ are arrows in some to be category in such a way that the monomorphism in this category will be exactly immersion and epimorphism-submersion).

  • $\begingroup$ They are not equivalent. The second is correct. See Lee smooth manifolds page 156 for definition and page 157 example 7.3 for an example of an injective immersion which is not an embedding. $\endgroup$ – Seth Feb 6 '15 at 17:02
  • $\begingroup$ An easier example btw is on the wikipedia page en.wikipedia.org/wiki/Immersion_(mathematics) $\endgroup$ – Seth Feb 6 '15 at 17:08
  • 1
    $\begingroup$ The correct definition is an immersion which is a topological embedding. I am assuming that's what you meant by the second definition but it isn't completely clear. $\endgroup$ – Seth Feb 6 '15 at 17:10
  • $\begingroup$ @PVAL: A proper injective immersion is always an embedding, but not conversely. E.g, the inclusion of an open subset into $M$ is a smooth embedding, but generally not proper. $\endgroup$ – Jack Lee Feb 6 '15 at 17:25
  • $\begingroup$ Proper means the inverse image of each compact set is also compact? $\endgroup$ – truebaran Feb 6 '15 at 17:29

The definition is a combination of the two options you give. A smooth map $f:M\to N$ is an embedding if it is an immersion, and an embedding in the topological sense.

If we omit the requirement that $f$ is an immersion, we get for example that $$f:\mathbb{R}\to\mathbb{R}^2,\quad x\mapsto(x^2,x^3)$$is an embedding. In other words, the image of $f$, which we call $c$, is a submanifold of $\mathbb{R}^2$. This is "bad" because the tangent space at the point $(0,0)$ does not behave as we want it to. The curve $c$ is a $1$ dimensional submanifold, and for $p\in c$ we expect $T_pc$ to be a $1$ dimensional subspace of $T_p\mathbb{R}^2$. However, if $\gamma:(-\epsilon,\epsilon)\to c$ is smooth as a map into $\mathbb{R}^2$ and satisfies $\gamma(0)=(0,0)$, then necessarily $\dot{\gamma}(0)=0$.

Edit, written a while after the original answer: There is another very good reason to require that an embedding will always be an immersion. Such embeddings have the following nice property:

Let $i:M\hookrightarrow N$ be an embedding of smooth manifolds, and let $f:L\to N$ be a smooth map, such that $\mathrm{Im}(f)\subset\mathrm{Im}(i)$. Then the induced function (in category of sets) $\tilde{f}:L\to M$ is a smooth map of manifolds.

If we omit the requirement that $i$ is an immersion, the above property does not necessarily hold. Note that in the category of topological spaces and continuous functions, the topology of a subspace is defined in the way that makes the corresponding property hold. When playing with smooth manifolds, we need an additional assumption which makes embeddings well-behaved.

  1. They are not. However, it is clear that the second one always implies the first one.
  2. The second one is more apt, if you think about its topological implications.
  3. The first one implies the second if the map is in addition a proper map. I don't think a lot of people call an injective immersion an embedding. (It is however a local embedding.)
  • $\begingroup$ Thank You, apparently the first definition which I met had a hidden assumption: the context of closed manifolds in which in this equivalent to the second. $\endgroup$ – truebaran Feb 6 '15 at 17:45

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