# Embedding, immersion

Could someone please explain what "embedding" means? (Maybe a more intuitive definition) I read that the Klein bottle and real projective plane cannot be embedded in ${\mathbb R}^3$ but is embedded in ${\mathbb R}^4$. Aren't those 2 things 3D objects? If so why aren't they embedded in ${\mathbb R}^3$? Also, I have some across the word "immersion". What is the difference between "immersion" and "embedding"?

Thanks.

-
"Embedding" something means to "stick it inside"; intuitively, if $A$ embeds into $B$, that means that there is a way to "stick" $A$ "inside" $B$, or that $B$ 'contains' a copy of $A$. There is no way to put a copy of the Klein bottle inside $\mathbb{R}^3$ because it is not really a 3-dimensional object: you cannot have a Klein bottle in $\mathbb{R}^3$ without self-intersection, even though the Klein bottle doesn't have any (just like the Moebius strip cannot be done in $\mathbb{R}^2$). "Immersion" is usually synonymous with "embedding", but used in certain fields more than in others. – Arturo Magidin Sep 28 '11 at 16:47
@Arturo: "Immersion" just means that the derivative is injective on the tangent space at each point. The manifold is allowed to self-intersect. – Grumpy Parsnip Sep 28 '11 at 16:54
So I'm saying the two terms are NOT synonymous. – Grumpy Parsnip Sep 28 '11 at 16:54
@ArturoMagidin: Thanks! I think I am not sure why the Klein bottle is not a 3D object...? I can see why the Mobius strip is not 2D though. Would you mind explaining a bit please? – surface Sep 28 '11 at 16:55
@surface: Essentially for the same reason that the Moebius strip is not a 2D object even though it is a 2-dimensional manifold. If you try to construct a Klein bottle in 3-space, you end up necessarily having to self-intersect in order to get the "twist" necessary to make it have only one "side" (i.e., no interior and no exterior); this "twist" can only happen in 4-space if you don't want the bottle to intersect itself, just like the twist of the Moebius strip has to happen in 3-dimensions if you don't want the strip to intersect itself in 2 dimensions. – Arturo Magidin Sep 28 '11 at 16:58

Basically an abstract surface has, at every point two independent directions along the surface. Or even better, there is an entire circle's worth of rays coming out of each point.

An immersion is, roughly, a map of the surface into a bigger manifold (such as $\mathbb R^n$) where there are still two dimensions worth of rays emanating out of each point. So for the usual embedding of a Klein bottle into $\mathbb R^3$, at the circle of self-intersection, each sheet still retains its two dimensional character. So it is an immersion. If you were to instead map the Klein bottle into $\mathbb R^3$ by mapping everything to a point, that would not be an immersion.

-

Understandably there are a lot of answers, but if you still have any further questions maybe this will help.

An embedding of a topological space $X$ into a topological space $Y$ is a continuous map $e \colon X \to Y$ such that $e|_X$ is a homeomorphism onto its image.

Both the Klein bottle ($f \colon I^2 / \sim \to \mathbb{R}^3$) is not embedded into $\mathbb{R}^3$, because it has self-intersections; this means that the immersion of the Klein bottle is not a bijection, hence not a homeomorphism, so not an embedding.

As I understand it, an immersion simply means that the tangent spaces are mapped injectively; i.e. that the map $D_p f \colon T_pI^2 \to T_{f(p)}\mathbb{R}^3$ is injective. In the Klein bottle example, at the self-intersection, any point of intersection has two distinct tangent planes, hence this map is injective.

I hope this makes some sense!

-

An immersion is precisely a local embedding – i.e. for any point x ∈ M there is a neighbourhood [sic], U ⊂ M, of x such that f : U → N is an embedding, and conversely a local embedding is an immersion.

So, an immersion is an embedding, i.e. an isomorphic (homeomorphic) copy, at each point, and vice versa, though the entire image may not be a homeomorphic copy.

But, later, the same article says:

If M is compact, an injective immersion is an embedding, but if M is not compact then injective immersions need not be embeddings; compare to continuous bijections versus homeomorphisms.

I wish I could give an example of a non-compace imbedding/embedding, or continuous bijections versus homeomorphisms, but though I understand the two ideas, roughly, I am not sure what conditions make them conflict...

-