What is the difference between "immersion" and "embedding"? 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 two things 3D objects? If so, why aren't they embedded in ${\mathbb R}^3$? Also, I have come across the word "immersion". What is the difference between "immersion" and "embedding"?
Thanks.
 A: via Wikipedia on Immersions

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-compact imbedding/embedding, or continuous bijections versus homeomorphisms, but though I understand the two ideas, roughly, I am not sure what conditions make them conflict...
A: In simple intuitive terms, the immersion of a topological object is a reasonably smooth injection or mapping of it into some containing space, while its embedding adds the stiffer condition that the mapping be bijective, i.e. point-to-point with no overlaps.
As a simple example, a topological disc can be embedded in the Euclidean plane but a Moebius strip cannot because it must have a crossing point where the mapping is not bijective. Such an injection of the Moebius strip or, say, a folded disc, is an immersion but not an embedding. The Moebius strip can be embedded in Euclidean space.
Similarly a Klein bottle cannot be embedded in Euclidean 3-space because it must have a crossing line, while the projective plane cannot because it must have a triple-point where three regions of it intersect. Both can be immersed in Euclidean 3-space and both can be embedded in Euclidean 4-space.
Topologically they are, like the disc and Moebius strip, 2D manifolds or surfaces. However where the disc and Moebius strip are manifolds with a boundary, the Klein bottle and projective plane have no boundary. The projective plane in particular is often regarded as a 2D space in its own right. It is only when one tries to embed them locally into some containing space that they become "4D." The only reason we might think of them as 3D is because we tend to think in 3D and they are therefore often first presented to us in textbooks in the form of their 3D immersions.
A: Reading a paper by Claudio Gorodski (2012), there is a section he called ``Riemann submanifolds and isometric immersions'', where he considers a Riemann manifold $(M,g)$ and an immersed submanifold $\iota :N \rightarrow M.$ Thus $N$ is a smooth manifold and $\iota$ is an injective immersion. Riemann metric $g$ induces a Riemann metric $g_N$ in $N$. So if $p \in N$, the tangent space $T_pN$ can be viewed as a subspace of $T_pM$ via the injective map $d_{\iota_p}: T_pN \rightarrow T_\iota(p)M.$ We may define $(g_N)_p$ to be the restriction of $g$ to this subspace:
$$(g_N)_p(u,v)=g_{\iota_p}(d_{\iota_p}(u),d_{\iota_p}(v)),$$
where $u,v\in T_pN.$ So $g_N$ is a Riemann metric. Indeed $g_N$ is an induced Riemannian metric in $N$, so that $(N,g_N)$ is a Riemannian submanifold of $(M,g.$
It is worthy to note that $g_N$ makes sense even if $\iota$ is an immersion that is not necessarily injective. $g_N$ is said to be the pulled-back metric, $g_N =\iota^*g,$ and $\iota : (N,g_N)\rightarrow(M,g)$ is an isometric immersion. But any immersion must be locally injective.
An important particular case is that of Riemannian submanifolds of Euclidean space. A regular curve or surface is an immersion whenever there is a parametrization which is assumed smooth. In such a case the parametrization is locally an embedding.   
A: 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!
A: 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 immersion 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. 
