# smooth embedding between manifolds

I'm sightly confused about the definition of an embedding between manifolds. (There seems to be several formulations and apparently they are meant to be equivalent.) From what I gather a smooth map $f: M \to N$ between manifolds is an embedding if it is (1) an immersion (smooth and derivative is injective), and (2) it is a topological embedding (homeomorphism onto image)

With this definition, the image of an embedding is a manifold (a submanifold of $N$) and $f$ is a diffeomorphism onto its image.

To fully understand this definition, can someone give me an example of (a) an injective immersion that is not an embedding (for example if the image is not a manifold), and (b) the necessity of the requirement of it being an immersion (for example a smooth injection whose image is not a manifold).

PS. I have seen an example of (a) that is the injective immersion from $\mathbb{R}$ to the figure 8 in $\mathbb{R}^2$ by taking $\pm \infty$ to the intersection of the figure 8 from top right/ bottom left. But I'm having trouble confirming why it's not a homeo onto image (it's clearly bijective but why is it not continuous wrt to the subspace topology of the figure 8 in $\mathbb{R}^2$?)

• For your PS: they have different fundamental groups. Any continuous mapping $\mathbb{S}^1\to \mathbb{R}$ is contractible to a point. There is an obvious mapping $\mathbb{S}^1 \to \mathbf{8}$ that is not contractible. – Willie Wong Oct 18 '16 at 14:52
• For your (b), take the cusp parametrization $t \mapsto (t^3, t^2)$, which is smooth and injective, but the image has no tangent space at the origin. – Andrew D. Hwang Oct 18 '16 at 14:54
• Thanks for both your comments. @WillieWong: I'm not that familiar with the fundamental group so will need to spend some time going over it. But just thinking about it in the "pedestrian way", it seems to me that the map $f: \mathbb{R} \to \mathbb{8} \subset \mathbb{R}^2$ defined above is continuous (any open set containing the intersection point does have a preimage that's open), so I'm led to believe that the problem is $f^{-1}$ is not continuous, i.e. f is not open. But it seems that any open interval does map to an open set in $\mathbb{8}$... so I'm not sure what goes wrong – AMFS Oct 18 '16 at 15:43
• Alex, in the case of the figure 8, the mapping is continuous, but it's not an open map (i.e., the inverse is not continuous). Consider what happens under the inverse mapping near the "crossing point" of the 8. – Ted Shifrin Oct 18 '16 at 16:37
• oh yes of course - sometimes I forget to think about things in terms of the subspace topology, and in this case I managed to fool myself into thinking that the one "branch" crossing the intersection point (the image of an open interval containing 0) is open! – AMFS Oct 18 '16 at 22:16

Let's call $\phi$ the parametrization of the 2-torus in $\mathbb{R}^4$, such that: $\phi: \mathbb{R}^2 \to T^2 =(\sin(x),\cos(x),\sin(y),\cos(y))$. Now, let $\pi$ be a line in $\mathbb{R}^2$ with irrational angolar coefficient, let's set it at $\sqrt{2}$. Let $\varphi:=\phi_{\vert \pi}$. Thus, the mapping is differentiable and it is, in fact an immersion, but is not an embedding, since the image is one-dimensional (yet, these sub-manifold is dense in the torus). Actually, in both these example and in yours, the patological behaviour is in some dense caused by the fact that those mappings are not "proper", they map too many points near infinity near to others. Formalizing this definition, we get the embedding' definition. It is necessary for it to be an immersion since, if it is not, thus the jacobian matrix won't have maximum rank somewhere, and thus the transformation won't be invertible