I'm trying learn the concept of a manifold. Wikipedia says a figure 8 is not a manifold but a klein bottle is. What distinguishes the two?
Is it because a Klein bottle is a 4-space object and not the 3-space projection of that object?
I'm trying learn the concept of a manifold. Wikipedia says a figure 8 is not a manifold but a klein bottle is. What distinguishes the two?
Is it because a Klein bottle is a 4-space object and not the 3-space projection of that object?
The comments already answered the question on a surface level, showing why the klein bottle is a manifold while the figure 8 isn't. But there is a deeper, conceptual similarity (or difference) here worth mentioning.
First consider a circle. Clearly, this is a one-dimensional manifold. Now we can embed the circle into $\mathbb R^2$ in the usual way, the image of this embedding will now be a submanifold. However we can also immerse it by the map $f : S^1\to \mathbb R^2$ defined $f(p)=(\sin(2t),\sin(t))$, where $p \in S^1$ has the coordinates $p = (\cos(t), \sin(t))$. The image will be exactly the figure 8. The map $f$ is locally injective and has nowhere vanishing differential, so one might hope it should also produce a submanifold. However note that $f$ is not (globally) injective, there is some self-intersection, namely $(-1,0)$ and $(1,0)$ get mapped to the same point. Hence it is not necessarily a sub manifold.
A similar thing happens with the klein bottle. The abstract definition, which could be a quotient construction as done here or the usual glueing construction (see below) gives a manifold.
And as with the circle before, we can immerse this manifold into $\mathbb R^3$, but we would necessarily get some points of self-intersection. This is illustrated by the famous models of klein bottles you have certainly seen somewhere already. And just as with the immersion of the circle onto the figure 8, the image (in $\mathbb R^3$) is not a manifold, altough the abstract object we started with is.
Furthermore, it's possible to have an injective map whose output is not a manifold. For example, you can take the line segment and map it onto the figure-8 without crossing itself (start your pen right next to the intersection point), which you know now isn't a manifold.
In general you can't say a whole lot about the output of a function just because it is/isn't injective.
– fish Mar 29 '25 at 21:37