Is number "8" topologically equivalent to a circle in the 2D space?

Question

We may treat the cross point of number "8" is just an overlap of two lines (instead of a "melting" point of two lines, so that it is not treated as two connected circle). To make it clear, for example, suppose we have a rope which are connected as a circle. Now I rotate part of the circle to make it a "8".

The question is, does this kind of "8" topologically equivalent to a circle in the two dimensional space?

We may say it is homeomorphic to a circle, because we can smoothly move from one to another. But I think they are still not equivalent in 2D because to smoothly change number "8" to a circle, we also need to rotate half part of "8" in the vertical direction. And the vertical motion is outside the the 2D, and is not allowed within the two dimensional space. Am I right?

Answer 1

No, they are not topologically equivalent. The fundamental group of the circle is the group $\mathbb{Z}$ of integers, and that of 8 is the free group on two symbols.

Answer 2

This really has nothing to do with $3$d or $2$d space.

If you remove any one point from a circle, the remaining space is connected.

That is not true for all points in the figure $8.$ Specifically, if you remove the cross point from the space, you end up with a space with two components.

Answer 3

They are not the same and you don't need winding numbers nor do you need fundamental groups.

There are two relevant facts:

1. A continuous function remains continuous if its domain is restricted.
2. Continuous functions take connected sets to connected sets.

If there was a homeomorphism from the circle to the figure eight, some point of the circle, call it $p$ gets mapped to the cross point on the figure eight. Remove that point from the domain. Then the homeomorphism would take the circle minus one point to the figure eight but, since the circle with a point removed is connected, the entire domain must be mapped to a connected component of the figure eight minus a point. Since that object is not connected the function can no longer be surjective so it could not have been a homeomorphism.

Answer 4

Perhaps you are thinking of the diagram '8' as a knot diagram: take your rope in $\mathbb{R}^3$, twist it once, and then project down onto $\mathbb{R}^2$. The result, which you can view as the symbol '8' (with a notion of over-under at the center point), is not itself homeomorphic to $S^1$, but the original knot certainly is.

So viewing the number '8' as a wedge of two circles is certainly not homeomorphic to $S^1$, but viewing it as a 4-valent graph with one vertex and two edges, as well as the information about overcrossing/undercrossing at the vertex, we get the standard 1-crossing diagram of the unknot.

However, as you note, this 'over-under' information encodes something that doesn't occur in the projection onto our page ($\mathbb{R}^2$), but occurs with our rope in space ($\mathbb{R}^3$).