How to prove that a given set is a torus

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In summary, the author is trying to solve an exercise and states that if C is a circular cylinder with S_1 and S_2 as its boundary circles and S_1 and S_2 are identified by mapping them both homeomorphically onto some third circle S, giving a map f: S_1 \cup S_2 \rightarrow S then (C - S_1 \cup S_2) \cup S with the identification topology (induced by the function that identifies all the points in S_1 and S_2 with corresponding points in S) is a torus. It is not enough to show that the topology on (C - S_1 \cup S_2) \cup S
  • #1
glmuelle
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Hi

I'm trying to solve this exercise

"Prove that if C is a circular cylinder with S_1 and S_2 as its boundary circles and S_1 and S_2 are identified by mapping them both homeomorphically onto some third circle S, giving a map [tex] f: S_1 \cup S_2 \rightarrow S[/tex] then [tex] (C - S_1 \cup S_2) \cup S[/tex] with the identification topology (induced by the function that identifies all the points in S_1 and S_2 with corresponding points in S) is a torus."

Is it enough to show that the topology on [tex] (C - S_1 \cup S_2) \cup S[/tex] is the subspace topology of R^3?

Thanks for your help
 
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  • #2
No, that is not enough. A point has the subspace topology on R^3 but a point is not a torus, is it?

The most direct way to prove that two topological spaces are the same (i.e are homeomorphic) is of course to construct an explicit homeomorphism between them. In this case, it should not be difficult to do so. Call T the usual torus in R^3 and T' the wannabe torus. I suggest that you find a continuous bijective map T'-->T. Since T' is compact and T is Hausdorff, it will follow that your map is a homeomorphism.

Also note that to show that your map is continuous, it suffices to show that it is continuous as a map from T' into R^3, since T has the subspace topology induced by R^3.
 
  • #3
Thank you very much, that answered my question!
 
  • #4
I don't understand you're question, but it's late and I'm probably just being completely mad, but can't you identify the two circles on a cylinder in a way to produce a Klein bottle instead of a torus? If you could show that it could also be embedded into R^3 then it would indeed be a torus because the Klein bottle cannot be embedded into R^3, only R^4; maybe this is what you meant at the end of your post?
 
  • #5
!

To prove that a given set is a torus, we need to show that it has the defining properties of a torus. One way to approach this problem is to use the definition of a torus as a surface of revolution.

First, we can see that the given set, (C - S_1 \cup S_2) \cup S, is a closed surface without boundary, which is one of the defining properties of a torus. This can be proven by showing that the set is homeomorphic to a closed disk, which is a well-known closed surface.

Next, we need to show that the set is a surface of revolution. This can be done by considering the mapping f: S_1 \cup S_2 \rightarrow S, which identifies the two boundary circles S_1 and S_2 onto a third circle S. This mapping can be thought of as a rotation of the cylinder C around the axis defined by S. Therefore, the resulting set is indeed a surface of revolution, which is another defining property of a torus.

Lastly, we need to show that the set has the topology of a torus. This can be done by considering the subspace topology induced by the function that identifies all the points in S_1 and S_2 with corresponding points in S. This topology is equivalent to the standard topology on the torus, thus proving that the given set is indeed a torus.

In conclusion, to prove that the given set is a torus, we need to show that it has the defining properties of a torus, namely being a closed surface without boundary, a surface of revolution, and having the topology of a torus. By following the steps outlined above, we can fully prove that the given set is indeed a torus.
 

Related to How to prove that a given set is a torus

1. How do you define a torus?

A torus is a three-dimensional geometric shape with the surface of a doughnut, where the inner and outer circles are parallel and the distance between them is the same at all points. It can also be described as a surface of revolution generated by revolving a circle in three-dimensional space about an axis coplanar with the circle.

2. What are the characteristics of a torus?

A torus has a continuous surface with no edges or corners, and it has a constant width and thickness throughout its shape. It is also a closed and compact surface, meaning it has no boundaries and can be described as a finite surface without any gaps.

3. How can you prove that a given set is a torus?

There are several ways to prove that a given set is a torus. One way is to show that the set satisfies the defining characteristics of a torus, such as having a continuous surface with a constant width and thickness. Another way is to use the parametric equations for a torus and show that the set satisfies these equations.

4. What are the parametric equations for a torus?

The parametric equations for a torus are:
x = (R + r * cos(u)) * cos(v)
y = (R + r * cos(u)) * sin(v)
z = r * sin(u)
where R is the distance from the center of the torus to the center of the tube, r is the radius of the tube, u is the angle of rotation around the axis of the torus, and v is the angle of rotation around the center of the tube.

5. Can a torus exist in higher dimensions?

Yes, a torus can exist in higher dimensions. In fact, a torus can be defined in any number of dimensions, as long as it satisfies the characteristics of a torus. For example, a 4-dimensional torus is commonly known as a hypertorus.

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