# Torus T^2 homeomorphic to S^1 x S^1

• Math Amateur
In summary, Crossley's example 5.43 discusses the maps f and g and how they are constructed to map S^1 x S^1 onto T^2. The logic behind their design and construction involves using the equations for S^1 and T^2 and substituting in specific values to check that they satisfy the equation for T^2. The number 2 is important because it is part of the definition of the torus as S^1 x S^1.
Math Amateur
Gold Member
MHB
I am reading Martin Crossley's book, Essential Topology.

Example 5.43 on page 74 reads as follows:

I am really struggling to get a good sense of why/how/wherefore Crossley came up with the maps f and g in EXAMPLE 5.43. How did he arrive at these maps?

Why/how does f map $S^1 \times S^1$ onto $T^2$ and how does one check/prove that this is in fact a valid mapping between these topological spaces.

Can anyone help in making the origins of these maps clear or perhaps just indicate the logic behind their design and construction? I am completely lacking a sense or intuition for this example at the moment ... ...

Definitions for $T^2$ and $S^1$ are as follows:

My ideas on how Crossley came up with f and g are totally bankrupt ... but to validate f (that is to check that it actually maps a point of $S^1 \times S^1$ onto $T^2$ - leaving out for the moment the concerns of showing that f is a continuous bijection ... ... I suppose one would take account of the fact that (x,y) and (x',y') are points of $S^1$ and so we have:

$x^2 + y^2 = 1$ ... ... ... ... (1)

and

$x'^2 + y'^2 = 1$ ... ... ... ... (2)

Then, keeping this in mind check that

$((x' +2)x, (x' +2)y, y'$ is actually a point on the equation for $T^2$, namely:

$x^2 + y^2 + z^2 - 4 \sqrt{x^2 + y^2} = -3$ ... ... ... (3)

So in (3) we must:

- replace x by (x' +2)x
- replace y by (x' +2)y
- replace z by y'

and then simplify and if necessary use (1) (2) to finally get -3.

Is that correct? Or am I just totally confused ?

Peter

#### Attachments

• Crossley - 2 - EXAMPLE 5.43.jpg
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• Equation of Torus - Crossley - page 25.jpg
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• S^1 - Crossley - page 32.jpg
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Look at the terms. What is ##(x'+2)x##? What is important about the number 2 here? (Hint: Look at your definition of the torus.)

I've always defined the torus as ##S^1\times S^1##. It's kind of "obvious" if you think about it. Just look at this picture:

1 person

## 1. What does it mean for a Torus T^2 to be homeomorphic to S^1 x S^1?

Homeomorphism is a mathematical concept that describes a mapping or transformation between two topological spaces that preserves their fundamental properties, such as shape and connectivity. In this case, it means that the Torus T^2 and the product space S^1 x S^1 have the same topological structure, even though they may appear different geometrically.

## 2. How can a Torus T^2 be visualized as a product space S^1 x S^1?

A Torus T^2 can be visualized as a product space S^1 x S^1 by imagining a circle (S^1) rotating around another circle (S^1) in a perpendicular direction. This creates a three-dimensional shape with a hole in the center, which is the torus.

## 3. What are some real-life examples of a Torus T^2 being homeomorphic to S^1 x S^1?

One example is a donut, where the hole in the center represents the S^1 x S^1 product space. Another example is a coffee mug, where the handle can be seen as an S^1 x S^1 space attached to the S^1 base of the mug.

## 4. How is the concept of homeomorphism useful in mathematics?

Homeomorphism is useful in mathematics because it allows us to study topological spaces and their properties by using simpler, more familiar spaces. It also helps us understand the relationships between different topological spaces and how they can be transformed into each other.

## 5. Can a Torus T^2 be homeomorphic to a space other than S^1 x S^1?

No, a Torus T^2 can only be homeomorphic to a product space S^1 x S^1. This is because the fundamental group (a topological invariant) of the Torus T^2 is isomorphic to the fundamental group of the product space S^1 x S^1.

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