Definition on an n-torus, two approaches.

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This wasn't originally a homework problem as such, so sorry if its confusing, but I thought I would ask it here;

Homework Statement

Show that the the two methods of creating the n-torus are equivalent.
1) The n-torus as the quotient space obtained from ℝⁿ by the relation x~y iff x-y $\in$ Zⁿ.
2) The n-torus given by the product of n circles Tⁿ= S¹×S¹×...×S¹.

Homework Equations

S¹ is the collection of equivalence classes [x]={y:y~x}.
An equivalence relation x~y iff x-y $\in$ Z

The Attempt at a Solution

Well, I suppose we want an inductive proof? In our second definition, when n=1, we just have S¹. This is clearly the same as the definition for S¹. So then we want to then look at n = k and n = k + 1...What's the best way to go about this?

( Note Z means the set of integers)

Thanks,

 

[lippyka]

Solution: We will use induction to show that the two methods of creating the n-torus are equivalent. Base Case: For n = 1, the n-torus is given by S1. This is the same as the definition of the n-torus in method 1, which is obtained by taking the quotient space of ℝ1 by the relation x ~ y iff x - y ∈ Z1. Therefore, the two methods of creating the n-torus are equivalent for n = 1.Inductive Step: Assume that the two methods of creating the n-torus are equivalent for n = k. We want to show that they are also equivalent for n = k + 1. The n-torus for n = k + 1 can be given by the product of k+1 circles, Tk+1 = S1 × S1 × ... × S1. Furthermore, we can take the quotient space of ℝk+1 by the relation x ~ y iff x - y ∈ Zk+1. Now, we note that Zk+1 = Zk × Z1 and so we can rewrite the relation as x ~ y iff (x - y) ∈ Zk × {0}. This implies that (x1,...,xk+1) ~ (y1,...,yk+1) iff (x1 - y1,...,xk - yk) ∈ Zk and xk+1 - yk+1 = 0. But this is precisely the condition for taking the product of k circles and identifying the last circle with the first one. Thus, we have shown that the two methods of creating the n-torus are equivalent for n = k + 1. By induction, we have shown that the two methods of creating the n-torus are equivalent for all n ∈ ℕ.