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I am going through section 2.8 of Aluffi's Algebra: Chapter 0. The exercise 8.10 asks us to find the quotient of $\Bbb R^2$ over $\Bbb Z^2$. The answer is of course a torus. However, I am wondering if the sphere $\Bbb S^2$ also works.

There should be a surjective group homomorphism from $\Bbb R^2$ to $\Bbb S^2$ by sending $(a,b)$ to $(r, 2\pi a, 2\pi b)$ for some fixed radius $r$. The kernel of this homomorphism should be $\Bbb Z^2$. By Isomorphism Theorem $\Bbb S^2$ would be isomorphic to the quotient space of $\Bbb R^2$ over $\Bbb Z^2$.

This shouldn't be right. Where did I do wrong? Any help is appreciated!

Daniele Tampieri
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Gladiator
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    The formula you gave defines a map onto the torus, not a sphere. Hence it proves your original statement, which is good. – Michal Adamaszek Jan 27 '20 at 09:55
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    I was trying to describe the circular movement horizontally and vertically around a sphere though. The second term refers to how much is turned horizontally around the sphere and the third term vertically. Does it make any sense? – Gladiator Jan 27 '20 at 10:09

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For starters the sphere $\mathbb{S}^2$ is not a topological group (no matter how you define the group operation). So topological group theory doesn't even apply here. Unless you mean non-topological group theory, but then you have to specify the group operation on $\mathbb{S}^2$. Such group operation won't be "nice" and surely won't arise naturally, unlike the torus case.

Note that in non-topological context the answer can be "yes, this is correct", depending on the group structure. Which isn't really surprising. Without topology torus and sphere are pretty much the same thing (just sets of cardinality $\mathfrak{c}$).

Secondly, the mapping $f:(a,b)\mapsto(r,2\pi a, 2\pi b)$ has different "kernel" (of course not in the algebraic sense, see here) on the sphere. Note that on the torus $f(\frac{1}{2}, \frac{1}{2})\neq f(0,0)$ while on the sphere the equality holds.

freakish
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