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Prove that $\mathbb{T}^2$ is homeomorphic to $S^1 \times S^1$. Here is my attempt on this. Is this correct? Let $C^*$ be the torus as an identification space of $[0,1] \times [0,1]$ under the quotient map $g:[0,1] \times [0,1] \rightarrow C^{*}$. Then $C^*=\{\{(x,0),(x,1)\}|0 < x < 1\} \cup\{\{(0,y),(1,y)\}|0 < y < 1\} \cup \{\{(x,y)\}|0 < x < 1, \ 0<y<1\} \cup\{\{(0,0),(0,1),(1,0),(1,1)\}\}$

Define the function $f:C^* \rightarrow S^1 \times S^1$ by $f(\{(x,y)\})=((\cos 2 \pi x,\sin 2 \pi x),(\cos 2 \pi y, \sin 2 \pi y)),0<x<1,0<y<1,f(\{(x,0),(x,1)\})=((\cos 2 \pi x,\sin 2 \pi x),(1, 0)),f(\{(0,y),(1,y)\})=((1,0),(\cos 2 \pi y, \sin 2 \pi y)),f(\{(0,0),(0,1),(1,0),(1,1)\})=((1,0),(1,0))$

$C^*$ is compact since it is the surjective image of the continuous function $g$. Similarly, $S^1 \times S^1$ is Hausdorff, since it is a subspace of the Hausdorff space $\mathbb{R}^2 \times \mathbb{R}^2$. Because of this it suffices to show $f$ is a continuous bijection, in order to prove it is a homeomorphism.

$f(\{(x,y)\}=f(\{(w,z)\} \implies ((\cos 2 \pi x,\sin 2 \pi x),(\cos 2 \pi y, \sin 2 \pi y))=((\cos 2 \pi w,\sin 2 \pi w),(\cos 2 \pi z, \sin 2 \pi z)) \implies x=w,y=z \implies \{(x,y)\}=\{(w,z)\}$, similarly $f(\{(x,0),(x,1)\})=f(\{(w,0),(w,1)\}) \implies \{(x,0),(x,1)\}=\{(w,0),(w,1)\}$ and $f(\{(0,y),(1,y)\})=f(\{(0,z),(1,z)\}) \implies \{(0,y),(1,y)\}=\{(0,z),(1,z)\}$ so that $g$ is injective.

Let $((\cos 2 \pi x,\sin 2 \pi x),(\cos 2 \pi y, \sin 2 \pi y)) \in S^1 \times S^1$. If $x=0,y \neq 0$ then $f(\{(0,y),(1,y)\})=((1,0),(\cos 2 \pi y, \sin 2 \pi y))$. If $x \neq 0,y=0$ then $f(\{(x,0),(x,1)\})=((\cos 2 \pi x,\sin 2 \pi x),(1, 0))$. If $x=0,y=0$ then $f(\{(0,0),(0,1),(1,0),(1,1)\})=((1,0),(1,0))$. Otherwise $f(\{(x,y)\}=((\cos 2 \pi x,\sin 2 \pi x),(\cos 2 \pi y, \sin 2 \pi y))$. So that the function is surjective.

Finally consider the quotient map $g:[0,1] \times [0,1] \rightarrow C^*$. $f$ is continuous if and only if $f \circ g$ is continuous by one of the well known theorems. Now for each $1 \leq i \leq 2$, the projection function $\pi_i \circ f \circ g:[0,1] \times [0,1] \rightarrow \mathbb{R}^2$ is continuous since $(\pi_1 \circ f \circ g)(x,y)=(\cos 2 \pi x,\sin 2 \pi x), (\pi_2 \circ f \circ g)(x,y)=(\cos 2 \pi y, \sin 2 \pi y)$ are continuous. Similarly $(1,0),(0,1)$ are continuous. So $f$ is continuous by one of the well known theorems and $f$ is a homeomorphism being a continuous bijection from a compact space to a Hausdorff space.

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ernesto
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1 Answers1

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So better let $f: [0,1]^2 \to S^1 \times S^1$ be defined by

$$f(x,y) = ((\cos 2 \pi x,\sin 2 \pi x),(\cos 2 \pi y, \sin 2 \pi y))$$

and check that $$f(x,y) = f(x',y') \iff (x,y) \sim (x',y')\tag{1}$$ under the equivalence relation that identifies the edges in the right way for the torus, i.e.

$$(x,y) \sim (x',y') \iff (x \sim_1 x') \land (y \sim_1 y')$$

where $\sim_1$ is an equivalence relation on $[0,1]$ with classes $\{0,1\}, \{\{x\}\mid 0 < x < 1\}$

Fact $(1)$ quite easily follows from the observation that $$(\cos 2\pi x = \cos 2\pi y) \land (\sin 2 \pi x = \sin 2 \pi y) \iff x-y \in \Bbb Z$$ applied to both components and $x-y \in \Bbb Z \land x,y \in [0,1]$ iff $x \sim_1 y$

Then $(1)$ gives continuity and 1-1-ness of $\tilde{f}([(x,y)] = f(x,y)$ by the universal property and simple set theory. $f$ is continuous, because it's even differentiable.

There is no need to define an $f$ on classes in the domain: we need an $f$ respecting classes (as $(1)$).

Henno Brandsma
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