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In my topology course we had some classical examples for homeomorphic topological spaces, e.g. $\mathbb{R}^2/{\sim_{\mathbb{Z}^2}} \cong S^1 \times S^1$ where $x \sim y$ if $x-y \in \mathbb{Z}^2$ and so on. We proved this statement via the quotient topology:

$$\overline{f} \colon X/\sim_f \rightarrow X/ \sim$$

$\overline{f}$ is a homeomorphism if it is continuous, surjective and open on saturated open sets.

Could you provide other similarly "simple" examples other than $\mathbb{R}^2/{\sim_{\mathbb{Z}^2}} \cong S^1 \times S^1$ and variations thereof?

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    1. You meant "e.g." rather than "i.e." (look up their meaning). 2. The plane $\mathbf R^2$ is not homeomorphic to $S^1 \times S^1$!! You'd better clarify what you are seeking.2017-01-30
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    "$\mathbb{R}^2 \cong S^1 \times S^1$" ?!??! You have perhaps misunderstood something, because those two spaces are not homeomorphic.2017-01-30
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    Perhaps you meant $S^1 \times S^1$ is homeomorphic to a *quotient* of the real plane?2017-01-30
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    Yup, I rushed a little upon writing the question, I am very sorry about that.2017-01-30

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Here are a couple of important examples that pop up in algebraic topology. For $n\geq 0$, let $$S^n := \{(x_1,...,x_{n+1})\in\mathbb{R}^{n+1}\ |\ x_1^2+\cdots+x_{n+1}^2=1\}$$ be the $n$-sphere, and $$D^{n+1}:= \{(x_1,...,x_{n+1})\in\mathbb{R}^{n+1}\ |\ x_1^2+\cdots+x_{n+1}^2\leq 1\}.$$ be the $n$-disk. Then, $S^{n} = \partial D^{n+1}\subseteq D^{n+1}$, and the space $D^{n+1}/S^{n}$ is homeomorphic to $S^{n+1}$. The cone of a space $X$ is the quotient $$CX:= \frac{X\times I}{X\times\{1\}}.$$ We have another homeomorphism $CS^n\cong D^{n+1}$. Similarly, the suspension of a space $X$ is defined by: $$\Sigma{X}:=\frac{X\times I}{X\times\{1\}\cup X\times\{0\}} \cong \frac{CX}{X\times\{0\}}.$$ Then, putting the previous two together it shouldn't be too hard to believe that $\Sigma{S^n}\cong S^{n+1}$.

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Here is one example of homeomorphism. $$ $$ Let us consider the annulus $ A=\{(x,y) \in \mathbb{R}^{2}|1 \leq x^{2}+y^{2}\leq 4 \}$ and Consider the Cylinder in $ \mathbb{R}^{3} $ given by $$ $$ $ C=\{(x,y,z)\in \mathbb{R}^{3}x^{2}+y^{2}=1, 0\leq z \leq 1 \}$ . Then there exists continuous functions $$ $$ $ f:A \rightarrow C \ and \ g:C \rightarrow A $ defined by $$ $$ $ f(x,y)= \left(\frac{x}{\sqrt {x^{2}+y^{2}}},\frac{y}{\sqrt {x^{2}+y^{2}}},\sqrt {x^{2}+y^{2}}-1 \right) $ and $ g(x,y,z)=\left( (1+z)x,(1+z)y \right) $ . Little more calculation shows $ f \circ g=g\circ f=I $ . Hence f and g are homeomorphism.