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In the book "Topology" from Boto von Querenburg I read the following example for product spaces:

"The product space of a circumference and an interval $[a,b]$ with $0<a<b$ is homeomorphic to the annulus $\left\lbrace (x,y) \in \mathbb{R}^2 \colon b \geq x^2+y^2 \geq a \right\rbrace$."

I think this statement looks evident by intuition. But how can I see this by formal calculating?

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circumference = circle? In that case we get a cylinder, not an annulus.. – Henno Brandsma Apr 16 '14 at 9:22
@HennoBrandsma same thing (in this case). – kahen Apr 16 '14 at 9:23

3 Answers 3

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Consider $f:\left\lbrace (x,y) \in \mathbb{R}^2 \colon a \geq x^2+y^2 \geq b \right\rbrace \rightarrow S^1\times[a,b] $ that sends $(x,y)$ to $((\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}),\sqrt{x^2+y^2})$. Verify that $f$ is a continuous bijection and use the following theorem after proving it:

Fact: Let $X,Y$ be compact Hausdorff spaces. If $f:X\rightarrow Y$ is a continuous bijection, then it is a homeomorphism.

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You can drop Hausdorff-ness on $X$ and compactness on $Y$ and it's still true. – kahen Apr 16 '14 at 9:27
@kahen Yes. I just wanted to type less. ${}{}{}{}$ – Amr Apr 16 '14 at 9:28

The same way you always show spaces to be homeomorphic: By exhibiting a homeomorphism.

You may find the following helpful:

The Greatest Trick of Elementary Topology (GTET): A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

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Let $A=\{(x,y)\in\mathbb R^2: a\ge x^2+y^2\ge b\}$, $I=\big[\sqrt{b},\sqrt{a}\big]$ and $S=\{(x,y)\in\mathbb R^2: x^2+y^2=1\}$.

Then the map $f:S\times I\to A$, with $$ f(s,t)=ts, $$ is a homeomorphism.

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Why $I=\big[\sqrt{b},\sqrt{a}\big]$ and not $I=\big[a,b]$? Because I want the product space of $S \times [a,b]$. – user138765 Apr 16 '14 at 9:29
It would be $I=[a,b]$ if $A$ was defined as $A=\{(x,y): a^2\ge x^2+y^2\ge b^2\}$. – Yiorgos S. Smyrlis Apr 16 '14 at 9:34

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