Let $A \subset \mathbb{R}^{2}$ be the annulus

$A = \{(x,y) \in \mathbb{R}^{2} \colon 1 \leq x^{2} + y^{2} \leq 4 \}$.

Prove that $A$ is homeomorphic to $S^{1} \times I$, where $I = [0,1]$ is the unit interval.

Define $f \colon S^{1} \times I \rightarrow A$, and $g \colon A \rightarrow S^{1} \times I$ by the following maps.

$$f(x,y,t) = ((t+1)x,(t+1)y)$$ and

$$g(u,v) =\left(\frac{u}{\sqrt{u^{2}+v^{2}}},\frac{v}{\sqrt{u^{2}+v^{2}}},\sqrt{u^{2}+v^{2}}-1\right).$$

$f$ and $g$ are clearly well-defined. Moreover, they and are continuous as their components are continuous. Lastly, we must show that $f \circ g = I_{A}$ and $g \circ f = I_{S^{1} \times I}$, where $I$ is the identity function.

\begin{equation*} \begin{split} f(g(u,v)) &= f\bigg(\frac{u}{\sqrt{u^{2}+v^{2}}},\frac{v}{\sqrt{u^{2}+v^{2}}},\sqrt{u^{2}+v^{2}}-1\bigg) \\ &= \bigg( (\sqrt{u^{2}+v^{2}}-1 + 1)\frac{u}{\sqrt{u^{2}+v^{2}}},(\sqrt{u^{2}+v^{2}}-1 + 1)\frac{v}{\sqrt{u^{2}+v^{2}}} \bigg) \\ &= (u,v). \end{split} \end{equation*} \begin{equation*} \begin{split} g(f(x,y,t)) &= g((t+1)x,(t+1)y) \\ &= \bigg(\frac{(t+1)x}{\sqrt{((t+1)x)^{2} + ((t+1)y)^{2}}},\frac{(t+1)y}{\sqrt{((t+1)x)^{2} + ((t+1)y)^{2}}}, \\ &\qquad \sqrt{((t+1)x)^{2} + ((t+1)y)^{2}} -1\bigg) \\ &= \bigg( \frac{(t+1)x}{\sqrt{(t+1)^{2}(x^{2}+y^{2})}}, \frac{(t+1)y}{\sqrt{(t+1)^{2}(x^{2}+y^{2})}}, \sqrt{(t+1)^{2}(x^{2}+y^{2})} -1\bigg) \\ &= \bigg(\frac{x}{\sqrt{x^{2}+y^{2}}}, \frac{y}{\sqrt{x^{2}+y^{2}}}, (t+1)\sqrt{x^{2}+y^{2}}-1\bigg) \\ &= (x,y,t). \end{split} \end{equation*}

I might look stupid after this, but I can't see why the last component on the second to last line is not working out. Unless, my maps weren't defined correct? Can someone see the problem?

Using the comments below, since $(x,y) \in S^{1}$, its norm $\sqrt{x^{2}+y^{2}} =1$. Thus, the last line in the chain of equalities in the above equation holds.

  • $\begingroup$ Thank you for fixing my formatting!! $\endgroup$ – Jack Feb 3 '16 at 7:04
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    $\begingroup$ $\sqrt{x^2+y^2}=1$ $\endgroup$ – Stefan Hamcke Feb 3 '16 at 7:09
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    $\begingroup$ Since $(x,y)\in S^1$, its norm is $1$. $\endgroup$ – Stefan Hamcke Feb 3 '16 at 7:14
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    $\begingroup$ You take the square root of the sum of the squares of the coordinates, and that is the norm of $(x,y)$. And this norm is equal to $1$ as the vector is from the unit sphere. $\endgroup$ – Stefan Hamcke Feb 3 '16 at 7:18
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Jack Feb 3 '16 at 7:20

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