Product space of Circle and interval is homeomorphic to annulus Given a circle $C = \{ (x,y)\in\mathbb{R}^2\colon x^2+y^2=r^2\}$ and a closed interval $I=[a,b],\ 0\le a\le b$, the product topology on the product space $(C\times I, \mathcal{O}_{C\times I})$ shall be considered.
The annulus is defined by the set $A = \{(x,y)\in\mathbb{R}^2\colon a\le x^2+y^2\le b\}$.
I want to find a homeomorphism $f\colon (C\times I, \mathcal{O}_{C\times I})\rightarrow (A,\mathcal{O}_A)$.
As I have no experience in topology, I am unsure if my reasoning is correct.
Define $$ f((x,y),t) = (\frac{xr}{\sqrt{t}},\frac{yr}{\sqrt{t}}) $$
Then for $(u,v) = f((x,y),t)$ it holds that $$u^2+v^2=r^2\Leftrightarrow \frac{x^2r^2}{t}+ \frac{y^2r^2}{t}=r^2\Leftrightarrow x^2+y^2=t \Leftrightarrow a\le x^2+y^2 \le b $$
It must follow that $f(C\times I)=A$. As I can find an inverse by equivalence transformations we have a bijective map. Finally, since the map and its inverse is continuous in every component, it follows that $f$ is a homeomorphism.
EDIT: Ok I think what I wrote was nonsense. Here is my proposed fix:
Define $$f((x,y),t)=(\frac{x\sqrt{t}}{r}, \frac{y\sqrt{t}}{r})$$
$f$ is surjective: For every point $(u,v)\in A$ we have that
$$a\le u^2+v^2\le b$$ This means there must exist a $t\in[a,b]$ such that
$$ u^2+v^2=t$$ From this it follows that
$$\frac{x^2t}{r^2}+\frac{y^2t}{r^2}=t \Leftrightarrow x^2+y^2=r$$ meaning we can find a preimage on $C\times I$.
$f$ is injective: From $$\frac{x\sqrt{t}}{r}=\frac{x'\sqrt{t'}}{r}$$ and analogue for $y$ and $y'$ it follows that
$$\frac{x^2t}{r^2} +\frac{y^2t}{r^2}=\frac{x'^2t'}{r^2}+\frac{y'^2t'}{r^2}$$
and using $x^2+y^2=r^2$ therefore $t=t'$. This implies $x=x'$ and $y=y'$.
That $f$ is a topological map is not hard to see, since every component is a composition of continuous maps and likewise for the inverse.
 A: You have the correct idea, but I think your arguments are not adequate. [Edited: After your edit they are adequate, but I leave my answer as it is]. Here is a suggestion:
Define
$$f : C \times I \to \mathbb R^2, f((x,y),t) = \left(\frac{x\sqrt{t}}{r},\frac{y\sqrt{t}}{r} \right) .$$
This is clearly a continuous map. For $(x,y) \in C$ and $t \ge 0$ we have
$$\left(\frac{x\sqrt{t}}{r},\frac{y\sqrt{t}}{r} \right)  \in A \Leftrightarrow a \le \frac{x^2t}{r^2}+ \frac{y^2t}{r^2} = t \le b .$$
This shows that $f(C \times I) \subset A$. Next define
$$g : A \to \mathbb R^2 \times \mathbb R,g(u,v) = \left(\frac{ur}{\sqrt{u^2 + v^2}},\frac{vr}{\sqrt{u^2 + v^2}},u^2 + v^2 \right) .$$
This is a continuous map such that $(\frac{ur}{\sqrt{u^2 + v^2}},\frac{vr}{\sqrt{u^2 + v^2}}) \in C$ and $u^2 + v^2 \in I$. Thus $g(A) \subset C \times I$. But for $(u,v) = f((x,y),t)  =\left(\frac{x\sqrt{t}}{r},\frac{y\sqrt{t}}{r} \right)$ we have $u^2 + v^2 = t$, thus
$$g(f((x,y),t)) = g\left(\frac{x\sqrt{t}}{r},\frac{y\sqrt{t}}{r} \right) = \left(\frac{x\sqrt{t}}{r}r/\sqrt{t},\frac{y\sqrt{t}}{r}r/\sqrt{t},t \right) = ((x,y), t) $$
and
$$f(g(u,v)) = f \left(\frac{ur}{\sqrt{u^2 + v^2}},\frac{vr}{\sqrt{u^2 + v^2}},u^2 + v^2 \right) = \left(\frac{ur}{\sqrt{u^2 + v^2}}\sqrt{u^2 + v^2}/r,\frac{vr}{\sqrt{u^2 + v^2}}\sqrt{u^2 + v^2}/r\right) = (u,v) .$$
This shows that $f$ and $g$ are inverse to each other.
