Suppose that $Z=\{(x,y)\in \mathbb{R}^2:1\leq x^2+y^2\leq 2\}$ annulus on the plane. Prove that $S^1\times [1,2]$ is homeomorphic to $Z$.
Consider two mappings $f:S^1\times [1,2]\to Z$ defined by $f((x,y)\times \{z\})=(\sqrt{z}x,\sqrt{z}y)$ and $g:Z\to S^1\times [1,2]$ defined by $g(x,y)=\left( \frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)$.
Using the following theorem we see that $g$ is continuous. Also $f$ is bijective. But I have trouble to prove that $f$ is continuous. I was trying to prove it directly taking some open set in $Z$ considered $f^{-1}(Z)$ but no results.
Can anyone show to me how to prove it, please?
EDIT: I want to show that the mapping $f:S^{1}\times [1,2]\to Z$ defined by $f((x,y)\times \{z\})=(\sqrt{z}x,\sqrt{z}y)$ is continuous. Suppose that $U$ is open in $Z$ then $U=(U_1\times U_2)\cap Z$ then taking some point $(x_0,y_0)\times \{z_0\}$ from $f^{-1}(U)$ we need to show that there exists open set $V$ in $S^{1}\times [1,2]$ such that $$(x_0,y_0)\times \{z_0\}\subset V\subset f^{-1}(U).$$
Since $(x_0,y_0)\times \{z_0\}\in f^{-1}(U)$ then $f((x_0,y_0)\times \{z_0\})=(\sqrt z x_0, \sqrt{z} y_0)\in U=(U_1\times U_2)\cap Z$ then easy to see that $z_0\in [1,2]$ and $(x_0,y_0)\in S^1$. Also $\sqrt{z_0}x_0\in U_1$ and $\sqrt{z_0}y_0\in U_2$.
But I am not able to complete this proof.
You can avoid checking the continuity of $f$. Once you establish the continuity of $g$ (which, as you pointed out, is much easier), check that $f \circ g$ and $g \circ f$ are the respective identities. This means that $g$ is a continuous bijection. But the domain of $g$ is compact and the codomain Hausdorff, so $g$ is a homeomorphism.
Alternatively, you can avoid using that theorem and instead use its proof to prove that $f$ is continuous, which is your question. This is equivalent to proving that $g$ is a closed map, since $g=f^{-1}$. But a closed set in $Z$ must be compact, and thus its image under $g$ is compact. But a compact subset of $S^1 \times [1,2]$ must be closed, so $g$ is a closed map. Hence $f$ pulls closed sets back to closed sets, so is continuous.
EDIT: I'm assuming that my $g$ is your $g$ and that my $f$ is then $g^{-1}$. However, I'm not sure that's true for the $f$ and $g$ you've given. For example, does your $g$ even land in the stated codomain?