Show that the cylinder $Y=\{(x,y,z)\in \mathbb{R}^3: x^2+y^2=1, 1\leq z\leq 2\}\subset \mathbb{R}^3$ is homeomorphic to $S^1\times [1,2]$.
Remark: In this question all subsets of $\mathbb{R}^n$ are given the subspace topology coming from the standard topology of $\mathbb{R}^n$. All products are given product topologies.
Proof: Let's consider two mappings:
1) $f:Y\to S^1\times [1,2]$ defined by $f(x,y,z)=(x,y)\times \{z\}$
2) $g: S^1\times [1,2]\to Y$ defined by $g((x,y)\times \{z\})=(x,y,z)$
Note that $g=f^{-1}$. It's easy to see that $f$ is bijective.
Also using the above theorem (from Munkres book) we note that $f$ is continuous because functions $f_1(x,y,z)=(x,y)$ and $f_2(x,y,z)=z$ are continuous.
But If i am not mistaken we cannot use this theorem in order to prove that $g$ is continuous.
Would be very grateful if anyone can explain how to prove that $g$ is also continuous.
EDIT: Take any open set in $Y$ namely $U:=\left(\prod \limits_{i=1}^{3}U_i\right)\cap Y$, where $U_i$ are open in $\mathbb{R}$.
Consider the preimage of this set under $g$, namely $g^{-1}(U)$ and take some point from it, namely $(x_0,y_0)\times \{z_0\}\in g^{-1}(U)$. Then $$g((x_0,y_0)\times \{z_0\})\in U=\left(\prod \limits_{i=1}^{3}U_i\right)\cap Y.$$ In other words, $(x_0,y_0,z_0)\in \left(\prod \limits_{i=1}^{3}U_i\right)\cap Y$ $\Rightarrow$ $(x_0,y_0,z_0)\in U_1\times U_2\times U_3$ and $(x_0,y_0,z_0)\in Y$.
Then $(x_0,y_0)\in S^1$ and $z_0\in [1,2]$. Since $U_i$ is open $\mathbb{R}$ then there exists basis elements such that $$(x_0,y_0,z_0)\in (x_0-\epsilon,x_0+\epsilon)\times (y_0-\epsilon,y_0+\epsilon)\times (z_0-\epsilon,z_0+\epsilon)\subset \prod \limits_{i=1}^{3}U_i$$
Denote $V_1:=(x_0-\epsilon,x_0+\epsilon)\times (y_0-\epsilon,y_0+\epsilon)$ and $V_2:=(z_0-\epsilon,z_0+\epsilon)$.
Then $(x_0,y_0)\in V_1\cap S^{1}$ and $z_0\in V_2\cap [1,2]$. Then $$(x_0,y_0)\times \{z_0\}\subset (V_1\cap S^{1})\times (V_2\cap [1,2])\subset g^{-1}(U).$$