Let $X:=[-1,1]\times \mathbb{S}^1$, where each space is taken with the usual euclidean topology. Let $S:=\{0\}\times \mathbb{S}^1 \subseteq X$ and call $H:=X/S$.
Is $H$ normal?
My only idea is the following, but I'm stuck on it, plus I would like a more 'direct' approach.
Calling $C:=\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2=z^2, z \in [0,1]\}$ and $f: X \rightarrow C$, $f(x,y,z)=(x|z|,y|z|,z)$, we have that $f$ is continuous and surjective and it makes the same identifications as the projection from $X$ to $X/S=H$, so If I knew it is an identification I would get that $C$ and $H$ are homeomorphic.
Since $C$ is a subspace of a $T4$ and $T1$ space it is itself $T4$ and $T1$ $\implies $ normal, and $H$ would be itself normal.
My questions are: how can I show that $f$ is an identification? Is there a more direct approach, that requires no homeomorphisms? (I'm not allowed to use the fact that closed identifications normal spaces map into normal spaces)
$H$ is a normal space if it is $T1$ and $T4$ (where $T4$:= closed disjoint non empty sets can be separated by disjoint open sets)
$X$ is compact and $C$ is Hausdorff, thus $f$ is a closed map. Closed surjective maps are identifications.