Show that the quotient space $[0,2]/]1,2]$ is not homeomorphic to an interval of $\Bbb{R}$.

Question

My idea is to see the domain as $[0,1] \cup [2]$, where $[2]$ is the equivalence class under the quotient map. I suppose I have to assume for a contradiction that $[0,1] \cup [2]$ is mapped homeomorphically by $f$ to an interval $[a,b]$ of $\Bbb{R}$, but I don't know how to go further.

Answer 1

Assume that $[0,2]/(1,2]$ is homeomorphic to an interval $\langle a, b \rangle$, where $\langle$ stands for $[$ or $($, similarly $\rangle$ for $]$ or $)$.

Then you get a quotient map $p : [0,2] \to \langle a, b \rangle$. Now $p((1,2])$ is a single-point subspace of $\langle a, b \rangle$, say $p((1,2]) = \{ x \}$. Then $(1,2] = p^{-1}(\{x \})$ which must be closed in $[0,2]$ by continity of $p$. This is a contradiction.