Quotient spaces not homeomorphic

Question

$A=[-2,-1]\cup[1,2],\, f:A\to[1,\infty)$ is a continuous map such that $f(x)=f(-x)$, $X=\bigcup\left\{I((x,0),(0,f(x))):x\in A\right\}$ where $I(a,b)$ denotes line segment between $a$ and $b$. $K=X\cap\left(\{0\}\times\mathbf{R}\right)$, $L=X\cap\left([-2,0]\times\mathbf{R}\right)$. I need to show that if $f(1)<f(2)$ then $X/K$ and $X/L$ are not homeomorphic. Any hints how do I proceed? And I can't quite see yet why the assumption $f(1)<f(2)$ is necessary.

Answer 1

In $X/K$, $K$ is always a cutpoint. I.e. $X/K$ is connected, but $X/K \setminus \{K\}$ is disconnected.

On the other hand, $X/L$ often does not have a cutpoint. If $f(1) < f(2)$, then $I((1, 0), (0, f(1))$ and $I((2, 0), (0, f(2))$ are two disjoint paths in $X$ connecting $K$ and $[1, 2] × \{0\}$. It follows that $X/L$ has no cutpoint in this case.

It seems that $X/L$ would have a cutpoint only when all the lines $I((x, 0), (0, f(x))$ for $x ∈ [1, 2]$ intersect at one point not on the $y$-axis. And even in this situation $X/K$ and $X/L$ would not be homeomorphic ($X/K$ would have three cutpoints while $X/L$ only one). I'm not sure about these last observations but it seems that the spaces are in fact never homeomorphic.