Munkres Example 22.6 This is a slight variation of Example 6 on pg 143 of Munkres. 
Let $X = \cup_{n\in \mathbb{N}} [0,1]\times \{\frac{1}{n}\}$ and $Z = \{x\times (\frac{x}{n}) | x\in[0,1], n\in\mathbb{N}]\}$ be subsets of $\mathbb{R}^2$. Then is the map $g:X\to Z$ given by $g(x\times \frac{1}{n}) = x\times \frac{x}{n}$ a quotient map?
The example in Munkres considers $X$ to be $\cup_{n\in \mathbb{N}} [0,1]\times \{n\}$ and shows that the above map is not a quotient map in that case. I was trying to figure out what exactly causes this map to not be a quotient map and came up with the above problem instead.
 A: Let $X=\cup_{n\in \Bbb N}([0,1]\times \{1/n\})$ and $X'=\cup_{n\in \Bbb N}([0,1]\times \{n\}).$
Let $T_{\Bbb R^2}$ be the standard topology on $\Bbb R^2.$ Let the topologies on $X$ and $X'$ be the standard ones : $T_X=\{U\cap X:U\in T_{\Bbb R^2}\}$ and $T_{X'}=\{U\cap X': U\in T_{\Bbb R^2}\}.$
Let $f(x,1/n)=(x,n)$ for $(x,1/n)\in X.$ Then $f:X\to X'$ is a homeomorphism. 
Let $Z=\{(x,x/n): x\in [0,1]; n\in \Bbb N\}.$ Let $g(x,n)=(x,x/n).$ 
If $g:X'\to Z$ is a quotient map then the topology on $Z$ is the strongest topology such that $g$ is continuous.  Consider the topology $T$ on $Z$ generated by the base (basis) $A\cup B$ where 
(i). $A=\{Z\cap U:(0,0)\not \in U\in T_{\Bbb R^2}\}.$
(ii).  Let $J$ be the set of all $h:\Bbb N\to \Bbb R^+$ and for $h\in J$ let $V(h)=\cup_{n\in \Bbb N} \{(x,x/n):1\ge x\in [0,h(n)\}.$ Now let $B=\{V(h):h\in J\}.$
Remark: If $J^k$ is the set of constant functions in $J$ then $A\cup \{V(h):h\in J^k\}$ is a base for the standard topology $T_Z=\{U\cap Z: U\in T_{\Bbb R^2}\}$ on $Z.$
Now $T$ is strictly stronger than the standard topology $T_Z$ on $Z.$ That is, $T\supsetneqq T_Z.$ For example if $h(n)=1/n$ for each $n\in \Bbb N$ then $V(h)\in T\setminus T_Z.$ And if we give $Z$ the topology $T$ then $g:X'\to Z$ is continuous.
So with the standard topology on $Z,$ the function $g:X'\to Z$ is not a quotient map . 
Let $e(x,1/n)=(x,x/n)$ for $(x,1/n)\in X.$ We have $e=g\circ f.$ Now if $U\in T$ then $V=g^{-1}U\in T_{X'}$  so (since $f$ is continuous)  $e^{-1}U =f^{-1}V\in T_X.$  So $e:X\to Z$ is continuous with the $T$ topology on $Z.$ So with the standard topology on $Z,$ the function $e:X\to Z$ is not a quotient map.
Remark. $T$ is in fact the strongest topology on $Z$ such that $g$ is continuous.
