TITLE: Prove that a map is continuous and surjective, but is not a quotient map
QUESTION [0 upvotes]: Could someone help me with the following problem?
Let $C_{n}$ be the subspace of $\mathbb{R}^2$ defined by
\begin{equation}
C_{n}=\left\{x \times y: \left(x - \frac{1}{n}\right)^2 + y^2 = \left(\frac{1}{n}\right)^2\right\}.
\end{equation}
Let $Y$ be the subspace
\begin{equation}
Y = \cup_{n \in \mathbb{Z}_{+}} C_n
\end{equation}
of $\mathbb{R}^2$, and let $X$ be the subspace $C_{1} \times \mathbb{Z}_{+}$ of $\mathbb{R}^{2} \times \mathbb{R}$. Define $g : X \to Y $ by the equation
\begin{equation}
g((x \times y) \times n) = \left( \frac{x}{n} , \frac{y}{n} \right).
\end{equation}
How could you prove that $g$ is continuous and surjective, but $g$ is not a quotient map?

REPLY [2 votes]: Each $g_n :D_n = C_1 \times \{n\} \to Y, g_n(x,y,n) = (x/n,y/n)$, maps $D_n$ homeomorphically onto $C_n$. The $D_n$ are disjoint open subsets of $D = C_1 \times \mathbb Z_+$. Thus the map
$$g : D = \bigcup_n D_n \to Y = \bigcup_n C_n, g(\zeta) = g_n(\zeta) \text{ for } \zeta \in D_n$$
is a continuous surjection.
The set $X  = \{(2/n,0) \mid  n \in \mathbb Z_+\}$ is not closed in $Y$ because $\overline X \setminus X = \{(0,0)\}$. But $g^{-1}(X) = \{(2,0)\} \times \mathbb Z_+$ is closed in $D$, thus $g$ is no quotient map.
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