Example of 2nd-countable T2 quotient of topological manifold that is not a topological manifold

Question

What is an example of a topological manifold (i.e., second-countable, locally Euclidean $T_2$-space) $M$ and a quotient space $M/\sim$ of $M$ that is not again a topological manifold even though it is second-countable and $T_2$?

An example with $M$ compact would be especially welcome.

Answer 1

All you need for the purpose of cooking up an example is a compact topological manifold $M$, a Hausdorff space $X$, and a continuous surjective function $f : M \to X$. The universal properties of quotient maps then produce an equivalence relation $\sim$ on $M$ such that $M/\sim$ is homeomorphic to $X$: the equivalence relation is simply $x \sim y \iff f(x)=f(y)$; and the homeomorphism from $M / \sim$ to $X$ is induced by $f$.

Simple examples abound.

For instance, take $M=S^1$ to be the unit circle in $\mathbb R^2$, a compact 1-manifold. Let $X$ be the cross $$X = \bigl([-1,+1] \times \{0\}\bigr) \cup \bigl(\{0\} \times [-1,+1] \bigr) $$ Define the surjective continuous function $f : S^1 \to X$ in a piecewise fashion: subdivide $S^1$ into octants; map each octant homeomorphically into the closest branch of the cross. For example, on the octant $0 \le \theta \le \pi/4$ define $$f(e^{2\pi i \theta}) = \bigl(1 - \frac{4}{\pi} \theta, 0 \bigr) $$ and on the octant $-\pi/4 \le \theta \le 0$ define $$f(e^{2\pi i \theta}) = \bigl(1 + \frac{4}{\pi} \theta, 0 \bigr) $$ As one can from these formulas, one portion of the equivalence relation is $$\theta \sim -\theta \quad\text{if $-\pi/4 \le \theta \le \pi/4$} $$ The other three portions are \begin{align*} \theta \sim \pi-\theta &\quad\text{if $\pi/4 \le \theta \le 3 \pi/4$} \\ \theta \sim 2\pi - \theta &\quad\text{if $3\pi/4 \le \theta \le 5 \pi/4$} \\ \theta \sim 3 \pi - \theta &\quad\text{if $5\pi/4 \le \theta \le 7 \pi/4$} \\ \end{align*}

Answer 2

For something that's not a manifold with boundary, and looks a bit pathological, consider any strange-enough looking Peano continuum. For any such continuum $X$ there exists a continuous surjection from $[0, 1]$ to $X$, hence, composing the projection from $S^1$ to $[-1, 1]$ with homeomorphism $[-1, 1]\cong [0, 1]$ and then surjection $[0, 1]\to X$, we obtain that any Peano continuum is a quotient map of the circle $S^1$.

For example, the Wazewski's universal dendrite: