# If a topological space $S$ is second-countable, must necessarily every quotient space of $S$ be second-countable?

Let $S$ be a second countable topological space. Let $S^*$ be a quotient space of $S$ with quotient map $\pi$. If $\pi$ is open, it's easy to show that it transfers a basis of $S$ into a basis of $S^*$. So, in that case $S$ being second countable space implies that $S^*$ is second countable.

Does this hold for general quotient maps?

• Let $X$ be the disjoint union of countably infinite many copies of $[0,1]$ and identify the left-hand endpoints. If $Y$ is the resulting space, it's not first countable at the identified point. – David Mitra Aug 6 '14 at 10:20

No. Take $X$ is the reals in the usual topology, and let $R$ be the equivalence relation that has $\mathbb{Z}$ and all $\{x\}, x \notin \mathbb{Z}$ as its classes (i.e. we identify the integers to a point). Then $X / R$, the corresponding quotient space, with map $q$, is not first countable at $q(0)$.