Is pointwise locally connectedness preserved under a quotient map?

Question

Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces. $X$ is locally connected at $x \in X$, if for each $U \in \mathcal{T}_X(x)$ there exists $U' \in \mathcal{T}_X(x)$ such that $U'$ is connected and $U' \subset U$. Here $\mathcal{T}_X (x)$ denotes the open subsets of $X$ which contain $x$. Let $X$ be locally connected at $x$ and $f : X \to Y$ be a quotient map.

Is $Y$ locally connected at $f(x)$?

Notes

• I can prove that the result holds when $f$ is a continuous open map. This includes projection from a product space to its factor.

Answer 1

I think K. Jiang's answer gives an example which does more than it needs to in order to answer the OP's question ($X^*$ I think has no locally connected points at all!) .

Let

$$ X = [0,0] \cup \bigcup_{n=1}^{\infty} [\frac{1}{2n},\frac{1}{2n-1}] \subseteq [0,1]. $$

Then clearly $1 \in X$ is a point where $X$ is locally connected: if $U\in \mathcal T_X(1)$, then there is an $\epsilon>0$ such that $(1-\epsilon,1+\epsilon) \cap X \subseteq U$, and thus taking $\epsilon_1 = \min \{\epsilon,1/4 \}$ we see that $V = (1-\epsilon_1,1+\epsilon_1)\cap X \subseteq U$ and $V = (1-\epsilon_1,1]$ is connected.

Now consider the quotient map $q\colon [0,1]\to S^1$ given by identifying $0$ and $1$. It restricts to a quotient map from $X$ to $Y =q(X)$, and because $X$ is compact, the quotient map $q_{|X}$ is even closed. However if we set $e=q(0)=q(1)$, then $Y$ is not locally connected at $e$, despite the fact that $X$ is locally connected at $1$ and $e=q(1)$. Indeed there is no neighbourhood of $e$ in $Y$ which is connected: the reason is just that the same is true of $X$ at $0$ -- any neighbourhood of $0$ contains infinitely many connected components of $X$, and any neighbourhood $V$ of $e$ yields a neighbourhood of both points in $q^{-1}(\{e\})$, and so in particular must be a neighbourhood of $0$.

Answer 2

No. Let $Q = \mathbb{Q} \cap [0, 1]$, and consider the space $X$ (as a subspace of $\mathbb{R}^2$ in the usual topology) defined as the union of the line segments $L_j$ joining the point $(0, 1)$ to each of the points $(q_j, 0)$, where $q_j \in Q$.

Consider the quotient space $X^*$ formed from $X$ that results by identifying the point $a = (0, 1)$ with the point $b = (0, 0)$; let $p$ be the corresponding quotient map.

While $X$ is locally connected at $a$ (indeed, $a$ is the only point at which $X$ is locally connected), the space $X^*$ is not locally connected at $p (a) = \{ a, b \}$, which we presently show.

Consider the open set of $X$ formed by the intersection of $X$ with the union of the two open "balls" $B_\infty \left( a, \frac{1}{4} \right)$ and $B_\infty \left( b, \frac{1}{4} \right)$. Denoting this set as $A$, we note that since $A$ is saturated with respect to $p$ and contains $a$, the image $p (A)$ is a neighborhood of $p (a) = \{ a, b \}$ in $X^*$. We claim that $p (A)$ does not contain a connected neighborhood of $p (a)$.

To show this, consider the (connected) components of $p (A)$. One is $C_0 = p \left( X \cap \left( B_\infty \left( a, \frac{1}{4} \right) \cup \left( \{ 0 \} \times [0, \frac{1}{4}) \right) \right) \right)$. Each of the others can be expressed as $C_k = p \left( X \cap B_\infty \left( b, \frac{1}{4} \right) \cap L_k \right)$, with $L_k$ again representing the segment joining $a$ to $(q_k, 0)$, with $q_k \in Q \backslash \{ 0 \}$, given that the set $C_k$ is non-empty.

Suppose that $V$ is a connected neighborhood of $p (a)$ contained in $p (A)$. Since $V$ is open in $X^*$, its preimage $U = p^{-1} (V)$ must be open in $X$. We know that $a$ is an element of $U$, so there must exist a basis element $B_a = X \cap B_\infty (a, \varepsilon_a > 0)$ contained in $U$. Additionally, $b$ is an element of $U$, so there must exist a basis element $B_b = X \cap B_\infty (b, \varepsilon_b > 0)$ contained in $U$. The set $B_a \cup B_b$ is contained in $U$, and so $p (B_a \cup B_b) = p (B_a) \cup p (B_b)$ is contained in $V$. But $p (B_a)$ intersects $C_0$, while $p (B_b)$ intersects some $C_k \ne C_0$, which contradicts the hypothesis that $V$ is connected.