Is projection of locally connected compact subset locally connected?

Question

Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces, $Z = X \times Y$, $\mathcal{T}_Z$ be the product topology on $Z$, $f : Z \to X$ be defined by $f(x, y) = x$, and $C \subset Z$ be compact and locally connected. Is $f[C]$ locally connected?

Background

A space $(Z, \mathcal{T}_Z)$, where $\mathcal{T}_Z$ is a topology on $Z$, is locally connected, if for each $z \in U \in \mathcal{T}_Z$ there exists a connected $V \in \mathcal{T}_Z$ such that $z \in V \subseteq U$.

Locally connected subset whose image is not locally connected

The following shows that some restrictions are necessary for the subset $C$. Let $X = Y = \mathbb{R}$, and $Z' = \{(0, 1)\} \cup \{(1/n, 0) : n \in \mathbb{N}^{> 0}\}$. Then $Z'$ is locally connected, but not compact, and $f[Z'] = \{0\} \cup \{1/n : n \in \mathbb{N}^{> 0}\}$ is not locally connected.

Holds when $f\restriction C$ is a quotient map

Suppose $f\restriction C$ is a quotient map. Quotient maps preserve local connectedness. Therefore $f[C]$ is locally connected.

This question provides conditions for $f\restriction C$ being a quotient map. However, as shown there, $f\restriction C$ is not always a quotient map.

Non-quotient strategy

There exist maps which are continuous, surjective, and preserve local connectedness, but are not quotient; in the linked example $X$ and $Y$ are both locally connected. If the claim does hold, then a general solution to this problem may need a stronger theorem for preservation of locally connectivity which includes these maps.

Edit: Since there were no answers, I asked this question also in Mathoverflow:

https://mathoverflow.net/questions/416561/is-projection-of-locally-connected-compact-subset-locally-connected

Answer 1

The question was answered in the negative by user "Taras Banakh" at Mathoverflow:

https://mathoverflow.net/questions/416561/is-projection-of-locally-connected-compact-subset-locally-connected

Here are some of my own observations.

As mentioned in the question, the result holds when

• $C$ is compact, closed, and locally connected, or
• $C$ is open and locally connected.

In both of these cases the result follows because the restriction is then a quotient map, as shown in the linked question.

Projection of intersection of compact, closed, locally connected subset and an open subset may not be locally connected

One may wonder whether the result holds when $C$ is the intersection of a compact, closed, locally connected subset $A \subset Z$ and an open subset $U \subset Z$. In this case the projection restriction may not be a quotient map. However, that still leaves open whether the projection is locally connected or not.

The following counterexample shows that the projection of such $C$ may not be locally connected:

Let

• $X = Y = \mathbb{R}$,
• $A = \{(t, t\sin(1/t) : t \in (0, 1]\} \cup \{(0, 0), (0, 1)\}$,
• $U = \{(x, y) \in \mathbb{R}^2 : y^2 < x^2 / 4\} \cup \{(x, y) \in \mathbb{R}^2 : (y - 1)^2 + x^2 < (1/2)^2\}$.

Then $A \subset Z$ is compact, closed, and locally connected, and $U \subset Z$ is open. Let $C = A \cap U$. Then $f[C]$ is not locally connected; it consists of a union of disjoint open intervals which converge to $0$, together with $0$.