This question was originally at Math Stackexchange, but had no answers:

### Problem

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 $z \in V \subseteq U$.

A space is *compact* if every open cover has a finite subcover.

### 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.