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


<https://math.stackexchange.com/questions/4348707/is-projection-of-locally-connected-compact-subset-locally-connected>


### 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](https://math.stackexchange.com/questions/2426794/a-quotient-of-a-locally-connected-space-is-locally-connected). Therefore $f[C]$ is locally connected.


[This question](https://math.stackexchange.com/questions/4350112/is-restriction-of-a-projection-to-a-compact-subset-a-quotient-map) 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](https://math.stackexchange.com/questions/259001/a-continuous-surjection-that-is-not-a-quotient-map) 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.



Is projection of locally connected compact subset locally connected?
A counterexample to this question can be constructed as follows.


Let $X=[0,1]$ be the closed interval with the standard Euclidean topology.


Let $Y=\omega$ and $\mathcal T\_Y$ be the topology on $Y$ consisting of the sets $W\subseteq Y$ satisfying two conditions:


$\bullet$ if $0\in W$, then $W=\omega$;


$\bullet$ if $1\in W$, then $\omega\setminus W$ is finite.


The definition of the topology $\mathcal T\_Y$ implies that $\omega\setminus\{0,1\}$ is an open discrete subspace of $(Y,\mathcal T\_Y)$.


It is easy to see that the space $(Y,\mathcal T\_Y)$ is not locally connected at $1$.


Choose any sequence $(U\_n)\_{n=1}^\infty$ of pairwise disjoint nonempty open sets in $X=[0,1]$ and let $g:X\to Y$ be the function defined by
$$g(x)=\begin{cases}n&\mbox{if $x\in U\_n$ for some $n\ge 1$};\\
0&\mbox{otherwise}.
\end{cases}
$$
The definition of the topology $\mathcal T\_Y$ ensures that the function $g:X\to Y$ is continuous.


Let $C=\{(x,g(x)):x\in X\}\subseteq X\times Y$ be the graph of the function $g$. It is clear that $C$ is homeomorphic to $[0,1]$ and hence is compact, connected and locally connected. But the projection of $C$ onto $Y$ is not locally connected.


