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?


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:


  • $\begingroup$ More interesting than I first thought. If the claim does hold, it must be somehow because we are dealing with the restriction of a projection. If $X$ is KC (e.g. Hausdorff) the claim indeed holds by the closed-map implies quotient map argument. $\endgroup$ Jan 4 at 17:28
  • $\begingroup$ Yes. The first thought is to restrict the projection, but that does not work in general for quotient maps or open maps. So I'm thinking perhaps there is some sneaky way, or perhaps there is a more general class of mappings which preserve local connectedness and which allow the restriction. What is KC? $\endgroup$
    – kaba
    Jan 4 at 17:32
  • 1
    $\begingroup$ A space $X$ is called KC if all compact subsets of $X$ are closed in $X$. It's an obscure separation axiom strictly between $T_1$ and $T_2$. $\endgroup$ Jan 4 at 17:37
  • $\begingroup$ @kaba so by "compact" you mean "compact Hausdorff"? Because your claim in "holds when $X$ is Hausdorff" requires $C$ to be Hausdorff as well. Otherwise there is a counterexample I think. $\endgroup$
    – freakish
    Jan 5 at 8:35
  • $\begingroup$ No, there is no Hausdorff requirement for $C$; the claim works without it. $\endgroup$
    – kaba
    Jan 5 at 8:36

1 Answer 1


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



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