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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 $z \in V \subseteq U$.

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.

  • $\begingroup$ More generally, the provided link shows that $f \restriction C$ is a quotient map when $X$ is Hausdorff, or $C$ is closed. $\endgroup$
    – kaba
    Feb 19, 2022 at 20:59
  • $\begingroup$ No, the Hausdorffness of $X$ is insufficient, $Y$ should be Hausdorff! $\endgroup$ Feb 19, 2022 at 21:41
  • $\begingroup$ In the question, the projection is onto $X$. In your answer the projection is onto $Y$. The claim is correct when projecting onto $X$. $\endgroup$
    – kaba
    Feb 19, 2022 at 22:55
  • $\begingroup$ You are right. Somehow I thought that you project on $Y$ not on $X$. Usually maps have $X$ as a domain and $Y$ as the rangle, not vice versa. But this is not so important. $\endgroup$ Feb 19, 2022 at 23:17

1 Answer 1


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.

  • $\begingroup$ Why is $Y$ not locally connected at $1$? Let $Y' = Y \setminus \{0\}$ and $\mathcal{T}_{Y'}$ consist of the cofinite subsets of $Y'$ and $\emptyset$. Then $\mathcal{T}_Y = \mathcal{T}_{Y'} \cup \{Y\}$ is a topology on $Y$ which satisfies your requirements. But every set in $\mathcal{T}_Y$ is connected. $\endgroup$
    – kaba
    Feb 19, 2022 at 22:53
  • $\begingroup$ @kaba $Y$ is not locally connected at $1$ because any neighborhood of $1$ is homeomorphic to a convergent sequence $\{0\}\cup\{\frac1n:n\in\mathbb N\}$ which is disconnected. Observe that the set $\omega\setminus\{0,1\}$ is open and discrete in $(Y,\mathcal T_Y)$. $\endgroup$ Feb 19, 2022 at 23:13
  • $\begingroup$ @kaba My topology $\mathcal T_Y$ is not a topology, but the topology. It is uniquely determined by those two conditions and the subspace topology it induces on $\omega\setminus\{0\}$ is larger than the cofinite topology. $\endgroup$ Feb 19, 2022 at 23:19
  • $\begingroup$ Ok. I have trouble understanding what exactly your topology $\mathcal{T}_Y$ is. How do those two conditions define a unique topology, and why does not my example topology satisfy your requirements? $\endgroup$
    – kaba
    Feb 19, 2022 at 23:23
  • $\begingroup$ Do you perhaps mean there is a maximal such topology? $\endgroup$
    – kaba
    Feb 19, 2022 at 23:25

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