# Is projection of locally connected compact subset locally connected?

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. 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.

• More generally, the provided link shows that $f \restriction C$ is a quotient map when $X$ is Hausdorff, or $C$ is closed.
– kaba
Feb 19, 2022 at 20:59
• No, the Hausdorffness of $X$ is insufficient, $Y$ should be Hausdorff! Feb 19, 2022 at 21:41
• In the question, the projection is onto $X$. In your answer the projection is onto $Y$. The claim is correct when projecting onto $X$.
– kaba
Feb 19, 2022 at 22:55
• 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. Feb 19, 2022 at 23:17

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.

• 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.
– kaba
Feb 19, 2022 at 22:53
• @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 $(Y,\mathcal T_Y)$. Feb 19, 2022 at 23:13
• @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. Feb 19, 2022 at 23:19
• 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?
– kaba
Feb 19, 2022 at 23:23
• Do you perhaps mean there is a maximal such topology?
– kaba
Feb 19, 2022 at 23:25