# Continuous image of a locally connected space which is not locally connected

The question is pretty much in the title, I'm looking for an example of a locally connected space and continuous mapping such that the image is not locally connected.

Thanks!

EDIT: Corrected the phrasing to the intended meaning.

• Did you want some condition on the image of the map? Otherwise, pick any non-locally connected space and any map from a single point to this space. Jun 25, 2013 at 15:40
• The identity from $\Bbb R$ with the discrete topology to the Sorgenfry Line. Jun 25, 2013 at 15:42
• @James: The image $f(U)$ need not be open unless the map $f$ is an open mapping. Jun 25, 2013 at 15:46
• See my answer to the question Is the image of a path or arc locally path/arc connected?. It shows that even the continuous image of the unit interval need not be locally connected. Jun 25, 2013 at 15:53
• May be the following will work. Let $X=[0;1]$ and $f:X\to\mathbb C$ such that $f(x)=xe^{i/x}$ for each $x\in X$. Jun 25, 2013 at 15:55

Boring example: any space $X$ is the continuous image of the discrete topology on $X$ (using the identity and noting that any function with a discrete domain is continuous). A discrete space is trivially locally connected (all singleton sets). Now let $X$ be any non-locally connected space.

• perfect answer. Apr 13, 2015 at 13:59

Consider the following variant on the topologist's sine curve.

This space $X$ consists of the graph of $y = \sin(\pi/x)$ for $0<x<1$, together with a closed arc from the point $(1,0)$ to $(0,0)$. Note that $X$ is not locally connected at $(0,0)$.

However, there exists a continuous surjection $f\colon [0,2)\to X$. Specifically, $f(0) = (0,0)$ and $f(1) = (1,0)$, with $f(t)$ following along the bottom curve for $0\leq t\leq 1$. For $t>1$, the function follows along the sine curve, i.e. $$f(t) \;=\; \left(2-t,\sin\left(\frac{\pi}{2-t}\right)\right)\qquad\text{for }t> 1.$$

• Another nice example. I really liked that graph, is there a simple way to create such graphs? :) Jun 25, 2013 at 17:23
• I borrowed the picture from a previous answer of mine. I don't recall how I made it, but I would guess Mathematica. Jun 25, 2013 at 18:03
• why is X not connected in (0,0)? isnt it that neighbor of X is not separable?
Apr 4, 2022 at 8:19

For example, consider $$X = \mathbb{N}\cup \{0\}$$ under discrete topology and $$Y = \{0\}\cup\{\frac{1}{n}|n\in \mathbb{N}\}$$ as subspace of real line $$\mathbb{R}$$. Define $$f:X\to Y$$ by $$f(0)=0,f(n)=\frac{1}{n},n\in \mathbb{N}$$.Then $$f$$ is a bijection. and $$f$$ is continuous. Therefore, $$Y$$ is continuous image of $$X$$. Note that $$X$$ is being a discrete space is locally connected but $$Y$$ is not locally connected.

• Welcome to math.SE! Please, learn some typesetting. Moreover, if $f:X\to Y$, why define it in $0$?
– user228113
May 13, 2015 at 17:58

The graph-parametrisation of the graph of $\sin \frac{1}{x}$ for $x>0$ ? Instead of letting the graph trail off to the right as $x\to\infty$, just turn it around and let the curve run along the interval $[-1,1]$ on the $y$-axis. Then every point of this interval will have the property that local connectivity fails.

• Maybe this is more reasonable as a comment? Jun 25, 2013 at 15:54
• What? That graph is locally connected. Jun 25, 2013 at 15:56
• It seems that the graph of $\sin \frac{1}{x}$ for $x>0$ is homeomorhic to $\mathbb R$. Jun 25, 2013 at 15:57
• I added the part of the curve along the $y$-axis. Jun 26, 2013 at 7:17