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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.


EDIT: Corrected the phrasing to the intended meaning.

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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. – Dan Rust Jun 25 '13 at 15:40
The identity from $\Bbb R$ with the discrete topology to the Sorgenfry Line. – David Mitra Jun 25 '13 at 15:42
@James: The image $f(U)$ need not be open unless the map $f$ is an open mapping. – Stefan Hamcke Jun 25 '13 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. – Stefan Hamcke Jun 25 '13 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 $x\in X$. – Alex Ravsky Jun 25 '13 at 15:55
up vote 3 down vote accepted

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

enter image description here

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. $$ f(t) \;=\; \left(2-t,\sin\left(\frac{\pi}{2-t}\right)\right)\qquad\text{for }t> 1. $$

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Another nice example. I really liked that graph, is there a simple way to create such graphs? :) – Serpahimz Jun 25 '13 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. – Jim Belk Jun 25 '13 at 18:03

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.

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perfect answer. – David Chan Apr 13 '15 at 13:59

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.

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Maybe this is more reasonable as a comment? – Dan Rust Jun 25 '13 at 15:54
What? That graph is locally connected. – Chris Eagle Jun 25 '13 at 15:56
It seems that the graph of $\sin \frac{1}{x}$ for $x>0$ is homeomorhic to $\mathbb R$. – Alex Ravsky Jun 25 '13 at 15:57
I added the part of the curve along the $y$-axis. – Mikhail Katz Jun 26 '13 at 7:17

For example, consider x = NU{0} under discrete topology and Y = {0}U{1/n|n€N} as subspace of real line R. Difine f:x------Y by f(0)=0,f(n)=1/n,n€N.then f is a bijection. and f is continues. Therefore,y is continous image of x. Note that x is being a discrete space is locally connected but y is not locally connected.

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Welcome to math.SE! Please, learn some typesetting. Moreover, if $f:X\to Y$, why define it in $0$? – G. Sassatelli May 13 '15 at 17:58

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