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