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The closed topologist's sine curve is the set

$\{(x,y)\in\mathbb{R}^2 : x=0, -1\le y\le 1\}\cup\{(x,y)\in\mathbb{R}^2 : y=\sin\frac{1}{x}, 0

Clearly it is not possible to continuously map the closed unit interval onto this set, as it fails to be path connected. But what about the disjoint union of two intervals?

I would guess not, I would hope the topologist's sine curve is more badly behaved, but I can't see how to show it's not possible. Is the continuous image of a locally (path) connected space, necessarily locally (path) connected under some reasonable assumptions?

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    @MarcelTuring: You can do it using a disjoint union of a closed and a half-open interval, if that helps. (Which implies you can also do it using a disjoint union of $\aleph_0$ closed intervals.)2012-05-10

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Maybe this is the more general insight you were looking for. If $X$ is a space with a finite number of path-components where each path-component is compact, and $f$ is a continuous map, then the path-components of $f(X)$ are also compact.

To see that this is true, note that the continuous image of a path-connected set is also path-connected. This means that any path-component of $f(X)$ must be a union of images of path-components of $X$. Since continuous images of compact sets are compact, and finite unions of compact sets are also compact, it follows that $f(X)$ can have only compact path-components.

The result you wanted now follows from the fact that the closed topologist's sine curve has a non-compact path-component.

Your conjecture about preservation of local (path-)connectedness fails very easily. An infinite discrete space is locally path-connected and maps continuously onto its one point compactification, which is not locally connected at the added point.