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My intuition is that the image $P$ of a path is locally path-connected because if neighborhood of some $p \in P$ that is open in $P$ is disconnected, then $P$ must oscillate quite wildly and become like the topologist's sine curve.

On the other hand, would it be correct to say that the image of an arc must be locally arc connected because it is the homeomorphic image of an interval, which is itself locally arc connected?

Stefan Hamcke
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Herng Yi
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    You are right on your second point. As a general comment on the first part, you will get no counterexamples in Hausdorff spaces. This is because paths in Hausdorff spaces are closed maps and, as such, preserve local (path) connectedness. – Miha Habič Apr 06 '13 at 13:55
  • @MihaHabič: That right, and more generally one can prove that all quotient maps preserve local (path-)connectedness. – Stefan Hamcke Apr 06 '13 at 14:15
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    If you have authoritative references for these, could you post them as answers? I've done my share of Googling but to no avail. – Herng Yi Apr 06 '13 at 14:47
  • Please use @Stefan in your comment, since I didn't get notified about your question. In Willard's General Topology you can find the proof of quotient maps preserving local (p.-)c. as well as many information about continua, Peano spaces etc. For example the continuous image of a unit interval into a Hausdorff space is a Peano space, that is a compact connected metric space (a continuum) which is also locally connected. Every Peano space is arc-connected, so every path connecting two points in a $T_2$ space contains an arc between them. In particular $T_2$ spaces are arc-conn. iff path-conn. – Stefan Hamcke Apr 06 '13 at 15:19
  • I don't have my copy of that book here, otherwise I would look if it contains an example of a path which is not locally connected. – Stefan Hamcke Apr 06 '13 at 15:25
  • I would like to suggest that you try to prove yourself that quotient maps preserve this property. It is not difficult. Use this characterization of local connectedness: Components of open sets are open. – Stefan Hamcke Apr 06 '13 at 15:30
  • @StefanH. I think I will try that tomorrow, thanks for the suggestion. – Herng Yi Apr 06 '13 at 15:52
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    @StefanH. It was easy to show that quotient maps preserve local (path-)connectedness, but how is a path a quotient map? The definition I'm using is a map $f : X \to Y$ such that any $U \subseteq Y$ is open in $Y$ iff $f^{-1}(U)$ is open in $X$. Consider the inclusion map path $p : [0, 1] \hookrightarrow \mathbb{R}^2$. $[0, 1]$ is obviously not open in $\mathbb{R}^2$ yet $p^{-1}([0, 1])$ is open in $[0, 1]$. – Herng Yi Apr 07 '13 at 02:56
  • Well, quotient maps are always assumed to be surjective, so we restrict the codomain to the image. The idea behind quotient maps is that we identify certain points in the domain $X$, which is why they are also called identification maps. The resulting set equipped with the final topology with respect to the identifying map, which is obviously surjective, is then the quotient space. So the inclusion of $I$ in $R^2$ is even an embedding, as an open interval in $I$ is open in the subspace $[0,1]$ of $R^2$. More generally continuous maps from compact to $T_2$ spaces are closed, thus quotient maps. – Stefan Hamcke Apr 07 '13 at 09:40
  • I see, thanks for explaining. The bounty's still up though, so if you have a good reference you can post it as an answer. I'll wait a few more days for answers. – Herng Yi Apr 07 '13 at 10:35

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After desperately trying to find a counterexample myself, I took a look in Counterexamples in Topology by Steen & Seebach (Unfortunately, it was borrowed from the library till today, so I couldn't get early enough for the bounty ;-) ), and found a path-connected, but not locally path-connected space. In fact, this is the only such space in the book which is not Hausdorff, which is a condition, since a Hausdorff space which is the continuous image of the unit interval $I$ must be locally path-connected. This is due to the fact that quotient maps preserve local (path-)connectedness.

The space I am talking about is the Integer Broom. It is the set $X$ of points with polar coordinates $\{(n,\theta)\}$ in the plane $\mathbb R^2$ where $n$ is a nonnegative integer and $\theta\in\{1/n\mid n\in\mathbb N\}\cup\{0\}$. The topology is generated by the basis consisting of all sets $U\times V$ where $U$ is open in the right order topology $\tau_{ro}$ on the nonnegative integers and $V$ is open in $\{1/n\mid n\in\mathbb N\}\cup\{0\}$ in the topology induced by the reals. The only neighborhood of the origin is $X$ itself. This space $X$ is also homeomorphic to the quotient space obtained by identifying all points $(0,\theta)$ in $(\mathbb N,\tau_{ro})\times\{1/n\mid n\in\mathbb N\}\cup\{0\}$ with euclidean coordinates . For a picture see http://en.wikipedia.org/wiki/Integer_broom_topology.

This space is not locally connected since $(1,0)$ does not have a local basis of connected sets. It is compact and not $T_1$ since the only neighborhood of the origin is $X$. Therefore, it makes a good candidate for an image of the unit interval.

Note that the space consists of countably many points, only one of which (the origin) is closed. We can find a closed subset of $I$ whose complement is a countably infinite union of disjoint open intervals, namely the sequence $S:=\{1/n\mid n\in\mathbb N\}\cup\{0\}$. We'll map $S$ to the origin. Let us assume the points except the origin are indexed $\{p_n\mid n\in\mathbb N\}$, and the components of $I\setminus S$ are indexed $\{C_m\mid m\in\mathbb N\}$. Then we just map $C_n$ to $p_n$ and $S$ to the origin, which determines a map $f:I\to X$. Then the preimage of each point $p_n$ is open and so is the preimage of each open set, thus $f$ is continuous. By construction, it is surjective, so $X$ can indeed be called a "path", though it looks quite different.

In the end we have found a path which is not locally path-connected, not even locally connected.

Stefan Hamcke
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    Great and elaborative answer! – Joachim Apr 15 '13 at 12:39
  • The only neighborhood of the origin is X itself?? What about $A\times B$, Is it a neighborhood of the origin? Since A is open in the right order topology on the nonnegative integers, B is open in the topology induced by the reals, $A={n| n\in\Bbb N }\cup{0}, B={1/n\mid n\in\mathbb N, n\gt1}\cup{0}$ – ziang chen Apr 09 '21 at 14:45
  • $B=(-1,2/3)\cap S$ is open , $S={1/n\mid n\in\mathbb N}\cup{0}$ – ziang chen Apr 09 '21 at 14:55