Is the image of a path or arc locally path/arc connected? 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?
 A: 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.
