Connected and locally connected, but not path-connected Allow me to use some non-standard terminology:


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*A h-contractible space is a non-empty topological space $X$ such that, for any topological space $T$ and any pair of continuous maps $f_0, f_1 : T \to X$, there exists a connected topological space $S$ with points $s_0, s_1$ and a continuous map $f : S \times T \to X$ such that $f (s_0, -) = f_0$ and $f (s_1, -) = f_1$. (Note that it is equivalent to require this only in this in the case where $f_0$ is a fixed constant map and $f_1 = \mathrm{id}_X$.)

*A h-path-connected space is a non-empty topological space $X$ such that, for any pair of points $x_0, x_1$, there exists a h-contractible space $I$ with points $i_0, i_1$ and a continuous map $p : I \to X$ such that $p (i_0) = x_0$ and $p (i_1) = x_1$.


It is not hard to see that any contractible space is h-contractible, and so any path-connected space is h-path-connected. Of course, every h-contractible space is h-path-connected; less obviously, every h-path-connected space is connected.
Question. Does there exist a connected (and preferably locally connected) topological space that is not h-path-connected?
The square $X = [0, 1] \times [0, 1]$ with the lexicographic order topology is connected, locally connected, and not path-connected, but unfortunately it is h-contractible: since $X$ is linearly ordered, the operation $\min : X \times X \to X$ is continuous and yields the required contracting "homotopy". This also shows that h-contractibility (resp. h-path-connectedness) is strictly weaker than contractibility (resp. path-connectedness). What I want to know is whether h-path-connectedness is strictly weaker than connectedness.
 A: All examples have to be fairly ugly, because while the category of topological spaces has examples of connected and locally connected spaces which are not path connected, many closely related categories that are equally good candidates at capturing topology do not.
Let's take a closer look at what $X$ being path connected means in terms. The closed interval is exponentiable in top (and should be in any replacement of Top), so we have a path space $I \to X$. Given any path $p: I \to X$ in $X$, we can restrict it to a pair of points in $X \times X$, and this map is continuous. Being path connected is equivalent to the assertion that this continuous map is a surjection (i.e. an epi).
This is really a property of the product of topological spaces in this special case where it is exponentiable. And the main issue with topological spaces that make people look at subcategories (like compactly generated spaces) or modifications of it (like locales, subsequential spaces, or condensed sets) is that you typically want something that has the same colimits but with a different product so that it is cartesian closed. Some of those alternatives (such as locales for example) do have the property that all connected and locally connected spaces are path connected, if the last property is interpreted using the product and exponentials in that category.
So for locally connected and connected sober spaces for example, failing path connectedness means that the product of $X$ (viewed as a locale) with itself or its path space isn't spatial, so that $X$ is the shadow of a space that is really path connected but where this isn't picked up by point-set topology. So while it is possible to find counterexamples, there is an argument in favour of viewing them as a technicality rather than something that should be internalized. The real lesson from the different notions of connectedness is that local connectedness is very important, in any setting.
