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Allow me to use some non-standard terminology:

  • 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.

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migrated from math.stackexchange.com Oct 30 '13 at 15:24

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  • $\begingroup$ Is a topologist's sine curve h-path-connected? It's not locally connected, but it seems like it might not be h-path-connected. $\endgroup$ – Eric Wofsey Dec 2 '13 at 3:10

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