Here $X$ is a topological space, $I$ is the unit interval, $X^I$ is the function space and the map is evaluation at the end-points. According to the discussion at On the openness of the map X^I -> X * X., this map is open if and only if $X$ is locally path-connected. I'm interested in (comparable?) conditions on $X$ that guarantee it is open onto its image, so that (possibly in a convenient category of spaces) the induced equivalence relation on $X$ is effective.
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$\begingroup$ This conditions implies that every path connected component of $X$ is locally path connected. I have the impression that the converse is also true but I haven't been able to prove it, so it is just a guess. $\endgroup$– Simon HenryCommented Sep 20, 2017 at 19:39
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$\begingroup$ @simon It looks to me now like each path component being locally path-connected isn’t sufficient. Consider $X = \bigcup_{n \in \mathbb{N}} ([0,1] \times \{1/n\}) \cup \{y = x(x-1) : x \in [0,1]\} \subset \mathbb{R}^2$. Then $((0,0),(1,0)) \in X \times X$ isn’t open in the image of $X^I \to X \times X$, but its inverse image in $X^I$ is open. So we need a condition saying something like: ‘if we move the endpoints of a path a little bit, and we can connect the new endpoints, we should be able to do it by moving the whole path a only little bit’. $\endgroup$– Ged Corob CookCommented Sep 26, 2017 at 12:58
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$\begingroup$ Yes, you are right. So I'm not sure that there is a nice way to formulate this. $\endgroup$– Simon HenryCommented Sep 26, 2017 at 18:56
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