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Recall a topological space $X$ is path connected if for all $x,y \in X$ there is a continuous function $f\colon [0,1] \to X$ such that $f(0)=x$ and $f(1)=y$.

Say that $X$ is continuously path connected if there is a continuous function $f\colon [0,1] \times X \times X \to X$ such that $f(0,x,y) = x$ and $f(1,x,y) = y$

Is every Polish path connected space also continuously path connected.

(I'd like to know the answer for Polish spaces, but if it is positive, feel free to mention the answer for larger classes of spaces.)

My guess is that the answer is no, and a counterexample will involve constructing a space with uncountably many "forks in the road" where deciding whether one can go left or right is not continuous in the initial data. However, I assume something like this has a well-known counterexample, so I am asking it here.

(Feel free to change the tags. Is this homotopy theory?)

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  • $\begingroup$ Typo: I think you mean $f(0,x,y)$ and $f(1,x,y)$. $\endgroup$ – Joel David Hamkins Apr 19 '16 at 17:01
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    $\begingroup$ Actually I think the answer is already no for extremely nice spaces, e.g. $S^1$. $\endgroup$ – Qiaochu Yuan Apr 19 '16 at 17:05
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    $\begingroup$ Actually this is true if and only if the space is contactable. $\endgroup$ – Anton Petrunin Apr 19 '16 at 17:10
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    $\begingroup$ For completeness' sake, you prove @AntonPetrunin's remark simply by noting that $(t,x) \mapsto f(t,x,y_0)$ is a homotopy between the identity and the constant function with value $y_0$. $\endgroup$ – Omar Antolín-Camarena Apr 19 '16 at 17:56
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    $\begingroup$ With an extracondition $f(t,x,x)=x$ for all $t\in[0,1]$ and $x\in X$ such spaces are well-known in geometric topology as equi-connected spaces. Equiconnected spaces are contractible and locally contractible. Finite-dimensional metrizable equi-connected spaces are absolute retracts. $\endgroup$ – Taras Banakh Apr 19 '16 at 20:56
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As pointed out by Anton Petrunin, the condition is stated is equivalent to the space being contractible. Following Omar Antolín-Camarena, this can be seen since $(t,x) \mapsto f(t,x,y_0)$ constitutes a homotopy between the identity on $X$ and the constant function $y_0$, and this is the defining property of contractible spaces.

Contractible spaces are $n$-connected for any $n$, in particular we see that the circle is a very easy example of a path-connected space that is not contractible.

However, I believe that the definition of "continuous path-connected" is not the correct one. Instead, it should be the relativization of the following:

Definition Call $X$ effectively path-connected, iff there is a computable multi-valued function $P : X \times X \rightrightarrows \mathcal{C}([0,1], X)$ where $p \in P(x,y)$ iff $p(0) = x$ and $p(1) = y$.

So every path is a continuous function as it should be, but the way we attach paths to points may be multivalued. Now the circle becomes effectively path-connected, as we can simply non-deterministically decide whether to go round clockwise or counter-clockwise, if our points are closed to $\pi$ apart, and take the direct way otherwise.

It is still not true though that any path-connected space would be effectively path-connected relative to some oracle. For a counterexample, consider the Warsaw circle (described here). I do not know whether being effectively path-connected relative to some oracle matches some established notion in continuums theory.

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