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Let $X,Y$ be topological spaces. A homotopy is a continuous $H:X\times[0,1]\to Y$. Currying, we get $H:[0,1]\to X\to Y$. Is there a way to interpret $H$ as a continuous function of type $H:[0,1]\to (X\to Y)$, where $(X\to Y)$ is the space of continuous function from $X$ to $Y$?

There are a few questions.

  • Given topological spaces $X,Y$, is there natural space of continuous functions from $X$ to $Y$?
  • Given a function out of a product space $f:X\times Y\to Z$, is there a natural way to curry?
  • If the above doesn't work for general spaces, is it possible for $Y=[0,1]$?
extremeaxe5
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    I think https://math.stackexchange.com/questions/123950/is-top-provably-not-cartesian-closed is relevant: some restrictions on $X$ and $Y$ are required to get everything to work nicely. (Please ignore the pointless exchange of comments with someone who didn't understand the question.) – Rob Arthan Feb 03 '21 at 22:31
  • @RobArthan that doesn't seem to be quite what is being asked above. Extremeaxe5, what does 'natural' mean to you? For instance if you give $C(X,Y)=(X\rightarrow Y)$ the indiscrete topology, then any continuous function $Z\times X\rightarrow Y$ has a continuous adjoint $Z\rightarrow C(X,Y)$. There is always a finest topology with this property. It is aptly called the natural topology. The adjunction in the other direction is more problematic, however, and in general there is no coarsest admissible topology on $C(X,Y)$. – Tyrone Feb 03 '21 at 22:48
  • I just said it was relevant, which I believe it is (to get everything to work nicely, specifically the adjunction in the other direction). – Rob Arthan Feb 03 '21 at 22:57
  • Search for "exponential law" in this forum. See also https://ncatlab.org/nlab/show/exponential+law+for+spaces . – Paul Frost Feb 12 '21 at 00:43

2 Answers2

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Q: Given two topological spaces, $X$ and $Y$, is there a natural space of continuous functions from one space to the other?

A: Yes. It is the collection of all continuous functions from $X$ to $Y$. (From another vantage point, one sometimes encounters an arbitrary set $X$, a topological space $Y$, and a set ${\cal F}$ of functions from $X$ to $Y$. One can then ask, What minimal topology will make all the functions in ${\cal F}$ continuous?)

Since you call this collection of functions a ``space'', I suspect you might be looking to find out whether there is some natural (topological?) structure on this collection. If so, yes, there is: the compact-open topology. I think it leads to a general answer to your question about regarding a homotopy as a continuous (in some sense) mapping from $[0, 1]$ to the set of continuous mappings from $X$ to $Y$. An efficient source to grasp this from is the book "Homotopic topology" by Fomenko, Fuchs, and Gutenmaher.

avs
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The answer is in [1]. $\newcommand{\Top}{\mathbf{Top}} % topological spaces$

  • Yes, the compact open topology.
  • No. A topological space $X$ is exponentiable if the functor $-\times X: \Top\to\Top$ has a right adjoint $(-)^X$. [1] shows that $X$ is exponentiable iff it is core compact. If $X$ is Hausdorff, $X$ is core compact iff it is locally compact.
  • Again, no. What matters is $X$, not $Y$.

[1]: Escardó, Martín; Heckmann, Reinhold, Topologies on spaces of continuous functions., Topol. Proc. 26(2001-2002), No. 2, 545-564 (2002). ZBL1083.54009.

extremeaxe5
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