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If $X, Y$ are topological spaces, let $C(X,Y)$ denote the collection of continuous maps $f: X\to Y$, endowed with the compact-open topology.

Assume that we are given topological spaces $X,Y$ such that for all spaces $Z$ we have $C(X,Z) \cong C(Y,Z)$. Does this imply that $X\cong Y$?

Dominic van der Zypen
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    If your isomorphism is natural in $Z$, then this is just a special case of the Yoneda lemma. (Your natural homeomorphism then gives a bijection of underlying hom sets, which is all that matters). – Chris Schommer-Pries Jul 08 '15 at 08:18
  • I don't know whether my isomorphism is natural in $Z$. I think your argument applies for locally compact Hausdorff spaces, see also http://mathoverflow.net/questions/210903/coarsest-admissible-topology-on-textcontx-y/210914#210914 – Dominic van der Zypen Jul 08 '15 at 08:25
  • One way for the isomorphism to be natural would be if it is induced by composition with a map between $X$ and $Y$. If for all $Z$ you have a bijection of sets $C(X, Z) \cong C(Y,Z)$ which is natural in $Z$, then $X$ and $Y$ are homeomorphic. The homeomorphism and its inverse are constructed by taking $X = Z$ and $Y=Z$, respectively, and looking where the identity maps to. This is a completely general (and standard) categorical argument and doesn't need any topological assumptions nor any kind of topology on the mapping sets. – Chris Schommer-Pries Jul 08 '15 at 08:44
  • To add to what Chris is saying: if $1$ is the one-point space, then for any space $A$ the hom-set $\hom_{Top}(1, A)$ is the underlying set of $A$. In the case of $A = C(X, Z)$, no matter what topology we put on $C(X, Z)$ and no matter what the topological assumptions are on $X$, we get $\hom(1, C(X, Z)) \cong \hom(X, Z)$, since the points of $C(X, Z)$ are by definition (parametrized by) maps $X \to Z$. Applying $\hom(1, -)$ to a natural isomorphism $C(X, Z) \cong C(Y, Z)$, we then get a natural isomorphism $\hom(X, Z) \cong \hom(Y, Z)$, at which point Yoneda applies. – Todd Trimble Jul 08 '15 at 09:00

1 Answers1

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Let $Z$ be the Sierpinski 2-point space. Then the underlyying set of $C(X,Z)$ is naturally identified with the collection of open sets of $X$, and the specialization order from the compact-open topology is the inclusion order. Thus we can recover the poset of open sets of $X$ from $C(X,Z)$. So if you restrict your question to sober spaces, the answer is yes.

However, for arbitrary spaces the question seems more subtle. For instance, let's consider the case where $X$ and $Y$ are indiscrete. If $X$ is indiscrete, then for any $Z$, it is easy to see that $C(X,Z)$ looks like $Z$ except each maximal indiscrete subspace $A\subseteq Z$ has been replaced by $A^X$. In particular, if $X$ and $Y$ are sets such that $|A^X|=|A^Y|$ for all sets $A$, then $C(X,Z)\cong C(Y,Z)$ for all $Z$, where $X$ and $Y$ are equipped with the indiscrete topology. So the question reduces to the following: if $\kappa$ and $\lambda$ are cardinals such that $\mu^\kappa=\mu^\lambda$ for all cardinals $\mu$, must $\kappa=\lambda$? The answer is yes: if $\kappa<\lambda$ and $\mu$ is a strong limit cardinal of cofinality $\kappa^+$, then $\mu^\kappa=\mu$ but $\mu^\lambda>\mu$.

So the answer is also yes if $X$ and $Y$ are both indiscrete. However, it is noteworthy that this argument might require you to take a very large $Z$ to distinguish $X$ and $Y$. For instance, when $|X|=\aleph_0$ and $|Y|=\aleph_1$, if $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$ and GCH holds above $\aleph_1$, then $C(X,Z)\cong C(Y,Z)$ for all spaces $Z$ of cardinality less than $\aleph_{\omega_1}$. So if the answer is yes in general, this is a sign that proving it probably isn't going to be very easy.

Eric Wofsey
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  • Aren't you close to establishing that the problem is undecidable by some cardinal arithmetic trouble? – Andrej Bauer Jul 08 '15 at 08:51
  • @AndrejBauer: I don't think so; the only cardinal arithmetic problem I've come up with in connection with the problem has a solution in ZFC. Interestingly, though, the indiscrete case does show that the answer is independent of ZC. – Eric Wofsey Jul 08 '15 at 09:08
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    What if you try to mix your two constructions? What I mean is consider spaces Z which are like the Sierpinski 2-point space but where you replace the closed point by an indiscrete space B. Then continuous maps from $X$ into $Z$ are in bijection with pairs $(U,f)$ of an open set $U \subseteq X$ and a set-theoretical map $U \to B$. I think you can recover the poset of opens and the cardinality of each open. That is very close to the space itself. Maybe you can actually get the space? – Chris Schommer-Pries Jul 08 '15 at 09:29
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    I know I'm not supposed to just say "thanks" in the comments (without any more mathematical content) -- but let me do it anyway: @EricWofsey, I always enjoy your answers a lot! – Dominic van der Zypen Jul 08 '15 at 09:49
  • I was about to post the same sugestion as Chris: Eric argument show that two spaces satisfying the hypothesis have the same soberification, hence it suffice to show that the cardinal of all the fiber of the natural maps to the soberification are the same and I was thinking that maybe, using Eric proof of the indiscrete case one can gets this informations from the space of map to some "enlarged" Sierpinski space (i.e. a space whose soberfication is the Sierpinski space) – Simon Henry Jul 08 '15 at 10:46
  • @ChrisSchommer-Pries: I had that idea too, but it's not so simple because you don't have any naturality. If the poset of open sets has a lot of automorphisms, it's not so clear to me how to make the argument come together. Also, just knowing the cardinality of each open is not nearly enough (it's easy to construct counterexamples to that); as in Simon's comment, you need the cardinalities of all the fibers of the map to the soberification. Maybe there's some simple clever idea I'm missing, but actually recovering all those fibers in the absence of naturality seems quite hard. – Eric Wofsey Jul 08 '15 at 21:43
  • To maybe put it more clearly: if you want to use multiple different spaces $Z$ to identify $X$, lack of naturality is going to be a serious hurdle. Since just knowing the cardinality of the open sets is not enough, you can't just use an enlarged Sierpinski space; it seems like you need to use some complicated $Z$ that has been tailored to your $X$. Not only that, but because of the difficulties of naturality, you probably need to somehow tailor a $Z$ that can recognize all the fibers of the soberification at once, rather than just constructing a different $Z$ for each one. – Eric Wofsey Jul 08 '15 at 22:01