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There is a natural bijection $$ \operatorname{Map}(X\times Y,Z)\cong\operatorname{Map}(X,\operatorname{Map}(Y,Z)),\quad f\mapsto(x\mapsto(f(x,{-})). $$ If $X$, $Y$, $Z$ are topological spaces one can ask if the 'exponential law' $$ C(X\times Y,Z)\cong C(X,C(Y,Z)) $$ still holds. And for the compact-open topology on $C(-,-)$ it is well-known that if $Y$ is locally compact and Hausdorff (or just strongly locally compact) and $X$ is Hausdorff, then above bijection induces such a homeomorphism.

I'm asking for a counter-example to the exponential law in the case where $Y$ is still (strongly) locally-compact but no separation axioms are assumed.

Comments. AFAICS,

  1. If $Y$ is (strongly) locally-compact the set-theoretical exponential law always induces a bijection between LHS and RHS.
  2. This bijection maps open sets of the form $M(K_1\times K_2,U)$ to open sets of the form $M(K_1,M(K_2,U))$ (where $M(K,U)=\{f:f(K)\subset U\}$). If $X$ is Hausdorff the later sets form a subbase of the compact-open topology on RHS.

This explains where Hausdorff condition is used in the proof (in the form it's written e.g. in Hatcher's AT book) — but I'd still would like an explicit counter-example.

Stefan Hamcke
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Grigory M
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  • Related (but not answering the question): @ n-Category Cafe – Grigory M Mar 30 '15 at 15:29
  • Have you already tried low dimensional examples? E.g. Sierpinski together with discrete finite spaces? – johndoe Mar 30 '15 at 16:05
  • @johndoe Well, if either $X$, $Y$ or $Z$ is discrete the statement seems to be more or less obviously true. But slightly more generally... – Grigory M Mar 30 '15 at 16:14
  • ...the idea of taking finite $X$ and $Y$ is quite tempting — because finite topological spaces are locally-compact (and I don't know many example of non-Hausdorff locally-compact spaces). I've tried to play with some examples — but haven't succeeded. – Grigory M Mar 30 '15 at 16:15
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    See for discussion about the assumptions that are needed for exponencial law to hold: http://math.stackexchange.com/a/468535/87690. Namely, $Y$ don't need to be Hausdorff, only $X$ does (in fact preregularity is enough). – Adam Bartoš Mar 30 '15 at 16:16
  • @johndoe ...And Sierpinski set is, perhaps, a good choice for $Z$ — AFAIR, there is some theorem along the lines 'if the exponential law hold for $Z=\text{Sierpinski}$ it holds for all $Z$'. – Grigory M Mar 30 '15 at 16:19
  • @user87690 Right, thank you (if one assumes a strong enough version of local compactness — for non-Hausdorff spaces there are different versions of the definition — one needn't assume that $Y$ is Hausdorff — I've updated the question). – Grigory M Mar 30 '15 at 16:25
  • @Stefan (Re: last comment) No-no, I want a counter example where X is not Hausdorff (and $Y$ is locally compact). – Grigory M Mar 31 '15 at 11:07
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    In https://math.stackexchange.com/q/2970857/116530 I characterize when the forward map is continuous, the proof can give you a counterexample for X being the one point compactification of Q – davik Nov 09 '18 at 13:53

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