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Let $X, Y$ be topological spaces and let $\text{Cont}(X,Y)$ be the collection of continuous functions $f:X\to Y$. We say that a topology $\tau$ on $\text{Cont}(X,Y)$ is admissible if the evaluation map $$e: \text{Cont}(X,Y)\times X\to Y; \ (f,x)\mapsto f(x)$$ is continuous.

Is there an example of spaces $X,Y$ such that the intersection of all admissible topologies on $\text{Cont}(X,Y)$ is no longer admissible?

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  • $\begingroup$ Maybe I'm being dumb, but could you elaborate on why $\tau_{Cont}$ is admissible? $\endgroup$ – Eric Wofsey Jul 6 '15 at 7:25
  • $\begingroup$ Oops - it's not clear to me either now that I think about it...! Thanks for pointing it out, @EricWofsey. Possibly I have to reformulate the question $\endgroup$ – Dominic van der Zypen Jul 6 '15 at 7:37
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    $\begingroup$ Of course the compact open topology on $\mbox{Cont}(X,Y)$ is the coarsest admissible topology if $X$ is locally compact. Hence we are looking for a space $X$ that is not locally compact. $\endgroup$ – Stefan Geschke Jul 6 '15 at 8:12
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Here's a simple example. Let $X=\mathbb{N}^2\cup\{\infty\}$ topologized such every subset of $\mathbb{N}^2$ is open and the neighborhood filter at $\infty$ is generated by the sets $\mathbb{N}\times[n,\infty)\cup\{\infty\}$ for $n\in\mathbb{N}$. Let $Y=\{0,1\}$ topologized such that $\{1\}$ is open but $\{0\}$ is not. We identify $Z=Cont(X,Y)$ with the set of open subsets of $X$. For each $x\in X$, let $D_x=\{U\in Z:x\in U\}$, and for each $n\in\mathbb{N}$ let $E_n=\{U\in Z:\mathbb{N}\times[n,\infty)\cup\{\infty\}\subseteq U\}$. For each $N\in\mathbb{N}$, let $\tau_N$ be the topology generated by the $D_x$ for all $x\in X$ and the $E_n$ for all $n\geq N$.

I claim that each $\tau_N$ is admissible. Indeed, the fact that $D_{(m,n)}\in\tau_A$ implies that evaluation is continuous at any point of the form $(U,(m,n))\in Z\times X$, and the fact that $E_n\in\tau_A$ implies that evaluation is continuous at any point of the form $(U,\infty)\in Z\times X$ such that $\mathbb{N}\times[n,\infty)\cup\{\infty\}\subseteq U$. Since every neighborhood of $\infty$ contains $\mathbb{N}\times[n,\infty)$ for some $n\geq N$ and evaluation is automatically continuous at $(U,\infty)$ if $\infty\not\in U$, we conclude that evaluation is continuous everywhere.

Now let $\tau$ be the intersection of all the $\tau_N$. I claim that $\tau$ is not admissible. Indeed, suppose evaluation was continous with respect to $\tau$ at the point $(X,\infty)\in Z\times X$. This means that there is a neighborhood $V\subseteq X$ of $\infty$ and a $\tau$-neighborhood $F$ of $X$ such that for all $W\in F$, $V\subseteq W$. That is, there is an $n\in\mathbb{N}$ such that $F\subseteq E_n$. But for any $N>n$, it is easy to see that there is no nonempty $F\in\tau_N$ such that $F\subseteq E_n$.

With some work, this argument can be generalized to apply to any $X$ which is Hausdorff and not locally compact. In full generality, it can be shown that there is a smallest admissible topology on $Cont(X,Y)$ (for $Y$ the Sierpinski space, or equivalently for all $Y$) iff $X$ is core-compact. There is a nice brief overview of this (including the definition of core-compact) on nLab; full details can be found in this nice little paper.

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  • $\begingroup$ Note that if you want an example with a nicer $Y$, a very similar argument works for the same $X$ and $Y=[0,1]$. $\endgroup$ – Eric Wofsey Jul 6 '15 at 10:21
  • $\begingroup$ I was in the middle of constructing my answer before I saw your edit. $\endgroup$ – Todd Trimble Jul 6 '15 at 11:12
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This topic is pretty well-known in studies of function spaces and when they satisfy the appropriate adjointness condition (where the functor $- \times X: \mathbf{Top} \to \mathbf{Top}$ is left adjoint to $Cont(X, -)$; we call such spaces $X$ exponentiable). When one tries to apply the general adjoint functor theorem to construct a right adjoint $Cont(X, -)$, one is led pretty quickly to the condition that there should be a coarsest admissible topology, in order to get the correct topology on $Cont(X, Y)$).

Stefan Geschke made a very pertinent comment that (for Hausdorff spaces at least) it is local compactness of $X$ which is the decisive factor for this condition; more exactly, when $X$ is Hausdorff, $X$ is exponentiable iff it is locally compact.

This paper by Escardó and Heckmann gives a general analysis, showing that for more general (possibly non-Hausdorff) $X$, the decisive condition is something called core-compactness (if $X$ is Hausdorff, then core-compactness is equivalent to local compactness). Probably the most elegant way of formulating it is that $X$ is core-compact iff the topology of $X$ is a continuous lattice.

To be more explicit: For open subsets $U$ and $V$ of a topological space $X$, let us write $V\ll U$ to mean that any open cover of $U$ admits a finite subcover of $V$; this is read as $V$ is relatively compact under $U$ or $V$ is way below $U$. We say that $X$ is core-compact if for every open neighborhood $U$ of a point $x$, there exists an open neighborhood $V$ of $x$ with $V\ll U$. In other words, $X$ is core-compact iff for all open subsets $V$, we have $V = \bigcup \{ U | U\ll V \}$. The theorem is that $X$ is exponentiable iff it is core-compact.

For what it's worth, I did a little write-up at the $n$-Category Café some years back, giving a concrete counterexample much the same as Eric's but with $X$ the space of rationals $\mathbb{Q}$; it can be found here. This example was based on my reading of the paper mentioned above; it turns out that if $X$ is not core-compact, then a counterexample (to the assertion that the intersection of admissible topologies is still admissible) can be always be found by taking $Y$ to be Sierpinski space $\mathbf{2}$ (as in Eric's example).

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  • $\begingroup$ Since you seem to know this story better than I do, maybe you can answer the following question for me. It is clear that if $X$ is exponentiable, then there is a coarsest admissible topology (namely the exponential topology). Is there an easy way to prove the converse? If you knew that $-\times X$ preserved quotients, then it would be easy to show the coarsest admissible topology is an exponential topology (any map $Z\times X\to Y$ would induce an admissible topology on $Cont(X,Y)$ as a quotient of $Z$), but this is also equivalent to being exponentiable! $\endgroup$ – Eric Wofsey Jul 6 '15 at 11:29
  • $\begingroup$ @EricWofsey I'm afraid it's not immediately obvious to me. But in the key case where the base $Y$ is Sierpinski space, see section 4 of the paper by Escardó and Heckmann where they show that the Scott topology is the exponential topology. $\endgroup$ – Todd Trimble Jul 6 '15 at 17:22

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