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Given two topological spaces $X$ and $Y$, the set $Y^X=\{f:X\rightarrow Y : f \textrm{ is a function} \}$ is the set of all (not necesarily continuous) functions from $X$ to $Y$. The subsets $U^K=\{f\in Y^X:f(K)\subset U\}$ can be defined in $Y^X$ and with these, a subbase can be defined that generates the compact-open topology, $$\mathcal S_k=\{U^K: U\textrm{ is open in }Y\textrm{ and }K\textrm{ is compact in }X\}$$

A topology $\tau$ defined over $Y^X$ is called admissible over compacta iff the evaluation map \begin{equation} \begin{aligned} e:Y^X\times X &\rightarrow Y\\ (f,x)&\mapsto f(x) \end{aligned} \end{equation} restricted to compact subsets of $K$ of $X$ is continuous.

I have proven that if the set $Y^X$ is compact fot a given topology $\tau$ that is admissible over compacta then the topology $\tau$ coincides with the compact-open topology. I have also proven that in general, the compact-open topology is coarser than any admissible topology over compacta.

I want to find an example for a topology admissible over compacta that does not coincide with the compact-open topology. I have found some exapmles using the uniform convergence topology but not over $Y^X$ but over $\mathcal C(X,Y)$ (the set of continuous mappings) instead.

Is there any simple example of a topology over $Y^X$ that is admissible over compacta but doesn't coincide with the compact-open topology?

Mikel Solaguren
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    I see problems in the definition of a topology admissible over compacta. First, since the evaluation map is defined on the product $Y^X\times X$, in can be restricted onto subspaces of $ Y^X\times X$ rather than onto subspaces $K$ of $X$. Next, even when for a map $f:X\to Y$ we consider as its "restriction" the map $f|K$ then $ f|K\in Y^K$ and we have to define the "restriction" of the topology $\tau$ from $Y^X$ to $Y^K$. Finally, the continuity of the restricted evaluation map implies that the map $f|K$ is continuous, but $f$ can be an arbitrary map from $X$ to $Y$. – Alex Ravsky Apr 27 '25 at 05:12
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    @AlexRavsky The deffinition is taken from T.Husain Topology and Maps, Chapter 54 Definition 4. A topology $\tau$ on $Y^X$ where $X$ and $Y$ are topological spaces is said to be a topology of joint continuity (what other authors call admissible) if the evaluation map is continuous. A topology is said to be of joint continuity on compacta if for every compact subset $C\subset X$ the restriction map of $Y^X\times C$ into $Y$ is continuous. So in this context, ''restriction to $C$'' means restriction to the subset $Y^X\times C \subset Y^X\times X$ – Mikel Solaguren Apr 27 '25 at 11:38
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    Is it even true that the compact-open topology is admissible on compacta? It's certainly true if compact implies Hausdorff (so compacta are also local compacta), but I'm not sure if it holds otherwise. – Tyrone May 01 '25 at 12:37
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    @Tyrone I don't know what the necesary conditions are for the compact-open topology to be admissible on compacta but given a topology $\tau_a$ admissible on compacta, if $Y^X$ is compact for $\tau_a$ then it holds that this topology coincides with the compact-open. In fact, this is partly the motivation behind the question. I know that the compact-open topology is coarser than any topology that is admissible on compacta and i know a sufficient condition for a given admissible topology on compacta to be equal to the compact-open topology. I then want an example of one that is not equal. – Mikel Solaguren May 02 '25 at 08:45
  • @MikelSolaguren Are you sure that the authors don't write $Y^X$ to mean continuous maps from $X$ to $Y$? This is sometimes the case – Jakobian May 05 '25 at 16:15
  • @Jakobian The author makes a distinction between the set of all maps $Y^X$ from $X$ to $Y$ and the set of all continuous maps $\mathcal C (X,Y)$ from $X$ to $Y$. – Mikel Solaguren May 06 '25 at 12:31

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