Given non-empty sets $A, B, C$, set $B^A$ to be the set of all functions $f:A\to B$ there is a natural bijection $\Lambda: C^{A\times B} \to (C^A)^B$ defined in the following way: for $f:A\times B \to C$ let $\Lambda(f):B\to C^A$ be defined by $[\Lambda(f)(b)](a) = f(a,b) \in C$.


Let $X, Y$ be topological spaces; we denote by $C(X,Y)$ the collection of continuous maps $f: X\to Y$.


A topology on $C(X,Y)$ is said to be *admissible* if for any space $Z$ and any continuous map $g: Z\to C(X,Y)$ the map $\Lambda^{-1}(g): Z\times X \to Y$ is continuous (where $Z\times X$ carries the product topology.


Dually, a topology on $C(X,Y)$ is said to be *proper* if for any space $Z$ and any continuous map $g\in C(Z\times X, Y)$ the map $\Lambda(g): Z\to C(X,Y)$ is continuous.


It turns out that a topology on $C(X,Y)$ is admissible if and only if the evaluation map $\text{eval}: C(X,Y)\times X \to Y$ defined by $(f,x)\mapsto f(x)$ is continuous. (This is not difficult to prove.)


Is there a similar criterion for proper topologies on $C(X,Y)$? (By "similar" I mean a criterion not involving $Z$, and possibly the evaluation map.)



Admissible and proper topologies on $C(X,Y)$
First, another name for a proper topology on a function space is *splitting topology*, and other names for an admissible topology are *approximating topology* and *conjoining topology*. This may be of aid in further Google searches. 


Here are some basic facts: 


* Any topology that is coarser (smaller) than a proper topology is also proper. Any topology that is finer (larger) than an admissible topology is also admissible. (The proofs are easy.)
* Any proper topology on $C(X, Y)$ is coarser than any admissible topology on $C(X, Y)$. See Lemma 2.2 [here](http://www.cs.bham.ac.uk/~mhe/papers/newyork.pdf). It follows that any proper topology is contained in the intersection of all admissible topologies.
* The intersection of all admissible topologies is a proper topology, called the *natural topology* on $C(X, Y)$. (This is subsumed under Proposition 5.13 that is cited below, but see also [here](http://www.sciencedirect.com/science/article/pii/S0166864108002162).)


It follows that the natural topology is the unique maximal topology among all proper topologies: a topology is proper iff it is coarser than the natural topology. The natural topology is in general not admissible; it is admissible iff $C(X, Y)$ is an exponential object (i.e., represents the functor $\hom\_{\text{Top}}(- \times X, Y): \text{Top} \to \text{Set}$ in the usual sense of category theory: there is an isomorphism $\hom\_{\text{Top}}(-, C(X, Y)) \cong \hom\_{\text{Top}}(- \times X, Y)$, induced in Yoneda-wise fashion from a map $C(X, Y) \times X \to Y$ which is the evaluation map). 


So a reasonable answer to the question is given by understanding a little more concretely the natural topology on $C(X, Y)$. It may be described in terms of filter convergence as follows: let us say that a filter $\Phi$ on $C(X, Y)$ *converges continuously* to $f \in C(X, Y)$ if, whenever a filter $\mathcal{G}$ on $X$ contains the filter of neighborhoods of $b \in X$, then the filter on $Y$ generated by the filter base $\{F(G): F \in \Phi, G \in \mathcal{G}\}$ contains the filter of neighborhoods of $f(b)$, where we put $F(G) := \bigcup \{f(G): f \in F\}$. As usual when we have a filter convergence relation, we go on to define $U \in C(X, Y)$ to be *open* with respect to continuous convergence if $U$ belongs to any filter $\Phi$ that converges to a point in $U$. 


**Theorem** (see Proposition 5.13 [here](http://homepages.inf.ed.ac.uk/als/Research/Sources/comparing.pdf)): For any spaces $X, Y$, the topology of continuous convergence on $C(X, Y)$ coincides with the natural topology on $C(X, Y)$. 


In summary: a topology $T$ on a function space $C(X, Y)$ is proper iff every $T$-open is also an open of every topology for which the evaluation map is continuous; equivalently, is an open with respect to continuous convergence. 


