TITLE: Currying a continuous function
QUESTION [7 upvotes]: There is a mapping which I will call the "currying operator" $\hat\square:Z^{X\times Y}\to(Z^Y)^X$ which maps $f:X\times Y\to Z$ to $\hat f:X\to Z^Y$ defined as $x\mapsto y\mapsto f(x,y)$, or equivalently $\hat f(x)(y)=f(x,y)$. This mapping $f\mapsto\hat f$ is a bijection, and sufficiently "natural" that I expect it to be able to transport topological structure on the sets, in the sense of category theory.
So suppose that $X,Y,Z$ are topological spaces, and topologize $X\times Y$ with the product topology. What topology do I put on $Z^Y$ (this being the set of continuous functions $Y\to Z$) to ensure that $f$ is continuous iff $\hat f$ is continuous? My guess is the compact-open topology, since I've heard it called the "natural" topology for a function space, but does anyone have a reference for the proof?

REPLY [6 votes]: The excellent text Topology and Groupoids by Ronald Brown discusses this extensively in chapter 5.9 and contains complete proofs. The category of topological spaces is not Cartesian closed, so in general you don't have an exponential object for any pair of topological spaces. However, the category of compactly generated spaces (k-spaces) has the desired property, and contains many spaces of interest such as first countable spaces, locally compact spaces, metrizable spaces, manifolds, CW complexes, etc.