# Is currying a homeomorphism of array spaces?

Define an "index space" to be any topological space $I$, which is separable and locally compact Hausdorff. Define a "value space" to be any topological vector space $V$, which is separable, locally convex and complete.

Define a $V$-valued array (indexed by $I$) to be a continuous function $a : I \to V$. The space of such arrays is $V^I := C(I,V)$, equipped with the compact-open topology. An array space is also a value space, so it makes sense to consider arrays-of-arrays.

Define the curry operator $\operatorname{curry} : V^{I \times J} \to (V^J)^I$ by $\operatorname{curry}(a)(i)(j) := a(i,j)$. Is the curry operator a homeomorphism of array spaces? i.e., is it the case that $V^{I \times J} \cong (V^J)^I$?

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This is known as the exponential law for spaces. See, for example, this section at ncatlab.org/nlab/show/exponential+law+for+spaces –  Mariano Suárez-Alvarez Apr 28 '14 at 17:43
See the answers to mathoverflow.net/questions/35246/… –  Andy Putman Apr 28 '14 at 17:47
Thanks for the links. @AndyPutman, "compactly generated" seems like a reasonable hypothesis. Do you know if locally convex topological vector spaces are compactly generated? –  Tom LaGatta Apr 28 '14 at 18:02
I seem to remember that on Frechet spaces the topology induced on the compact subsets by the usual topology and by the weak topology is the same. If this is true the usual topology cannot be compactly generated. –  Denis Nardin Apr 28 '14 at 18:59
@TomLaGatta : I'm not sure. Definitely metric spaces (and, more generally, first countable spaces) are compactly generated, so Fréchet spaces are fine. But I don't know about the general case. –  Andy Putman Apr 28 '14 at 18:59