# Is Top provably not cartesian closed?

It is often said that the category $\sf Top$ of topological spaces and continuous mappings is not cartesian closed. E.g., in the Wikipedia article on compactly generated spaces and in an answer on this site. Can anyone point me at a proof that $\sf Top$ cannot be made into a cartesian closed category? I.e., that there is no way of putting a topology on the space $X\rightarrow Y$ of continuous functions between topological spaces $X$ and $Y$ that makes the natural "Currying" operation from $X \times Y \rightarrow Z$ to $X \rightarrow Y \rightarrow Z$ into a homeomorphism.

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That's not the definition of function space... – Zhen Lin Mar 24 '12 at 14:38
What I have stated is what would be required to make $\sf Top$ into a cartesian closed category. Follow the references in zulon's answer and my comment on it for more information. – Rob Arthan Mar 24 '12 at 23:23
No, you have not. You have mixed up the internal and external homs in your formulation. The correct definition simply asks for the functor $(-) \times Y$ to have a right adjoint $(-)^Y$. – Zhen Lin Mar 24 '12 at 23:32
The existence of such an adjunction is equivalent to what I stated. – Rob Arthan Mar 25 '12 at 3:58
No, it is not. If I put the discrete (or indiscrete) topology on all function spaces then there is a homeomorphism of the kind you claim. Please study the condition of being cartesian closed more carefully. – Zhen Lin Mar 25 '12 at 6:09

There is a sketch of a proof at the ncatlab: they define exponentiable spaces in the "examples" section, ie. the spaces for which there is an exponential. Then in the "Counterexamples" section, there is a counterexample in the category of exponentiable spaces. Basically they use local compactness and Hausdorffness to show that the exponential of two topological spaces is not necessarily exponentiable (but read the proof).

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Thanks. Working back through the references leads to what looks like the original reference, a paper by Fox, and a nice more recent result by Escardo and Heckman – Rob Arthan Mar 24 '12 at 15:04

Another proof is to show that for the usual function space there is a monoidal closed structure on Top, see

R. Brown, Function spaces and product topologies'', Quart. J. Math. (2) 15 (1964), 238-250.

Booth, P.I. and Tillotson, J., "Monoidal closed categories and convenient categories of topological spaces". Pacific J. Math. 88 (1980) 33--53.

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Read chapter 7 in the second volume of Borceux's Handbook; Proposition 7.1.1 shows some contraints on a monoidal closed structure on $\bf Top$:

1. If $U\colon \bf Top\to Set$ is the forgetful functor, then $U\circ(-\otimes-)=U\circ(-\times-)$.
2. $U(Y^X)=$the set of continuous maps $X\to Y$.
3. The unit for the monoidal structure must be the singleton set.

Proposition 7.1.2 explicitly proves that the category of all top spaces is not cartesian closed.

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