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.
Is Top provably not cartesian closed?
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1That's not the definition of function space... – 2012-03-24
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0What 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. – 2012-03-24
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3No, 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 $(-)^Y$. – 2012-03-24
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0The existence of such an adjunction is equivalent to what I stated. – 2012-03-25
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3No, 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. – 2012-03-25
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0I disagree. If you change the topology on the spaces of continuous functions, then you change the underlying set of $X \rightarrow Y \rightarrow Z$ but not that of $X \times Y \rightarrow Z$. See the paper by Escardo and Heckmann mentioned below for a nice discussion of this. – 2012-03-25
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0Your claim is correct only because you have fixed the underlying set of $Y^X$, but this is not valid in general. (Not every category is concrete.) For example, if I define $Y^X$ to be the one-point space for all spaces $X$ and $Y$ then there is always a homeomorphism $Z^{X \times Y} \cong (Z^Y)^X$, but there is no bijection from $\textrm{Hom}(X \times Y, Z)$ to $\textrm{Hom}(X, Z^Y)$. This is what I mean when I say you've mixed up the internal and external homs: it is simply not possible to state the definition of being cartesian closed without invoking the external hom. – 2012-03-25
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0let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/2894/discussion-between-rob-arthan-and-zhen-lin) – 2012-03-25
3 Answers
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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0Thanks. Working back through the references leads to what looks like the original reference, [a paper by Fox](http://www.ams.org/journals/bull/1945-51-06/), and a nice more recent result by [Escardo and Heckman](http://www.cs.bham.ac.uk/~mhe/papers/newyork.pdf) – 2012-03-24
Read chapter 7 in the second volume of Borceux's Handbook; Proposition 7.1.1 shows some contraints on a monoidal closed structure on $\bf Top$:
- If $U\colon \bf Top\to Set$ is the forgetful functor, then $U\circ(-\otimes-)=U\circ(-\times-)$.
- $U(Y^X)=$the set of continuous maps $X\to Y$.
- 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.
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.
See also, and dealing with the non Hausdorff case,
Booth, P.I. and Tillotson, J., "Monoidal closed categories and convenient categories of topological spaces". Pacific J. Math. 88 (1980) 33--53.