# 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.

• 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).

• 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

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.

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

[1] R. Brown, "Ten topologies for $$X \times Y$$. Quart. J. Math. (2) 14 (1963), 303-319.

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

See also, and dealing with the non Hausdorff case,

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

These papers give an exponential law $$X^{Z \times_S Y}\cong (X^Y)^Z$$ where the topology on $$Z \times _S Y$$ is defined by the property that a function $$f: Z \times Y \to W$$ is continuous if and only if

1 $$f|\{z\}\times Y$$ is continuous for all $$z \in Z$$, and

2 $$f \circ (1 \times g)$$ is continuous for all maps $$g: A \to Y$$ of compact Hausdorf spaces $$A \to Y$$,

This product is associative but not commutative, as shown in [1].

The idea of the category of Hausdorff $$k$$-spaces being "adequate and convenient for all purposes of topology" is mentioned in the Introduction of [1]. For more information see the n-cat-lab.

This question already has an accepted answer, but I'll add mine (based on my answer to this question, itself adapted from Ronald Brown's 'Topology and Groupoids') since it's a self-contained and relatively simple proof.

Let $$\mathbb{Z}$$, $$\mathbb{Q}$$ and $$\mathbb{R}$$ have their usual topologies. We will exhibit a colimit that the functor $$-\times\mathbb Q$$ does not preserve, and then use this to show that $$\mathbf{Top}$$ is not cartesian closed. Let $$i:\mathbb{Z}\hookrightarrow\mathbb{R}$$ be the usual inclusion, and define $$j :\mathbb{Z}\to\mathbb{R}$$ by $$j(n) = i(n+1)$$. Our example of failed preservation will be that the canonical map $$\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q})\to\mathrm{coeq}(i,j)\times\mathbb{Q}$$ is not a homeomorphism.

Since the forgetful functor $$\mathbf{Top}\to\mathbf{Set}$$ preserves both limits and colimits, the underlying function of this map is indeed a bijection. The underlying set of both spaces is the quotient of $$\mathbb{R}\times\mathbb{Q}$$ by the equivalence relation that relates $$(r,q)$$ to $$(r',q)$$ whenever both $$r$$ and $$r'$$ are integers. The reason this bijection is not a homeomorphism is that there are open sets in $$\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q})$$ whose images are not open in $$\mathrm{coeq}(i,j)\times\mathbb{Q}$$.

To construct such an open set, consider the graphs of two continuous functions $$f,g:\mathbb{R}\to\mathbb{R}$$ with the following properties:

• Both $$f(x)$$ and $$g(x)$$ are strictly positive for all $$x$$, but tend to $$0$$ as $$x$$ tends to $$+\infty$$ and $$-\infty$$.

• We have $$f(x)=g(x)$$ iff $$x$$ is an integer, and in this case $$f(x)$$ and $$g(x)$$ are irrational.

For example we could take $$f(x)=\frac{\pi+\sin(\pi x)}{1+x^2}$$ and $$g(x)=\frac{\pi-\sin(\pi x)}{1+x^2}$$. Now define $$U$$ to be the image under the quotient map of the subset of $$\mathbb{R}\times\mathbb{Q}$$ containing the points $$(r,q)$$ for which $$q$$ is either less than both $$f(r)$$ and $$g(r)$$ or greater than both $$f(r)$$ and $$g(r)$$.

Then $$U$$ is open in $$\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q})$$ since its preimage under the quotient map is open in $$\mathbb{R}\times\mathbb{Q}$$. But it is not open in $$\mathrm{coeq}(i,j)\times\mathbb{Q}$$ since every neighbourhood of $$0$$ in $$\mathrm{coeq}(i,j)$$ contains arbitrarily large nonintegers and hence every open rectangle around $$(0,0)$$ in $$\mathrm{coeq}(i,j)\times\mathbb{Q}$$ meets the area between the graphs of $$f$$ and $$g$$.

Now we will use this to show that $$\mathbf{Top}$$ is not cartesian closed. Let $$\mathbb S$$ be the Sierpiński space. This is the space with two points, which we'll call $$\bot$$ and $$\top$$, such that $$\{\top\}$$ is open but $$\{\bot\}$$ is not. It's easy to show from this definition that the functor $$\mathrm{Hom}_\mathbf{Top}(-,\mathbb S)$$ is precisely the contravariant functor that sends a space $$X$$ to its set of opens $$\mathrm O X$$, with a map $$f:X\to Y$$ being sent to the function $$U\mapsto f^{-1}U$$ from $$\mathrm O Y$$ to $$\mathrm O X$$.

In particular, the above proof shows that the canonical function

$$\mathrm{Hom}_\mathbf{Top}(\mathrm{coeq}(i,j)\times\mathbb{Q},\mathbb S)\to\mathrm{Hom}_\mathbf{Top}(\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q}),\mathbb S)$$

is not a bijection, since we exhibited an open in $$\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q})$$ that was not the preimage of an open in $$\mathrm{coeq}(i,j)\times\mathbb{Q}$$. Since contravariant $$\mathrm{Hom}$$ functors send colimits to limits, this is the same as saying that the canonical function

$$\mathrm{Hom}_\mathbf{Top}(\mathrm{coeq}(i,j)\times\mathbb{Q},\mathbb S)\to\mathrm{eq}(\mathrm{Hom}_\mathbf{Top}(i\times\mathbb{Q},\mathbb S),\mathrm{Hom}_\mathbf{Top}(j\times\mathbb{Q},\mathbb S))$$

is not a bijection. This shows that the functor $$\mathrm{Hom}_\mathbf{Top}(-\times\mathbb Q,\mathbb S)$$ does not send colimits to limits, which means that it cannot itself be isomorphic to a contravariant $$\mathrm{Hom}$$ functor. But the definition of cartesian closed requires precisely that there is a space $$\mathbb S^\mathbb Q$$ such that $$\mathrm{Hom}_\mathbf{Top}(-\times\mathbb Q,\mathbb S)$$ is naturally isomorphic to $$\mathrm{Hom}_\mathbf{Top}(-,\mathbb S^\mathbb Q)$$. So $$\mathbf{Top}$$ is not cartesian closed.