TITLE: Is Top provably not cartesian closed?
QUESTION [15 upvotes]: 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.

REPLY [7 votes]: 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.