In "Handbook of categorical algebra Vol 2" from Francis Borceux, the author gives a proof that $Top$ is not cartesian closed. It seems to me that this proof can be adapted to show that the category $\omega$-$QTop$ (the name is not standard) of quotient of countably based topological spaces is not cartesian closed either.

But in the paper 'Comparing cartesian closed categories of (core) compactly generated spaces' from Martín Escardó, Jimmie Lawson, Alex Simpson, it is stated that the category of quotient of countably based topological spaces, is a cartesian closed category.

So there is probably something that I don't understand well somewhere...

I apologize for the size of my question. But I think this is necessary that I reproduce here my own adaptation of the proof from Francis Borceux's book (almost everything is identical):

Let $[0,1]$ and $\mathbb{Q}$ be topological spaces with their well-known topologies (the both spaces are second countable so there are in $\omega$-$QTop$). Let us suppose that $[0,1]^\mathbb{Q}$ exists in $\omega$-$QTop$. Then the topology on $[0,1]^\mathbb{Q}$ is in particular the coarest countably-based-quotient topology making the evaluation map $ev:[0,1]^\mathbb{Q} \times \mathbb{Q} \rightarrow [0,1]$ continuous.

The proof consists in showing that the fact right above implies that $\mathbb{Q}$ is locally compact, which is a contradiction, since it is not:

  • Consider now the constant function $\delta:\mathbb{Q} \rightarrow [0,1]$ mapping everything to 0.

  • Take any point of $q$ of $\mathbb{Q}$. Since evaluation is continuous we have open neighborhood $V$ of $q$ and $U$ of $\delta$ such that $ev(U \times V) \subseteq [0,\frac{1}{2})$

  • We will now show that $\overline{V}$ is compact.

  • Take an open cover $\bigcup W_i$ of $\overline{V}$. Add $\mathbb{Q} \setminus \overline{V}$ to the open cover.

  • Then we create of a topology $T$ on $C(\mathbb{Q}, [0,1])$ making the evaluation map continuous: This is the topology generated by open sets of the form $$T(A,B)=\{f|\ f(\overline{A} \subseteq B)\}$$ for $A$ basic open of $\mathbb{Q}$ such that $\overline{A}$ is included in one $W_i$ and $B$ basic open of $[0,1]$. This topology is second coutable and so $(C(\mathbb{Q}, [0,1]), T)$ is in $\omega$-$QTop$.

  • Sine $[0,1]^\mathbb{Q}$ is structured with the coarest topology making the evaluation map continuous, there is an open $O=T(A_1,B_1) \cap \dots \cap T(A_n,B_n)$ of $T$ such that $\delta \in O \subseteq U$.

We now show that $V \subseteq \overline{A_1} \cup \dots \cup \overline{A_n}$.

  • Suppose not, then take $x$ such that $x \in V$ and $x \notin \overline{A_1} \cup \dots \cup \overline{A_n}$. Since $\mathbb{Q}$ is completly regular, one can find a continuous function $f$ such that $f(\overline{A_1} \cup \dots \cup \overline{A_n})=0$ and $f(x)=1$.

  • Since $f$ coincides with $\delta$ on each $\overline{A_i}$ we have that $f \in O \subseteq U$ but $ev(f,x) \notin [0,\frac{1}{2})$ which is a contradiction.

  • So $V \subseteq \overline{V} \subseteq \overline{A_1} \cup \dots \cup \overline{A_n} \subseteq W_{l_1} \cup \dots \cup W_{l_n}$.

  • Thus $V$ is compact. Since we can do the same for any $q \in \mathbb{Q}$ we have that $\mathbb{Q}$ is locally compact. which is a contradiction.

  • So $\omega$-$QTop$ is not cartesian closed...

  • $\begingroup$ I might have one answer, reading the phd thesis "Admissible Representations for Continuous Computations" from Matthias Schröder, It appears that $\omega$-$QTOP$ should maybe seen as the category of quotient of second countable topological spaces, with functions sequentially continuous with respect to the convergence relation induced by the topology. Since I'am not very confortable with sequential continuity and sequential spaces, I'am not sure wether or not the two notions are equivalent on those spaces, but if it is not then my argumentation is probably wrong... $\endgroup$ Commented Dec 19, 2011 at 18:09
  • $\begingroup$ Archimondian: are you using the ordinary topological product in your argument, or are you using the product in the category of sequential spaces? Because they don't coincide. I'm not absolutely certain, but it looks like you might be using the topological product in your second bullet point. $\endgroup$
    – Todd Trimble
    Commented Dec 19, 2011 at 19:31
  • $\begingroup$ Thanks, indeed, my cartesian product is certainly wrong with respect to the category... $\endgroup$ Commented Dec 19, 2011 at 23:34

2 Answers 2


Let $C$ be the category of quotients of countably based spaces, and continuous maps between them. The product $X \times Y$ of $X$ and $Y$ in $C$ is computed as the sequentialization of the usual topological product $X \times Y$. This means that we enlarge the topology of $X \times Y$ by all sequentially open subsets (a subset is sequentially open iff every convergent sequence whose limit is in the subset is eventually in the subset). As Todd mentions, there is a problem with your second bullet point. There exists an open neighborhood of $(q,\delta)$ in the space $\mathbb{Q} \times [0,1]^\mathbb{Q}$. But how do you know that the neighborhood may be assumed to be of the form $V \times U$?

The paper Topological and Limit-space Subcategories of Countably-based Equilogical Spaces by Alex Simpson and Matias Menni (Math. Struct. in Comp. Sci., 12:739-770, 2002) contains all the proof and more. I think it will help you see what is going on.

  • $\begingroup$ Thank you very much. I had the wrong product in the category... I'm looking at the paper now, it seems to explain everything I need. $\endgroup$ Commented Dec 19, 2011 at 23:41

Have a look at

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

This may not answer your question but does put it in a general framework.


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