# Is the defining bijection for a pullback of topological spaces a homeomorphism?

I work in the category of CGWH spaces enriched over themselves. If a space $P$ is the pullback of $A \rightarrow B \leftarrow C$, then for every space $T$ the canonical map

$$Top(T,P) \rightarrow Top (T,A) \times_{Top(T,B)} Top (T,C)$$

is a bijection of sets. Now both sides are equipped with a topology, so I'm wondering whether the map is an homemorphism.

It is clear that the map continuous (because it is induced by continous maps), so the question boils down to asking whether the inverse map, which takes two compatible maps into $A$ and $B$ and builds the map into $P$ is continuous.

P.S.: I will gladly change the title of the question, if someone comes up with a better idea.

Yes. The functor $Top(T,-)$ preserves limits because it is a right adjoint.

• Thanks, but why does it preserve limits in an enriched way. To be more precise, let $F: I \to Top$ be a diagram of topological spaces. The canonical map $Top(T,lim_I F) \to lim_I Top (T,F)$ is a continuous bijection of sets, why is it a homeomorphism? Does it follow from the fact that the natural adjunction bijections $Top(A \times B,C) \to Top (A,Top(B,C))$ are homeomorphisms, or is it something one has to check (or accept). – Alexander Körschgen Dec 9 '14 at 1:46
• @AlexanderKörschgen The answer above answers your question. It says that $Top(T,-)$ is right adjoint to $T\times -$ as endofunctor of the category CGWH. In other terms, it is because CGWH is cartesian closed. – Philippe Gaucher Dec 10 '14 at 14:56
• Yes, sorry, I did not see the wood for the trees. – Alexander Körschgen Dec 13 '14 at 1:57

Yes. The proof that the canonical map is a bijection is constructive.

I have been asked to say more, which I do gladly. In constructive mathematics it is impossible to exhibit a discontinuous map (for instance, to define a discontinuous real function one needs excluded middle). We can even assume that all functions are continuous without getting into trouble, but we need not do that here.

In the situation at hand we proceed as follows. First we note that there is a model of constructive mathematics which includes the topological spaces. For instance filter spaces, limit spaces, or equilogical spaces -- almost any locally cartesian category extending topological spaces will do. For the argument at hand a cartesian closed category is probably enough (and this is what Alexander is doing when he cuts down to CGWH spaces, which I decipher as "Compactly generated Weak Hausdorff", that's a ccc). Thus, if we make an argument about topological spaces which is constructive, we can interpet it in this model to obtain a corresponding fact about topological spaces.

Inside the model everything looks like a set. We can thus treat pullbacks as in the category of sets, and there the canonical map $$P^T \to A^T \times_{B^T} C^T$$ is easily seen to be a bijection, constructively (actually I'd be curious to see a natural non-constructive proof). We can write down the inverse explicitly. Thus we get two maps, which interpreted in our model yield continuous bijections.

The trouble with this method is that one has to learn to think constructively and to switch between external (outside the model) and internal (inside the model) reasoning. This is a big upfront cost, but once you've paid for it, you can pretend a lot of the time that everything is a set (when in reality it is a space, or a sheaf, or some such).

• Can you say more? – Tom Goodwillie Dec 4 '14 at 2:52
• It would be good to know why I deserved a -1 for my answer. My answer is homotopic to Tom's. – Andrej Bauer Dec 4 '14 at 7:20
• @Andrej: my guess is that many people think of "constructive => continuous" just as a heuristic, and don't know there are formal statements; so your un-elaborated answer looked (to the downvoter) like invoking a heuristic as a proof. – Peter LeFanu Lumsdaine Dec 4 '14 at 10:03
• @AndrejBauer I am not familiar with constructive mathematics: why is it impossible to exhibit a discontinuous map in constructive mathematics ? If I define $f:\mathbb{R}\to \mathbb{R}$ by $f(x)=0$ if $x<0$ and $f(x)=1$ if $x\geq 0$, this map is not continuous on $0$. And for me, $f$ looks like something constructive. I can easily write a program calculating $f$. Where do I use excluded middle ? I guess that I am missing something. – Philippe Gaucher Dec 4 '14 at 15:01
• You used excluded middle when you assumed that $f$ is defined everywhere on $\mathbb{R}$, because for that to be true you need $\forall x \in \mathbb{R} . x < 0 \lor x \geq 0$. That's excluded middle. And no, you cannot implement a program for calculating $f$. Floating points are not reals. A real $x$ is represented on a computer as a function which takes $k \in \mathbb{N}$ as input and produces a rational number $p/q$ which is within $2^{-k}$ from $x$. How would you implement your $f$ with this representation of reals? – Andrej Bauer Dec 4 '14 at 17:01