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Consider the category whose objects are topological spaces and whose morphisms are the open maps (or closed maps, open continuous maps, closed continuous maps ... that is, one whose isomorphisms are precisely the homeomorphisms). How does such a category compare with the usual one whose objects are topological spaces and whose morphisms are continuous maps? For example, what limits and colimits exist?

I'm probably missing something obvious, but why don't products typically exist in the category with open maps? The projections from the usual product (in the category with continuous maps) are open, yielding a canonical open map from the usual product to the putative unusual product.

Todd's observation is true enough: the product in the usual topology (contiuous maps) typically fails to realize the corresponding universal property in the unusual topology (open maps). Nevertheless, some other object might realize that universal property. Is it even clear that the if such a space exists its underlying set should be naturally identifiable with the underlying set of the factors? After all, while one point spaces are still terminal, maps out of such objects tend not to be open: it seems one would thereby only extract the subset of isolated points. In any event,
treats the special case of squares. The appropriate space is $X\times X$ with the weakest topology (stronger than the usual) which makes the diagonal embedding open. This construction is clearly not available for products of distinct spaces. My question concerns whether there isn't (as that post suggests there isn't) some devious workaround.

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The isomorphisms in the category of topological spaces and continuous maps are precisely the homeomorphisms. Did you mean to say "bijection" instead of "isomorphism"? – Zhen Lin Dec 25 '12 at 0:17
I think he means what he wrote, namely, to what extent these three categories differ, taking into account that they have the same objects ands the same isomorphisms. – Pietro Majer Dec 25 '12 at 11:35
Regarding the category whose morphisms are open maps, there are a few remarks here: For example, it is asserted that all coproducts exist, that not all products exist, and that "something analogous to powers of a single space" exists. – Adam Epstein Dec 26 '12 at 20:36
I'm probably missing something obvious, but why don't products typically exist in the category with open maps? The projections from the usual product (in the category with continuous maps) are open, yielding a canonical open map from the usual product to the putative unusual product. After this I am stuck. – Adam Epstein Jan 3 '13 at 20:10
If you're asking why doesn't the usual cartesian product with the usual projections give you cartesian products with respect to open maps, consider this example. Open maps $Z \to X \times X$ ought then to be in natural bijection with pairs of open maps $f: Z \to X$, $g: Z \to X$ given by composing with projections. In particular, there ought to be an open map $X \to X \times X$ corresponding to the pair $f = 1_X$, $g = 1_X$; set−theoretically this would be the diagonal map. But the diagonal isn′t open unless X$ is discrete. So the projections, while open, don't realize the universal property. – Todd Trimble Jan 3 '13 at 21:13

This seems difficult to answer in any kind of uniform way, i.e., it seems we have to work on a case-by-case basis. Let's start with $\mathrm{Top}_{\mathrm{open}}$, the category of topological spaces and open continuous maps.

A useful heuristic is to let oneself be guided by locales instead of topological spaces. (For general information on locales, see for instance Mac Lane-Moerdijk, Sheaves in Geometry and Logic, chapter IX.) Recall that the category of locales is (by definition) the category opposite to the category of frames (sup-lattices in which finite meets distribute over arbitrary joins); the frame associated with a locale (by viewing the same object in the opposite category) should be thought of as its topology. Now a frame automatically carries Heyting algebra structure, and it turns out that open continuous maps of locales correspond to frame maps that are also Heyting algebra maps (i.e., Heyting algebra maps which are sup-preserving). Thus, the category of locales and open continuous maps is opposite to the category of cocomplete Heyting algebras, which is a category of algebras of an infinitary algebraic theory. It follows that the category of locales and open continuous maps is complete. Moreover, limits of cocomplete Heyting algebras are computed just as they are in $Set$, and so the calculation of colimits in locales and open continuous maps will be similarly straightforward.

Guided by this heuristic, one might at the very least expect that colimits in $\mathrm{Top}_{\mathrm{open}}$ are calculated just as they are in $\mathrm{Top}$. Indeed, one can easily check this directly for arbitrary coproducts (where the coproduct topology on a disjoint union $X = \sum_\alpha X_\alpha$ is the largest one such that all the canonical inclusions $X_\alpha \to X$ are continuous; notice they are open). As for coequalizers, if two maps

$$f, g: R \stackrel{\to}{\to} X$$

are open and continuous, and if we consider the coequalizer $q: X \to Y$ in $\mathrm{Top}$ (where the topology on $Y$ is the usual quotient topology induced by $q$), then we claim that $q$ is also open and this is the coequalizer in the category of open continuous maps. For, if $V$ is open in $X$, then $q(V)$ is open in $Y$ iff $q^{-1}(q(V))$ is open in $X$, and clearly this holds since

$$q^{-1}(q(V)) = V \cup \bigcup_n (g f^{-1})^n(V) \cup (f g^{-1})^n(V)$$

where the right side is open because $f$ and $g$ are open and continuous.

Pausing briefly to consider the category of topological spaces and closed continuous maps, it is certainly true that finite coproducts exist and are computed just as they are in $\mathrm{Top}$. Also, it is true that quotients of equivalence relations are computed as in $\mathrm{Top}$. That is to say: if $E \subseteq X \times X$ is an equivalence relation and the two projection maps $\pi_1: E \to X$ and $\pi_2: E \to X$ are closed, then the coequalizer $q = \mathrm{coeq}(\pi_1, \pi_2)$ in $Top$ is the coequalizer in the category of closed continuous maps. Indeed, if $C$ is closed in $X$, then $q(C)$ is closed because

$$q^{-1}(q(C)) = \pi_1(\pi_2^{-1}(C))$$

is closed.

Limits are trickier, and I believe do not always exist in $\mathrm{Top}_{\mathrm{open}}$ (see below). Even when they do exist, one should not expect to calculate limits as one would in $\mathrm{Top}$ (for example, Cantor space as a countable product of copies of the discrete two-element space could not possibly be the correct product in $\mathrm{Top}_{\mathrm{open}}$ since Cantor space has no isolated points), nor even as in $\mathrm{Set}$, so this makes limits a little weird.

Here's a partial result: if we have a diagram $F: D \to \mathrm{Top}_{\mathrm{open}}$ and all the spaces $F(d)$ are sober (i.e., of the form $\mathrm{Pt}(\mathcal{O}(X))$, where $\mathrm{Pt}: \mathrm{Frame}^{op} \to \mathrm{Top}$ is right adjoint to $\mathcal{O}: Top \to \mathrm{Frame}^{op}$), then the limit of $F$ in $\mathrm{Top}_{\mathrm{open}}$ can be constructed as $\mathrm{Pt}(\mathrm{colim} \; \mathcal{O}(F d))$ where the colimit is computed in the category of cocomplete Heyting algebras, provided that colimit exists. This uses the fact that for any topological space $X$, the set of open continuous maps $X \to \mathrm{Pt}(H)$ ($H$ a cocomplete Heyting algebra) is in natural one-one correspondence with cocomplete Heyting algebra maps $H \to \mathcal{O}(X)$, together with elementary categorical manipulations.

I conjecture that finite colimits of cocomplete Heyting algebras exist (so that finite limits of sober spaces in $\mathrm{Top}_{\mathrm{open}}$ exist). But inasmuch as the countable coproduct of copies of the free cocomplete Heyting algebra on one generator does not exist (it would be the free cocomplete Heyting algebra on countably many generators, which is known not to exist), it seems to me that $\mathrm{Top}_{\mathrm{open}}$ does not have countable products.

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Hm, the assertion that cocomplete Heyting algebras are monadic over $Set$ is certainly wrong. This casts some doubt on a large chunk of the answer; I'll have to see if it can be repaired. – Todd Trimble Jan 1 '13 at 19:41
I fixed what I could, although there is some tentativeness toward the end. – Todd Trimble Jan 1 '13 at 20:33

The category can be drastically different. For example, suppose your morphisms are precisely local homeomorphisms. Call this category $Top^{et}$. This category is locally a topos, which is something certainly not the case for the category $Top$ of all continuous maps. Moreover, $Top^{et}$ lacks a terminal object (which is something which will happen for many variants). In fact, $Top^{et}$ behaves much more like a category of sheaves on a single space, then a category of spaces; to see this, if we let $\mathfrak{Top}^{et}$ denote the bicategory of (etale) topological stacks and local homeomorphisms, this is equivalent to the bicategory of stacks on some (filtered colimit of) etale topological stack(s). The bicategory $\mathfrak{Top}^{et}$ contains $Top^{et}$ as a full subcategory, and (if we restrict to a set of topological spaces) is a $2$-topos, so has all the limits, colimits etc. you can imagine. Using this, it can be shown that $Top^{et}$ does have at least binary products, but they behave very differently than in $Top.$ For example, instead of $Top$, consider the category $Mfd$ of smooth manifolds. Given an $n$-manifold $N$ and an $m$-manifold $M,$ their product $N \times^{et} M$ in $Mfd^{et}$ is empty if $n \ne m,$ and if $n=m,$ their product is a highly non-Hausdorff smooth $n$-manifold. This is discussed in

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On the other hand, pullbacks in $Top^{et}$ are the same as in $Top$, unlike $Top_{\text{open}}$, where they seem to be awfully complicated. – მამუკა ჯიბლაძე Sep 22 '15 at 21:03

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