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Given any commutative ring $A$ let $\operatorname{Spec} A$ be the space of prime ideals of $A$. Can we interpret the Zariski topology as an initial (or final) topology with respect to some canonical maps from $\operatorname{Spec} A$ (to $\operatorname{Spec} A$)?

I am asking this because we all know that in general topology you have to know only one topology: the initial topology. (Implicit is the claim that any "canonical" topology is an initial or final topology. It would be very sad if the Zariski topology is not of this form.)

user26857
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Zlatan der Zechpreller
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    because we all know that in general topology you have to know only one topology: the initial topology. Well, I didn't know that. – Daniel Fischer Jul 10 '14 at 18:43
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    @DanielFischer: well if you know the initial topology then you know (and understand conceptually) the subspace topology, product topology, weak topology on a normed space etc etc etc. the list goes on. and you don't have to know the final topology because it is dual, and then you know the quotient topology, disjoint union topology, the topology on the space $\mathcal{D}$ (distribution theory, you know) etc etc. I don't know how to interpret your comment: are you saying that you disagree with this statement? – Zlatan der Zechpreller Jul 10 '14 at 18:44
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    If we take that quote literally: yes, I disagree. After all, you need to have a topology on the target space(s) to define the initial topology, so if all you know is IT, you have gotten yourself in a loop. On another note, there are topologies that are nicer to describe in other ways than as an initial topology (the topology induced by a metric, for example). I would not for a moment be tempted to deny that the initial topology is muy importante, but it's not the be-all-and-end-all. – Daniel Fischer Jul 10 '14 at 18:59
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    But all of those are topologies coming from some topology already given on some space. It is a long way from that to having a topology "appear" from nothing (this is a reply to Zlatan, not to Daniel). – Tobias Kildetoft Jul 10 '14 at 19:00
  • @DanielFischer: I agree, I forgot the metric topology (and some "toy" topologies, which are however not very important). But apart from that none comes to my mind, without perhaps the Zariski topology. Yes my statement should not be taken too literally, but I don't think the argument about the topology on the target space isn't very convincing (I think you are pointing out the obvious here), since it could be any topology (unless we are in a specific case like the weak topology on a normed space). but perhaps I should have used the word "way to construct a topology" instead. – Zlatan der Zechpreller Jul 10 '14 at 19:12

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It depends on what you mean by "canonical". The Zariski topology is the weakest such that every $A\to A/I$ induces the inclusion of a closed subset. We could describe this as an initial topology by considering maps from $\operatorname{Spec} A$ to the Sierpinski space $S$ (the unique 2-point space that is $T_0$ but not Hausdorff) that send $V(I)$ to the closed point.

Generally, every topological space $X$ has the initial topology with respect to the maps $X\to S$ mapping an open set $U$ to the open point of $S$, and we can simplify this description with a nice basis for $X$. If you like thinking categorically, this can even be a good way to think of metric spaces. By Urysohn's Lemma, we can describe any normal topology as an initial topology with respect to maps to $[0,1]$. Since metric spaces are normal, every metric space is an initial topology in the category of metric spaces.

Note that the final topology is not dual in any precise sense, because the category of topological spaces is not self-dual. I doubt that it is so easy to construct a general topology in a nontrivial way as a final topology.

Addendum: You mention maps $\operatorname{Spec}A \to \operatorname{Spec}A$. If you prefer to think of only these maps, you can do the same construction as above by giving the target various appropriate topologies—this is not very useful, since describing these topologies categorically requires the Sierpinski space anyway, but I thought I'd mention it.

Andrew Dudzik
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    When you say "describing these topologies categorically requires the Sierpinski space anyway" do you mean by require that there is no other way, or that you don't know a different way? Of course giving a description which holds for any topological space is almost the contrary of what is meant by "canonical". Also I don't know how this is a useful way of thinking abou the Zariski topology – Zlatan der Zechpreller Jul 10 '14 at 20:11
  • @ZlatanderZechpreller I'm pretty confident in my assertion. If you work out the details, you'll see that the necessary topology on $\operatorname{Spec} A$ has $S$ as its $T_0$ quotient. In other words, there is a natural map to $S$, with respect to which $\operatorname{Spec} A$ has the initial topology. Trying to describe what is interesting about this topology categorically is precisely identical to identifying categorical properties of the Sierpinski space. Since spectra can be quite general, topologically speaking, this is about as good as you can expect to do. – Andrew Dudzik Jul 10 '14 at 22:18
  • @ZlatanderZechpreller As for usefulness, well, that's up to you—I'm just describing what the situation looks like from the perspective of category theory. Generally, one works with principal opens $\operatorname{Spec} A_f \hookrightarrow \operatorname{Spec} A$ as a nice basis, and this is a good perspective from which to identify the key features that set $\operatorname{Spec} A$ apart from an arbitrary topological space, as well as its connection to the ring itself. – Andrew Dudzik Jul 10 '14 at 22:22
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    @ZlatanderZechpreller But here's a concrete application: look at the valuative criterion for separatedness/properness in scheme theory. Under certain conditions (typically something like noetherianness), one can deduce a huge amount by looking at the relationship of a scheme to rings whose spectrum is homeomorphic to $S$, i.e. discrete valuation rings. – Andrew Dudzik Jul 10 '14 at 22:24