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Stream: learning: questions

Topic: partial continuous maps

Nathaniel Virgo (Jan 17 2022 at 04:09):

I want to consider partial continuous maps, i.e. partial maps in Top.

I know two ways to construct these. One is via the definition at [[partial function]], where a partial map is a span $A\xleftarrow{\iota}D\xrightarrow{f}B$ such that $\iota$ is a monomorphism. The other is via the maybe monad.

Both of these seem to make sense in Top, but the maybe monad is much easier for me to think about. The question is whether they are equivalent - am I missing anything if I think of partial continuous maps as defined by the maybe monad instead of as being defined as spans?

David Egolf (Jan 17 2022 at 04:24):

That link seems to describe the left map (to $A$) as being a monomorphism, not an isomorphism.

Nathaniel Virgo (Jan 17 2022 at 04:26):

Ah, I thought one word and typed another, I'll correct that

Zhen Lin Low (Jan 17 2022 at 04:46):

What do you mean by the Maybe monad? Taking the disjoint union with the point probably won't give you as many examples of partial maps as you would like.

Zhen Lin Low (Jan 17 2022 at 04:47):

On the other hand, your definition using spans will probably give you too many partial maps. I think it would be better to restrict to strong monomorphisms if you want to work in Top.

Mike Shulman (Jan 17 2022 at 05:57):

(Since strong monomorphisms in Top are subspace inclusions.)

Mike Shulman (Jan 17 2022 at 07:08):

With the "maybe monad" defined as a disjoint union $B+1$, I think one will get the partial maps whose domain is clopen. But one could try to topologize $B+1$ differently to get a larger class of partial maps.

Jon Sterling (Jan 17 2022 at 08:25):

I was thinking about splitting this into two problems: finding a classifier for strong monomorphisms, and checking if the partial product that is needed to define its partial map classifier exists. The nlab suggests that TOP does have a classifier for strong monos in a remark at [[quasitopos]], given by the two point space with the indiscrete topology; but TOP is not locally cartesian closed, so it doesn't yet follow that we would have the needed partial product.

Mike Shulman (Jan 17 2022 at 10:13):

It seems unlikely to me that that partial map classifier exists in Top, but I don't have a counterexample at the moment.

Mike Shulman (Jan 17 2022 at 10:14):

Although, @Nathaniel Virgo, if you're not wedded to traditional open-set-based topological spaces, you could consider using a similar category that is a quasitopos instead, and then everything would work swimmingly.

Fawzi Hreiki (Jan 17 2022 at 11:08):

If it did have a partial map classifier, would the operation assigning to each object its partial map classifier be a monad?

Nathaniel Virgo (Jan 17 2022 at 13:24):

Thank you for the replies. It took me a while to see the problem with the maybe monad (using the coproduct topology), but I see it now.

A much more sensible notion of partial continuous map seems to arise if we give $X+1$ the topology where the open sets are

• subsets of $X$

• the whole space $X\cup\{\bot\}$

• nothing else

This seems to correspond to saying that a partial continuous map $A\to B$ consists of an open subset of $A$ to which values of $B$ are assigned. The special element $\bot$ is assigned to all other points in $A$, and the map is continuous on the open set on which it's defined. That seems like a fairly reasonable notion of what "partial continuous map" ought to mean.

Zhen Lin Low (Jan 17 2022 at 13:34):

So you do not want to consider continuous maps defined on not-necessarily-open subspaces? I suppose that's a valid choice and could have interesting connections with sheaves...

Oscar Cunningham (Jan 17 2022 at 13:36):

I believe this is itself a monad, and the algebras are the pointed spaces where the point is in the closure of any nonempty subset.

Nathaniel Virgo (Jan 17 2022 at 13:42):

Actually that's a good point Zhen Lin Low - I probably do want to consider those. E.g. one example I had in mind was the map $f\colon \mathbb{R}^2\to \mathbb{R}$ given by $f(a,a)=a$ and $f(a,b)$ undefined if $a\ne b$.

I wonder if it's possible to get those with a monad, or if one really has to use a version of the span construction for that.

(Sorry for being slow to spot these things - topology isn't an area where I have much experience at all.)

Zhen Lin Low (Jan 17 2022 at 14:38):

OK. Then you need a different topology on your putative partial map classifier. The partial map classifier for 1 must also classify all the possible domains of definition of partial maps, so that's 2 with the indiscrete topology, as previously mentioned. That suggests that we might consider the topology on X + 1 where a subset is open iff it is empty or contains the new point and the remainder is non-empty open in X.

Mike Shulman (Jan 17 2022 at 15:56):

Zhen Lin Low said:

the topology on X + 1 where a subset is open iff it is empty or contains the new point and the remainder is non-empty open in X.

I don't think that's a topology; it's not closed under intersections if X contains disjoint nonempty opens.

Zhen Lin Low (Jan 17 2022 at 22:08):

Oops, right. Maybe that's an indication of which X do have partial map classifiers...

Jens Hemelaer (Jan 18 2022 at 08:16):

Elaborating a bit on the ideas above... For any point $x \in X$ an any subset $A \subseteq X$ there should be a continuous function $f : X \to X \cup \{\bot\}$ that sends $A$ to $x$ and $X-A$ to the new point $\bot$. This shows that any open set containing $x$ also contains $\bot$, otherwise the inverse image of this open set would be $A$, and then $A$ is necessarily open. Similarly, it follows that any open set containing $\bot$ also contains $x$. So $x$ and $\bot$ are topologically indistinguishable.

Because $x$ was arbitrary, $\bot$ must be topologically indistinguishable from any point of $X$, and because topological indistinguishability is an equivalence relation, this means that all points in $X \cup \bot$ are indistinguishable from each other... The only topology that makes this happen is the indiscrete topology. Since $X \subseteq X \cup \{\bot\}$ should be a subspace, this implies that $X$ has the indiscrete topology as well.

Nathaniel Virgo (Jan 18 2022 at 08:30):

That does seem to show that no topology on $X\cup \{\bot\}$ can do the sort of thing I was hoping it could do. That probably just forces me to use the span construction instead, which is probably no problem, it's just less intuitive for me. But still, just out of curiosity, I'm wondering if there could be a monad on Top that behaves suitably like the Maybe monad, where the underlying set of $\operatorname{Maybe} X$ is something other than $X+1$.

Zhen Lin Low (Jan 18 2022 at 09:10):

Hmmm. I suppose you mean to realise the category of partial maps as the Kleisli category of some monad. Well, that means the only possibility is that the "inclusion" of the category of total maps into the category of partial maps has a right adjoint, which implies that partial map classifiers exist.

Zhen Lin Low (Jan 18 2022 at 10:44):

Actually, it occurs to me that by the adjointness relation, the partial map classifier for X, if it exists, must have underlying set X + 1. So either there is a topology on X + 1 making it a partial map classifier for X, or there isn't a partial map classifier at all.

Notification Bot (Jan 18 2022 at 12:15):

This topic was moved here from #general: mathematics > partial continuous maps by Matteo Capucci (he/him)

Mike Shulman (Jan 18 2022 at 17:14):

One thing you can do, as I suggested above, is to consider a slightly different category than Top, which is a quasitopos and therefore has partial map classifiers. For instance, you can embed Top in the category of [[pseudotopological spaces]], which is a quasitopos. Then there will be a maybe-style monad on PsTop whose Kleisli category is equivalent to a category of partial maps of pseudotopological spaces with subspaces as domains of definition, which will contain the desired category of partial maps of topological spaces. (I'm not sure whether it will be a full subcategory; I don't know offhand whether every sub-pseudotopological space of a topological space is topological.)

Mike Shulman (Jan 18 2022 at 17:15):

Or, if all the spaces you care about are well-behaved, you can restrict to a subcategory of Top that is a quasitopos instead, such as that of [[subsequential spaces]].

Mike Shulman (Jan 18 2022 at 17:16):

Undergraduate topology courses (and often even graduate ones) sometimes give the impression that the open-set-based definition of "topological space" is "the correct notion of space", but in fact it's just one of a wide array of possibilities, and not a particularly well-behaved one at that.

Mike Shulman (Jan 18 2022 at 17:16):

(As this discussion shows!)