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  1. Sometimes I see something like "a mapping preserves the structures of its domain and of its codomain". From Wiki about morphisms in category theory:

    a morphism is an abstraction derived from structure-preserving mappings between two mathematical structures. The notion of morphism recurs in much of contemporary mathematics. In set theory, morphisms are functions; in linear algebra, linear transformations; in group theory, group homomorphisms; in topology, continuous functions, and so on.

    I was wondering why the structure-preserving mappings between two topological/measurable spaces are defined by the "inverse" of the mapping, while the structure-preserving mappings between two groups/vector spaces are not?

    Why are the structure-preserving mappings between two topological spaces chosen to be continuous mappings instead of open mappings?

  2. I also see that "a mapping preserves some property of subsets, points or whatever". Such as

    Continuous linear mappings between topological vector spaces preserve boundedness.

    According to Brian's reply to my earlier question, this quote should be understood as "under a continuous linear mapping, the image of any bounded domain subset is also a bounded codomain subset", not as "under a continuous linear mapping, the inverse image of any bounded codomain subset is also a bounded domain subset".

    I wonder why? It seems at first to me like how continuous mappings preserve topologies, but it is actually in the same way as how group homomorphisms preserve group structures.

Thanks and regards!

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  • Note that for number two, if the linear map is a bijection then both statements are true. Furthermore, bijective continuous linear maps between topological vector spaces are isomorphisms, which is what is typically meant by a map that "preserves structure". – Alex Becker Feb 25 '12 at 03:49
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    It's worth noting that while maps are often defined going backwards or forewards, other information often travels the opposite way. The continuous image of a compact set is compact, and the preimage of a normal subgroup is normal, for example. – Brett Frankel Feb 25 '12 at 03:51
  • @AlexBecker: My understanding is that the bijective case is for isomoprhism, but is not necessary for "preserving structures/properties". For example, homomorphisms and isomorphisms between two groups. – Tim Feb 25 '12 at 03:51
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    In order to view topological maps (i.e., continuous maps) as "preserving a structure", you really need to think of them in terms off preserving the notion of "closeness", not the notion of "open sets". It just so happens that the right way to say "$f$ sends points that are close-to-one-another to points that are close-to-one-another" is via inverse images when you consider open sets. To define it in terms of direct images, you consider instead the filter of neighborhoods of a point. – Arturo Magidin Feb 25 '12 at 03:52
  • If we think of the category of topological spaces as a generalization of the category of metric spaces, then maps can just as easily be defined "going forewards," ie. continuous maps are those that respect limits. Does anybody know which definition, historically, came first? – Brett Frankel Feb 25 '12 at 03:57
  • Related: http://mathoverflow.net/questions/22658/why-are-inverse-images-more-important-than-images-in-mathematics – Quinn Culver Feb 25 '12 at 04:06
  • @ArturoMagidin: (1) I am not quite able to understand what direct images are yet, and not soon either because of lack of prerequisite knowledge. (2) So are you saying preserving closeness between points is preserving convergence of a filter or a net of points? (3) Why are structure-preserving mappings between measurable spaces measurable mappings? – Tim Feb 25 '12 at 04:07
  • @Tim: By "direct image" I simply mean things like: if $f\colon X\to Y$ and $A\subset X$ (or $A\in X$), considering $f(A)$; vs. what is usually done in topology, which is to consider $B\subset Y$ and look at $f^{-1}(B)$, an "inverse image". – Arturo Magidin Feb 25 '12 at 04:11
  • @Tim: Actually, measurable mappings are not "structure preserving mappings" between measure spaces. For example, the Lebesgue-measurable functions $\mathbb{R}\to\mathbb{R}$ are not the ones that map Lebesgue measurable sets to Lebesgue measurable sets, nor are they the ones such that the inverse image of a Lebesgue measurable set is a Lebesgue measurable set. Measurable mappings have to do with integration, not with "structure". – Arturo Magidin Feb 25 '12 at 04:12
  • @Tim: I'm saying that if you abstract the notion of "preserving limits" (such as the one mentioned by Brett Frankel) to arbitrary topological spaces, then you end up discussing the filter of neighborhoods at each point in the domain, and how the "direct image" of neighborhoods of a point behave relative to neighborhoods of the image of the point. – Arturo Magidin Feb 25 '12 at 04:14
  • @ArturoMagidin: I guess you might understand a Lebesgue measurable mapping wrong (or I don't understand your comment?). It is defined as a measurable mapping between Lebesgue sigma algebra on domain and Borel sigma algebra on codomain, so it preserves structures just in the sense of the definition of a measurable mapping. See http://en.wikipedia.org/wiki/Measurable_function – Tim Feb 25 '12 at 04:19
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    @Tim: You're right about Lebesgue functions pulling Borel sigma-algebras back to Lebesgue sigma-algebras, but the analogy is not quite right here, since in category theory we want to be able to compose arrows. Compositions of Borel-measurable functions are Borel-measurable, but compositions of Lebesgue-measurable functions are generally not Lebesgue-measurable. – Brett Frankel Feb 25 '12 at 04:26
  • @Tim: Yes, I am aware of what Lebesgue measurable mapping is; but it's not a "structure preserving mapping" in the same sense as maps between groups, topological spaces, linear algebra, etc.: they are not closed under composition, even when this is possible, and the analogy simply breaks down too early. – Arturo Magidin Feb 25 '12 at 04:32
  • @BrettFrankel: Thanks! (1) Is it the definition of category about morphism-composition part that requires a mapping "preserving structures" to be still preserving structures after composition with another mapping of the same kind? (2) When composing a Lebesgue measurable mapping with another, the domain of one and the codomain of the other do not have the same structure and therefore are different objects, since one has Borel sigma algebra, while the other has Lebesgue sigma algebra, so we cannot compose the two measurable mappings in the same way as composing two morphisms in a category. – Tim Feb 25 '12 at 04:36
  • @Tim: Well, the idea behind categories is that structure must be preserved, but of course this isn't quite built into the definition. What we have is an axiom that says that "composition" of morphisms gives a new morphism. So Borel sigma-algebras (with Borel-measurable functions) on $\mathbb{R}$ form a category, but Lebesgue sigma-algebras (and Lebesgue-measurable functions) don't. – Brett Frankel Feb 25 '12 at 04:45
  • @BrettFrankel: Thanks! Since continuous mappings preserve convergence of nets/filters in a category of topological spaces, I wonder what is preserved by measurable mappings in a category of measurable spaces? – Tim Feb 26 '12 at 16:11

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Top is the category of topological spaces and continuous maps simply by definition; topology typically deals with continuous maps, making this category the most important one, and thus by convention it's the one meant when saying "the category of topological spaces".

(aside: other conventions on what Top or "the category of topological spaces" stands for are far more likely to disagree on what the objects are, rather than the morphisms. e.g. to make the objects be merely the compactly generated Hausdorff spaces)

You can, of course, make all sorts of other categories. The category of topological spaces and open maps is a perfectly reasonable category to make; it's just less useful.

It takes a bit to get used to, but category theory rejects the mindset that mathematics is about objects, with the mappings between objects being a derived notion. Instead, you need to consider objects and mappings as equals -- or even to consider the objects superfluous.

On that last point, my favorite example of a category whose emphasis is on the morphisms is matrix algebra. The set of all matrices, with composition defined by multiplication, form a category. (with addition, you get an Abelian category) The objects of this category really play no role beyond bookkeeping to say which matrix products are defined.

(This category is, of course, equivalent to the category of fintie-dimensional vector spaces and linear maps)

  • +1 Thanks! (1) Why in the category of matrices, the objects "matrices" are superfluous? (2) How about measurable mappings preserving sigma algebras? – Tim Feb 25 '12 at 04:28
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    In the category formed by matrix algebra, the matrices are the morphisms, not the objects. If you insist on there being objects, in the simplest version of this category, the objects would be numbers. So an $m \times n$ matrix is a morphism from $n$ to $m$ (or vice versa, depending on your conventions). –  Feb 25 '12 at 09:29
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The question actually boils down to:
"Why do we define continuous maps as the way we do?!?"

  • Most answers go like:
    If you consider continuity between metric spaces you can abstractize! it to the algebra of open sets. Not very satisfying!!!

  • Another approach goes about neighborhoods:
    Consider what neighborhoods should be like intuitively. What you get is Felix Hausdorffs definition of neighborhood systems. In this context, one would guess! what a function should be like when talking about continuity, that is for every neighborhood one can find another one who's image fits inside. That is what we always pictured in our mind when we were checking continuity in first year analysis course. Nice, but it still doesn't say why this should be continuity in abstract spaces.

  • Better approach might be a closeness relation:
    Consider a relation obeying some sort of axioms that seem intuitive for what we understand by being close. This again needs some sort of intuition, that is in first place not very satisfying since, well, math is about facts not intuition. However, it has at least the advantage that continuity pops up! almost evident in this context: If sth is closeby it stays closeby under continuous maps. Moreover, it is only in this context where the image is involved rather than the preimage!!!

Despite not being a complete answer, I hope this gives at least some insight.

Addendum: One could consider the preimage of closed sets which is super useless or open mappings which do not permit an inverse. See the post on open mappings.

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