3
$\begingroup$
  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!

  • 0
    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".2012-02-25
  • 1
    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.2012-02-25
  • 0
    @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.2012-02-25
  • 2
    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.2012-02-25
  • 0
    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?2012-02-25
  • 0
    Related: http://mathoverflow.net/questions/22658/why-are-inverse-images-more-important-than-images-in-mathematics2012-02-25
  • 0
    @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?2012-02-25
  • 0
    @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".2012-02-25
  • 0
    @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".2012-02-25
  • 0
    @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.2012-02-25
  • 0
    @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_function2012-02-25
  • 2
    @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.2012-02-25
  • 0
    @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.2012-02-25
  • 0
    @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.2012-02-25
  • 0
    @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.2012-02-25
  • 0
    @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?2012-02-26

2 Answers 2