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Is there a bijection between kinds of mathematical structures and morphisms that preserve all properties of this kind of structure?

For example:

Topologycal spaces $\leftrightarrow$ Homeomorphisms

Metric spaces $\leftrightarrow$ Isometries

Sets $\leftrightarrow$ Bijections

Groups $\leftrightarrow$ Isomorphisms of groups

Rings $\leftrightarrow$ Isomorphisms of rings

Modules $\leftrightarrow$ Isomorphisms of moodules


Being more precise, my question becomes in 2 particular questions in concrete categories:

  1. Given a mathematical structure in a set (concrete object), ¿this follows which functions are the morphisms?.

  2. Given a function between sets, ¿this follows which objects must be the sets?.

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I don't really get your question. Because for any structure you choose on the LHS you can select the morphisms that preserve that structure and put them on the RHS. Perhaps there won't be any such morphisms but this shouldn't be a problem. Now I don't see how do you derive the group structure (just to pick one) from the groups morphisms. To be more precise I don't see what do you mean by "Groups $\leftrightarrow$ Isomorphisms of groups" instead of "Groups $\rightarrow$ Isomorphisms of groups". In any case I would be very glad to read the answer of an expert!! – Giovanni De Gaetano Jul 20 '12 at 14:42
Are you looking for a way to map between categories, as in functors? Probably the most common example would be a forgetful functor that maps a group to its underlying set and the isomorphism becomes a set bijection. Of course, a functor doesn't necessarily have to be a bijection. – axblount Jul 20 '12 at 14:59
You have to be more precise about the two sets you want to find a bijection between. For example, I would be interested in your definition of 'the set of all kinds of mathematical structures'. Also your examples are totally unclear...what is the bijection between 'sets' and 'bijections'? – wildildildlife Jul 20 '12 at 17:07
Yes, there are problems with the question, but, also, it can be intepreted/rewritten in interesting ways. For example, yes, we can define "categories" in an object-free fashion, only in terms of morphisms, leaving the "objects" to a secondary role. Thus, knowing all isomorphisms (bijections) in "some category of sets" does indirectly tell us the isomorphism-classes of sets. So the ambiguity and problems with this question have potential interest! – paul garrett Jul 20 '12 at 17:38
Just for clarification, since I am also new to the field. The OP is suggesting that we create a set $A$ of all the structures and set $B$ of all the morphisms and then show that there is a one-to-one correspondence between $A$ and $B$ right? – Jayesh Badwaik Jul 27 '12 at 4:39
up vote 5 down vote accepted

Your question is a little unclear for the reason Giovanni De Gaetano described. Bracketing that, there are mathematical structures which are "preserved" by several types of isomorphisms. The website nLab has this to say about Banach spaces, emphasis added:

In functional analysis, the usual notion of ‘isomorphism’ for Banach spaces is a bounded bijective linear map $f:V\to W$ such that the inverse function $f^{−1}:W\to V$ (which is necessarily linear) is also bounded. In this case one can accept all bounded linear maps between Banach spaces as morphisms. Analysts sometimes refer to this as the “isomorphic category”.

Another natural notion of isomorphism is a surjective linear isometry. In this case, we take a morphism to be a short linear map, or linear contraction: a linear map f such that $∥f∥\leq 1$. This category, which is what category theorists generally refer to as $\mathbf{Ban}$, is sometimes referred to as the “isometric category” by analysts.

So "isomorphism of Banach spaces" is not clear. You might also find page 23 of The Joy of Cats (available for free here) interesting. More generally, if you are interested in questions of this type, you should do some reading on category theory.

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I don't know if it helps you, but one can define non trivial elementary embedding in ZFC, and the existence of such embedding is proved to be independant of ZFC : it's a possible axiom (see critical points for example). So I'm not sure if your question has a answer as you think it should...

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