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In many books I read I found isomorphism defined as a 'bijective homomorphism'.

I do not understand why is it that existence of inverse or order preserving requires the property of a homomorphism, i.e, $f(a+b)=f(a) \circ f(b)$ since $+$ and $\circ$ need not even be defined.

Surely I can just look at order preserving mapping over sets where $a \circ b$ is not even defined? Is that just a requirement in ring isomorphism? Many thanks

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    $\begingroup$ What do you mean by order preserving? $\endgroup$ – user99914 May 6 '15 at 4:03
  • $\begingroup$ for example i can have two partially ordered sets {a, b,c} ,{1,2,3} where a<b<c; 1<2<3 and f(a)=1,f-1(1)=a e.t.c. $\endgroup$ – user125988 May 6 '15 at 4:06
  • $\begingroup$ Your set has an ordering? And is also .... a group? $\endgroup$ – user99914 May 6 '15 at 4:08
  • $\begingroup$ it doesn't need to be, so let's say no. see, i can let a,b,c,d....=A,B,C,D...but i dont need to define what is a+b (or a*b in that matter) $\endgroup$ – user125988 May 6 '15 at 4:20
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    $\begingroup$ Just to make the misunderstanding explicit: were you reading about order isomorphisms in one book and then looked up isomorphism in some other source? If so you probably ended up looking at a definition of isomorphism for whatever algebraic structure that other source was talking about, not for order. $\endgroup$ – JHance May 6 '15 at 5:45
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In order to define a homomorphism both sets must have the algebraic structure you want to, otherwise it doesn't make sense to say that $f(a+b)=f(a)\circ f(b)$.

An homomorphism is a function that preserves the algebraic structure you're working with, so when you have a bijective homomorphism (isomorphism) between two groups, rings, modules, or whatever, you're basically saying that viewed as a group, ring, module, or whatever, they are essentially the same: as sets they are, and the algebraic operations among their elements are also equivalent.

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  • $\begingroup$ which is exactly why I am asking, in case where algebraic structure + and o do not exist, I can still define some sort of equivalence relationship, can't i? i can say that a, b,c is equivalent to A,B,C using a mapping that preserve order without defining what's a*b. $\endgroup$ – user125988 May 6 '15 at 4:14
  • $\begingroup$ Defining an homomorphism requires the structure to exist. Otherwise it would be just a function, or not even that. Depending on the kind of structures you're working with (groups, rings, partially ordered sets, etc.) you define functions that have the property of preserving the structure of your sets. For example, a group homomorphism is one that is defined only between two groups and preserves their algebraic structure. In the case of posets (which you mention) you'd ask that $f(a)\leq f(b)$ if $a\leq b$. $\endgroup$ – hjhjhj57 May 6 '15 at 4:18
  • $\begingroup$ Thank you very much, I only asked this since i was reading "introductory real analysis" by Kolmogorov and Fomin where they defined isomorphism as order-preserving 1-1 mapping between sets, which differs from what i usually see. $\endgroup$ – user125988 May 6 '15 at 4:27
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" In many books I read I found isomorphism defined as a 'bijective homomorphism'."

Throw away those books. An isomorphism is a homomorphism which admits a two-sided inverse homomorphism.

The statement that a bijective homomorphism is an isomorphism is not a definition but a theorem, which holds in some categories (and has to be shown for each of those categories!) but does not hold in any category. The first counterexamples one encounters might be topological spaces or differentiable manifolds ($x \mapsto x^3$...). I really hate linear algebra lectures, where an isomorphism is defined to be a bijective homomorphism.

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    $\begingroup$ By saying “which admits a two-sided inverse” and leaving out the part where it’s mentioned that this inverse should be a homomorphism as well, you’re hiding the whole point you are making. $\endgroup$ – k.stm May 6 '15 at 5:13
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    $\begingroup$ Also, I think this answer is missing the point of the question. $\endgroup$ – k.stm May 6 '15 at 5:14
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    $\begingroup$ Yes, indeed. I changed the wording. Nevertheless the message is the same and is very important I think. $\endgroup$ – MooS May 6 '15 at 5:19
  • $\begingroup$ Now I'll never forget this. $\endgroup$ – hjhjhj57 May 6 '15 at 5:27
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I want to stress MooS’ excellent point by saying a few more words regarding that.

I think what is important to note here, is that “isomorphism” is a term that is used for many things and the definition MooS has given captures their common essence (if you interpret “homomorphism” accordingly).

It is the basic idea of category theory which we are talking about: Considering the entirety of all mathematical objects of a certain structure and their structure-preserving maps. In category theory, the entirety is called category, the mathematical objects are called objects and the structure-preserving maps are called morphisms or arrows (and actually need not be set-theoretical maps at all).

In this context, an isomorphism is always a morphism with a two-sided inverse morphism. Take these examples:

  • If you’re talking linear algebra, you consider linear spaces as mathematical objects and linear maps as structure-preserving maps. Linear maps are also called “homomorphisms”. An isomorphism of linear spaces is a linear map which has a (two-sided) inverse that is also a linear map. It turns out that isomorphisms of linear spaces are exactly bijective linear maps.

  • If you’re talking order theory, you consider partially ordered sets¹ as mathematical objects and order-preserving maps as structure-preserving maps. An isomorphism of ordered sets is an order-preserving map which has a (two-sided) inverse that is also an order-preserving map. It turns out that isomorphisms of ordered sets are exactly bijective order-preserving maps.

  • If you’re talking topology, you consider topological spaces as mathematical objects and continuous maps as structure-preserving maps. An isomorphism of topological spaces is a continuous map which has a (two-sided) inverse that is also a continuous map. It turns out that there are bijective continuous maps between topological spaces which are not isomorphisms of topological spaces.


¹: … or lattices or totally ordered sets or preorders or …

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An isomorphism is a special kind of homomorphism. Every isomorphism is a homomorphism with some additional properties.

An isomorphism is an injective homomorphism. It is bijective with the range of the homomorphism.

If $f:G \to H$ is an injective homomorphism, then $G$ is isomorphic to $f(G)$; i.e., $G \cong f(G)$.

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