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When we have an isomorphism, between 2 groups or vector spaces let us say, then it is said to be structure preserving. An isomorphism exists when there is at least one mutually invertible morphism between sets (and this is arbitrary when a set is mapped on to itself). What do we mean intuitively, and perhaps somewhat analytically, when we say structure preserving? What structure do these sets, groups, vector spaces, etc . . . come with?

This is a question I'm looking for more intuition on than anything else but specific examples are very helpful. Here is are related threads

What does structure preserving mean?

Mathematical Structures

Preserving Structures

Isomorphisms: preserve structure, operation, or order?

I sought the answer throughout the forum but I never really found the questions asked/answer in a satisfactory way. Thanks everyone!

edit: For example, if we have sets A = {a, b, c} and B = {1, 2, 3} and we have a bijection $f$ between them such that

$f(a) = 1$

$f(b) = 2$

$f(c) = 3$

Is structure preserved or does that require an operation of some kind to also be present? If structure is preserved, what is it?

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    $\begingroup$ In the case of groups, the "structure" is given exactly by the operation on your set. In the case of topological spaces, it is given by the collection of open sets. In the case of metric spaces, it is given by a metric (i.e. a notion of distance between two points). You may wish to have a look at Wikipedia's page on abstract algebra. If you are really interested in this, you should also check out universal algebra, which is the formal study of algebraic structures. $\endgroup$ – A.P. May 24 '15 at 16:00
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    $\begingroup$ I had this dilemma once, but now I have the reverse: what are structures, other than that which is preserved by certain maps? Actually if I understand correctly, category theory tells us we can drop the objects altogether, and only deal with maps! $\endgroup$ – GPerez May 24 '15 at 17:26
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I'm going to start with your example and work towards a more abstract notion of structure throughout this writing. So let's see, the bijection you give is a function $f:A\rightarrow B$. But all we have are the sets $A,B$. No other information is given. So what does the bijection encode? Well, both sets have $3$ elements. Perhaps that is what we should look at. So, let $$M\overset{f}\longrightarrow N$$ be a bijection between sets. If we know $M$ is of finite cardinality, it is not too difficult to deduce from the pigeon hole principle that $N$ is also of finite, equivalent, cardinality. We use this notion for the infinite as well. Two sets have equivalent cardinality if, and only if, there exists a bijection between them. Thus, given the information $M,N$ are sets with $f$ a bijection between them we can really only deduce $M,N$ have the same cardinality (under some very technical assumptions if I remember correctly). For this reason, we would say $M,N$ are isomorphic as sets with $f$ a set isomorphism between $M$ and $N$.

Now let's take a look at something more general. Let's say $M$ is a set and $\tau$ is a structure on $M$. We can write this as $(M,\tau)$ if we would like to. Now suppose we have another set $N$ with a structure (of similar "type") called $\sigma$. Again, we can keep track of this set and its structure by the ordered pair $(N,\sigma)$. As before, if we are given a bijection, $$M\overset{f}\longrightarrow N$$ we can really only conclude $M,N$ are isomorphic as sets. However, let's say that $\tau$ gives $M$ a vector space structure (i.e. elements of $M$ can be added and scalar multiplication from a field $K$ is allowed). This is where the same "type" is important. If $\sigma$ is a topology on $N$, there isn't an immediate (to me it is not immediate) way to show $\tau$ and $\sigma$ are pretty much the same. So assume $\sigma$ is also a vector space structure but on $N$. We want to describe an isomorphism between $M$ and $N$ so that the only real difference between $(M,\tau)$ and $(N,\sigma)$ is how we labeled them. One way to accomplish this is to observe whenever $f(ka+b)=kf(a)+f(b)$ for $a,b\in M$ and $k\in K$ it does not matter if we compute the operations of $+, \cdot$ in $M$ first or in $N$, after we map with $f$. Since $f$ was assumed to be a bijection, we have that $M,N$ are the same size as sets (same cardinality), and all of the operations are the same. So $(M,\tau)$ and $(N,\sigma)$ are isomorphic as vector spaces and $f$ is a vector space isomorphism between them.

We can extend this idea above to any finite number of structures on a set $M$ - if we have $\tau_1, \tau_2,\dots, \tau_n$ we look at $(M,\tau_1,\dots, \tau_n)$ - but now this comes down to our idea of what a structure is. There probably isn't a definitive definition for this yet (maybe there is, but I would wager in a lot of cases this definition is from intuition built on many examples and uses). So let's say

Definition: A structure on a set $M$ is any extra information we are given about the behavior of elements, functions, or subsets of $M$. This captures the idea of an algebraic structure (a set $M$ with a binary operation $\phi$), a topological structure (a set $M$ with a collection of open subsets $\tau$), an ordering (a set $M$ with a relation $\preceq$), and many other ideas. We say that $(M,\tau_1,...,\tau_n)$ is an object $M$ with the structures $\tau_1,\ldots,\tau_n$.

This also motivates the

Definition: Two objects with given structure $(M,\tau)$ and $(N,\sigma)$ are said to be isomorphic under the given "type" of structure if $(M,\tau)$ and $(N,\sigma)$ are the same up to relabeling of elements.

Mathematically, these are pretty informal definitions. But they can be given rigor through set theory.


Edit: Let's show that there is some truth in these definitions (with or without justification).

We will say a structure is valid or consistent if it does not give rise to any illogical propositions. For the rest of this, all structures are assumed valid.

An object $(M,\tau_1,...,\tau_n)$ is a set $M$ with any finite number of structures $\tau_1,...,\tau_n$. We call $(U, \tau_1|_U,...,\tau_n|_U)$ a subobject if $U\subseteq M$ and $\tau_1|_U,...,\tau_n|_U$ are the induced structures on this restriction.

Given a map $f:M\rightarrow N$ of sets with associated structures $\tau$ and $\sigma$ respectively, we will call $f$ a morphism if $f$ induces a valid substructure in $\sigma$. More formally, if we denote by $\hat{f}(\tau)$ the substructure induced by $f$, then $(f(M),\hat{f}(\tau))$ is a subobject of $N$.

Proposition: If $(M,\tau_1,...,\tau_n)$, $(N,\sigma_1,...,\sigma_m)$ are objects and $f$ is an injective morphism (where an injective morphism means a morphism whose mapping is an injection on the level of sets) from $M$ to $N$, then $M$ is isomorphic to a subobject of $N$.

Proof. Since $f$ is a morphism, $(f(M),\hat{f}(\tau_1),...,\hat{f}(\tau_n))$ is a subjobect of $N$ (if $n\geq m$ it may be the case that $f$ "forgets" some of the structure). Since $f$ is an injection, for any $x\in M$ we have a distinct $y=f(x)\in N$. So we may relabel $x$ as $y$. Now we just need to check that structure is preserved. But this is true since $\hat{f}(\tau_i)=\sigma_j|_{f(M)}$ for some $i,j$ by definition.$\square$

In fact, this is a generalization of the first isomorphism theorem when $f$ is an injection.

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  • $\begingroup$ This is exactly what I was looking for. Thank you so much! $\endgroup$ – Aaron Zolotor May 24 '15 at 17:01
  • $\begingroup$ @az89 No problem, I'm enjoying this. $\endgroup$ – Eoin May 24 '15 at 17:02

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