What is a homomorphism and what does "structure preserving" mean?

Question

I am not a mathematician and have not formally studied mathematics so I hope someone will be able to explain this to me in a way that I can understand given my level of mathematical understanding.

I have read the other posts about this question but the answers seem to assume some knowledge that I don't have.

I am learning about multilinear algebra and topological manifolds.

It is said that "linear maps" (of which I understand the definition) between vector spaces are so called "structure preserving" and are therefore called "homomorphisms".

Could someone explain in both an intuitive way, and with a more formal definition, what it means for a "structure to be preserved"?

You might find my post here helpful. I discuss ring homomorphisms, but the scenario will be analogous to vector space homomorphisms with the "multiplication" now scalar multiplication. http://math.stackexchange.com/questions/2004755/how-to-visualize-ring-homomorphisms-isomorphisms/2004817#2004817 — Kaj Hansen, Jan 28 '17 at 08:12
In short, if we say that the function doesn't preserve structure, we mean that if you were to apply the structure (add vectors, scale them, etc) and then apply the function to the result, you will get a different result from first applying the function, and then applying the structure. — Arthur, Jan 28 '17 at 08:14
Context is important but for example group homomorphisms map identities to identities and so preserve structure in that sense. Similarly for inverses. — Karl, Jan 28 '17 at 08:54

score 26 · Accepted Answer · edited Apr 17 '21 at 08:35

There are many sorts of "structures" in mathematics.

Consider the following example: On a certain set $X$ an addition is defined. This means that some triples $(x,y,z)$ of elements of $X$ are "special" in so far as $x+y=z$ is true. Write $(x,y,z)\in{\tt plus}$ in this case. To be useful this relation ${\tt plus}\subset X^3$ should satisfy certain additional requirements, which I won't list here.

Assume now that we have a second set $Y$ with carries an addition ${\tt plus'}$ (satisfying the extra requirements as well), and that a certain map $$\phi:\quad X\to Y,\qquad x\mapsto y:=\phi(x)$$ is defined by a formula, some text, or geometric construction, etc. Such a map is called a homomorphism if $$(x_1,x_2,x_3)\in{\tt plus}\quad\Longrightarrow\quad\bigl(\phi(x_1),\phi(x_2),\phi(x_3)\bigr)\in{\tt plus'}\tag{1}$$ for all triples $(x_1,x_2,x_3)$.

In $(1)$ the idea of "structure preserving" works only in one direction: special $X$-triples are mapped to special $Y$-triples. Now it could be that the given $\phi$ is in fact a bijection (a one-to-one correspondence), and that instead of $(1)$ we have $$(x_1,x_2,x_3)\in{\tt plus}\quad\Longleftrightarrow\quad\bigl(\phi(x_1),\phi(x_2),\phi(x_3)\bigr)\in{\tt plus'}$$ for all triples $(x_1,x_2,x_3)$. In this case $\phi$ is called an isomorphism between the structures $X$ and $Y$. The elements $x\in X$ and the elements $y\in Y$ could be of totally different "mathematical types", but as far as addition goes $X$ and $Y$ are "structural clones" of each other.

Very clarifying. Is it the case that that if you generalize your example to other operators, that this gives a general definition of homomorphism and isomorphism? Secondly, if the above is the definition of homomorphism, can you prove that a linear map from a vector space to another is a homomorphism with respect to the vector space operators? — user56834, Jan 28 '17 at 10:27
By the way, is this why it is called a HOMOmorphism? Because the relation only goes one way? — user56834, Jan 28 '17 at 10:27

operatorerror · Answer 2 · 2017-01-28T08:17:38.697

12

It is, in short, highly context dependent. Isomorphisms constitute a renaming of points in a space and do not change any of the properties we care about for that particular space.

An isomorphism of vector spaces preserves the properties we care about in a vector space. If you are a linearily independent set in the domain, you will form a linearily independent set when mapped to the image. If you are a subspace here, you are a subspace there. As far as linear algebra is concerned, the two sets just have elements with different names.

An isomorphism of topological spaces, although never called this, is a homeomorphism. If I am an open set in the domain, I am an open set in the image. If I am connected in the domain, I am connected etc. As far as topology is concerned, we just renamed a bunch of points.

Same goes for isometries of metric spaces, diffeomorphisms on manifolds, group isomorphisms, ring isomorphisms and so on.

edited Jan 28 '17 at 08:17

answered Jan 28 '17 at 08:09

operatorerror

29,881

Thank you, this is going in the right direction. Couple of questions: 1) is there a formalization of "properties we care about"? Does that formally refer to the axioms of the mathematical structure that the homomorphism applies to (e.g. vector spaces)? 2) so if we say that mathematical structures A and B are homomorphic, this means exactly the same thing as saying that they are isomorphic? (I am thinking in particular about the definition of isomorphic as made in mathematical logic, which has as a corollary that all S-sentences are true in A iff they are true in B). – user56834 Jan 28 '17 at 08:18
so more generally, if we would do a complete search of all definitions and theorems published in mathematics papers that contain the words homomorphism or homeomorphism or isomorphism, and change them all to "isomorphism", then we wouldn't lose any information? They are completely synonymous, except for their customary differences in the contexts in which they are used?

user56834

Jan 28 '17 at 08:21

so more specifically, is it true that if we know a) that A and B are homomorphic structures of some kind (say vector spaces), and b) we know nothing else about them, do we then have no way of distinguishing them from eachother?

– user56834 Jan 28 '17 at 08:24

Peter Franek · Answer 3 · 2017-01-28T09:15:50.503

In a more abstract level, being equal "up to isomorphisms" says that you have two instances of the same type. Consider for example what is a square? The word square may refer to two things: square as a shape (type of shape) --- such as "we have square, rectangle, ball, line, etc". But it can also refer to a particular square that you draw into a plane. In the second sense, there are infinitely many squares, although each of them represents the shape "square".

It is reasonable to consider, what they all have in common. One approach is to say that you can map a square into any other with a "structure-preserving" bijection. In this case, the "structure"-preserving will probably mean that it takes "equally long line segments" to "equally long line segments" and right angles to right angles.

More generally, mathematical objects are usually described as a set and a structure---the structure may be various kinds of additions, multiplications, compositions, ability to take limits or derivatives, smoothness or whatever. A homomorphism is then a map that preserves this structure, whatever it is --- it takes a sum to a sum, a difference to a difference, a smooth vector field to a smooth vector field, a square to a square. The individual definitions differ from case to case, but the idea is always the same.

Isomorphism is maybe more intuitive then homomorphism. Homomorphism means that it preserves structure in the above sense, but some information may be lost (you may map everything to zero, for example).

score 6 · Answer 4 · answered Jan 28 '17 at 08:23

Let's start with a vector space. You probably know the precise definition of a real vector space $V$. But what is a vector space? The intuitive answer is short. A vector space is a structure in which you can add elements and you can multiply them by scalars. In short: If $\lambda,\mu\in \mathbb{R}$ and $v,w \in V$ then $\lambda v+\mu w\in V$. Another way to say this is that a vector space is closed under linear combinations. Now that's the essence of vector spaces.

A morphism of vector spaces should be compatible with the essence of vector spaces. Hence it shoud preserve linear combinations: $T(\lambda v+\mu w)=\lambda T(v)+\mu T(w)$. This is what is meant by structure preserving. It also follows from this definition that $\text{Im}(T)$ is itself a vector space, so this really motivates the term structure preserving.

The same holds for other structures such as groups, rings, fields, algebras, modules, topological spaces, manifolds,$\dots$ In each example you have to decide what the essence of the structure is, and morphisms should be compatible with that.

Thank you. You give the example of "structure preservation" in the example of vector spaces. However I'm still somewhat confused about how this generalizes to the general concept "structure preservation" (though all the answers here have helped). Would it make sense to define "structure preserving map" M between two structures A and B of type X (e.g. X = vextor space) as a map between A and B such that any X-operations done on elements of A and then mapped to B using M, give the same result as X-operations done on the elements of B that you get when applying M on those same elements of A? — user56834, Jan 28 '17 at 08:34
There is no definition for "structure preserving maps" but in each case you can find natural candidates. Most of the times it is desired that the collection of a particular kind of structure together with the proper morphisms between them forms a category (see wikipedia). But that doesn't tell you how to define the correct notion of morphism. — Mathematician 42, Jan 28 '17 at 09:23

What is a homomorphism and what does "structure preserving" mean?

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