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Why do homomorphisms get called "structure-preserving maps/functions" or get said to "preserve the multiplication" when they only induce structural preservation for a subalgebra of the algebra that they map to?

If the question doesn't come as clear here, consider the algebras described by the following operation tables

G  a  b
a  a  b
b  b  b

F  1  2  3
1  1  2  3
2  2  2  2
3  3  2  3.

Of course here, redundantly, we can say that $G: \{a, b \} \times \{a, b \} \to \{a, b\}$, $F: \{1, 2, 3\} \times \{1, 2, 3\} \to \{1, 2, 3\}$ just from those tables. Define a unary function $H:a \to 1, b \to 3$. One can check that for all $x, y$ in $\{a, b\}$, $HGab=FHaHb$ [or equivalently $abGH=aHbHG$ or equivalently $H(G(a,b))=F(H(a),H(b))$]. Thus, we have a homomorphism here (more specifically a monomorphism, since $H$ only qualifies as an injection). But, it comes as clear that $(\{a, b\}, H)$ satisfies for all $x$, $Hxx=x$, while $(\{1, 2, 3\}, F)$ does not satisfy for all $x$ $Fxx=x$, though the subalgebra $(\{1, 3\}, F)$ of $(\{1, 2, 3\}, F)$ does satisfy for all $x$ $Fxx=x$.

Why then do homomorphisms get called structural-preserving maps, or get said to preserve the multiplication? Does this consist of an error on people's part, or do they have something else in mind? Why not say that homomorphisms between algebras consist of sub-transferred maps, or sub-induced maps, or preserve some sub-multiplication?

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I slightly fixed the formatting. Please note that a paragraph starting with four blank spaces is interpreted as a code block. –  t.b. May 27 '11 at 15:11
    
Thanks! I'm new here. How did you get the operation tables to format like that? –  Doug Spoonwood May 27 '11 at 15:24
    
By putting them in a code block, that is I indented them with four blank spaces :) click on the 'edit' link right on the bottom left of your question to see what exactly I did. –  t.b. May 27 '11 at 15:26
    
To give a short, brief answer, a homomorphism is a mapping that takes true equations to true equations (and says nothing about inequalities). This, in essence, is Theo's answer below. –  Zhen Lin May 27 '11 at 17:50
    
@Theo Thanks! @Zhen I've changed the the middle row and the middle column to "2" instead of "3". The equation for all x, Hxx=x holds true for ({a, b}, G), but does not hold true for ({1, 2, 3}, F). So, how does a homomorphism take true equations to true equations? –  Doug Spoonwood May 29 '11 at 12:39

3 Answers 3

up vote 4 down vote accepted

A homomorphism $x \mapsto x'$ doesn't actually preserve anything, unless it's an endomorphism, that is, from a set to itself. In this case, we can write $(x \star y)'= x' \star y'$. I guess a better term would be respect instead of preserve. When the map is between different sets, a homomorphism preserves the corresponding relationships, as in $(x \star y)'= x' \star' y'$. There's no deeper meaning to the terms than that.

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I agree with you endomorphisms preserve structure from the domain. But, more specifically, don't epimorphisms (surjective homomorphisms here) also preserve the structure of the domain here, since all points of the co-domain get mapped to in such a case? If epimorphisms do preserve the structure here, I think the catch comes as that the structure mapped to (g, *') by the homomorphism H:f->g might have "more" structure than the structure which the homomorphism maps from (f, *). Not sure about all of that though, just a thought. –  Doug Spoonwood May 27 '11 at 16:39
    
@Doug, no I think structure is best preserved by injective homomorphisms. Surjection does not seem related to preserving structure but rather to imposing structure. But all this is too philosophical... –  lhf May 27 '11 at 16:43

The "structure" that is preserved by an algebra hom $\rm\:h\::\:A\to B\:$ is the equational algebraic structure of the domain $\rm\:A\:,\:$ i.e. how the elements of the domain are related to each other under the operations of the algebra. In particular, any equation holding true in the domain $\rm\:A\:$ is mapped to an equation holding true in its image $\rm\:h(A)\:.\:$ For example, if a ring $\rm\:A\:$ satisfies a polynomial identity so too will its image $\rm\:h(A)\:.\:$ Thus, in addition to the ring axioms, homs also preserve any other universal properties, e.g. commutativity $\rm\:\forall x,y\::\ xy = yx\:$ or Booleaness $\rm\:\forall x\::\ x^2 = x\:.$ If an element is a unit (invertible) then this is preserved: $\rm\: a\:b=1\ \Rightarrow\ h(a)\:h(b) = 1\:;\:$ if it's a root of some specific polynomial equation then this is preserved, e.g. an $n$'th root of unity remains so in the image. Such preservation properties follow simply by using "structural induction" to lift the hom's preservation of the basic algebra operations to all "polynomial" expressions composed of the basic operations.

In fact it is precisely the definition of a hom that serves to characterize what constitutes an innate algebraic property, and what doesn't. For example, internal set-theoretical representations of the elements of an algebra need not be preserved by an algebra hom. So such internal representational properties are not algebraic properties. From an algebraic perspective the elements of the algebra are atoms whose internal structure is not pertinent. Whether the elements of a ring are represented by matrices, differential operators, etc is not relevant to the isomorphism class of the ring. Thus complex numbers can be represented by pairs of reals (Hamilton) or 2x2 matrices, or polynomial expressions in $\rm\:\mathbb R[x]/(x^2+1)\:,\:$ etc. Algebraically, the structure (isomorphism class) depends only on the "holistic" property of how the elements relate to each other under the ring operations. Thus two rings are isomorphic iff they have the same addition and multiplication tables. Any deeper-level structure (such as element representation) is (intentionally) not captured algebraically.

Note that while algebra homs preserve universal $(\forall)$ statements, they need not preserve statements of more general logical form, e.g. statements involving disjunctions, existential quantifiers, etc. For example, the property of being an integral domain $\rm\:\forall x,y\::\ xy=0\ \Rightarrow x=0\ \ or\ \ y=0\ $ is not preserved by ring homs; indeed $\rm\:\mathbb Z\:$ is a domain but its image $\rm\:\mathbb Z/4 =\:$ integers $\rm\:(mod\ 4)\:$ is not. The precise relationship between syntax, semantics and preservation properties will become much clearer when one studies universal algebra and model theory.

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Great answer. Mine was too simplistic. –  lhf May 27 '11 at 17:44
    
1. How does the domain have an equational algebraic structure? Do you mean that the system (f, F_1, ..., F_n) where f denotes the set, and F_1, ..., F_n the operations, has an equational structure? 2. So, the algebraic structure only gets transferred from the system S, to a subsystem R' of R by a homomorphism in general, correct? It could get transferred to R if R=R', of course, but in general this doesn't hold, correct? 3. Do, in general, the universal statements get transferred from system S to the system R, or a subsystem R' of R? –  Doug Spoonwood May 28 '11 at 0:06
    
Thinking more on this, what precisely does the homomorphic image h(A) consist of? A bare, unstructured set. So, how can an equation (other than those of set theoretic equations) hold true in the homomorphic image H(A), since H(A) has no operations on it? Additionally, if you look at the tables in the post now (the second table has "2" in the middle row and middle column at present), the first table has universal property of what you call "Booleaness". The second table does not. But, H:a->1, b->3 does qualify as a homomorphism. So, how has "Booleaness" gotten preserved? –  Doug Spoonwood May 29 '11 at 12:48
    
@Doug (1) The equational structure is the mentioned algebraic equations that hold between polynomial terms. Equivalently it is the equations represented by the operation tables for the algebra, $\rm\:a*a = a\:$ etc.$\ $ (2,3) Yes, the preserved structure is only in the image $\rm\:h(A)\:.\:$ Elements in $\rm\:B\:$ but not in $\rm\:h(A)\:$ are not images of elements of $\rm\:A\:$ so one cannot employ the hom $\rm\:h\:$ to transfer properties of $\rm\:A\:$ to them. –  Bill Dubuque May 30 '11 at 18:44
    
@Doug The target of a hom isn't just a set but an algebra of the same type, so it has the same algebra operations. Perhaps you're confused by the common notational abuse of employing the same name for the algebra hom and its underlying set-theoretic map. The Boolean (ring) property is preserved in the image $\rm\:h(A)\:$ but not necessarily in all of $\rm\:B\:.\:$ Similarly, commutativity is preserved in the image of $\:\mathbb R\:$ as the constant diagonal matrix subring of the ring $\rm\:B\:$ of 2x2 real matrices; but the full ring $\rm\:B\:$ is noncommutative. –  Bill Dubuque May 30 '11 at 21:29

"Structure" refers to the domain, not to the range. Homomorphisms preserve structure in the domain when they map things into the range. That is, one should think of a homomorphism $f : A \to B$ as directed from $A$ to $B$, and in such a way that the structure in $A$ gets preserved "along the way" to $B$.

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What is confusing about saying that the structure in $A$ gets preserved is that it is obviously not true because something is lost if the map is not injective. For instance, $\mathbb Z \to \mathbb Z_6$ turns a domain into a ring that has zero divisors. The homomorphism theorem is then about what does get preserved. –  lhf May 27 '11 at 15:50
    
@lhf: I suppose. It's not perfectly preserved. –  Qiaochu Yuan May 27 '11 at 16:20
    
How could structure refer to the domain, when a homomorphism starts with the domain and ends with the co-domain? Examples like the above indicate that the structure doesn't end up in the co-domain, correct? So, how could the structure get preserved "along the way"? The structure would seem to either have to gradually disappear, or vanish immediately before reaching the co-domain, would it not? –  Doug Spoonwood May 27 '11 at 16:29
1  
@Doug: to be honest, I think you're overthinking this. Here's a basic example of what gets "preserved." Suppose in a ring $R$ you have some equation like $ax + b = c$. If $f : R \to S$ is a ring homomorphism, it follows that by applying $f$ you get another equation $f(a) f(x) + f(b) = f(c)$. In other words, the equality has been preserved. –  Qiaochu Yuan May 27 '11 at 16:45
    
Sure. But, look above. We have the equation G(x,x)=x, or (xGx) if you prefer infix notation, in system ({a, b}, G). In system ({1, 2, 3}, F) F(H(2), H(2)) cannot get claimed to obtain via the homomorphism, since F(H(2), H(2)) is undefined by the homomorphism and since H(2) is also undefined. As the operation table for F makes clear, I've made it so that idempotence fails. I could have also made it so that commutativity fails for some system F', since H here only defines F for the pairs (1, 1), (1, 3), (3, 1), and (3, 3). –  Doug Spoonwood May 27 '11 at 23:52

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