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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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    @Doug: That's not an equation, it's a quantified propositional formula. It's certainly true that $Gxx = x$ for all $x \in \{ a, b \}$. Applying $H$ we obtain that $F(Hx)(Hx) = Hx$ for all $x \in \{ a, b \}$. Notice the quantifiers haven't changed! If I now write $y = Hx$, we get $Fyy = y$ for all $y \in \{ Ha, Hb \}$. But that doesn't imply that $Fyy = y$ for all $y \in \{ 1, 2, 3 \}$, as you rightly observe.2011-05-29

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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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    @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...2011-05-27
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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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    @Doug As I said, it's a common notational abuse. One shouldn't worry to much about such *representational* matters when first studying these topics, i.e. don't lose site of the forest for the trees. Instead, focus on the *concepts*.2011-06-01
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"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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    @Doug: I still think you're overthinking this. I haven't made any claims about structure in the _range_, only structure in the _domain_, and in any case I think you're getting too hung up on some personal concept you have of the word "preserve" (or possibly "structure," I'm not sure). I don't really know what to say because I can't quite figure out what this personal concept is.2011-05-28