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A group homomorphism is usually defined to be a function $\phi$ such that if $x * y = z$, then $\phi x \times \phi y = \phi z$. However, this can be generalized. We could define that a group homomorphism is a relation $\phi$ such that if $x * y = z$ and $(x,x') \in \phi$, and $(y,y') \in \phi$ and $(z,z') \in \phi$, then $x' \times y' = z'$. This reduces to the usual definition in the case where $\phi$ is a function. However, I've never seen this generalization mentioned anywhere. Is there something wrong with it?

My thoughts are that it might have to do with functions having well-behaved preimages. In particular, if $f$ is a function then $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$. If $f$ were simply an arbitrary relation, this property could (would?) fail.

So my question is, why are homomorphisms usually assumed to be functions? Note that this is not a question about group homomorphisms specifically.

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The answer to "Why is X defined this way?" is always "because it is convenient and useful to do mathematics with that definition." I would speculate that having a deterministic relationship like a function is just the simplest, easiest to work with relationship that doesn't have any of the ambiguity of one-to-many relationships. –  rschwieb Jan 25 '13 at 17:38
    
@rschwieb I agree. However that's like saying that the answer to "Why did X happen?" is always "because the Standard Model of Particle Physics says so." I'm looking for something more. –  user18921 Jan 30 '13 at 15:30
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My point is that I think that there isn't anything more. Definitions are just shorthand for axioms. The reasons the axioms for fields are what they are is because they contain and/or imply all the features of fields that we consider important. –  rschwieb Jan 30 '13 at 20:47
    
At the risk of being a smart-arse: Assume for a contradiction that there isn't anything more. Then there's a theorem to that effect. But that theorem is something more. Contradiction! –  user18921 Jan 30 '13 at 21:06
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5 Answers

First, there are plenty of categories where the morphisms are in any way functions. Having said that, there are various reasons to primarily consider homomorphisms between algebraic structures to be functions and not relations (or other things). I'll illustrate just one. The fundamental group is an group associated to any topological space $X$ with a chosen base point $p\in X$. A continuous functions $(X,p)\to (Y,q)$ such that $f(x)=y$ induces a group homomorphism between the associated fundamental groups. Fundamental groups tell us a lot of the space and possible continuous functions between spaces. In this case, the reason we want group homomorphisms to be special functions is that that is what we get from the fundamental group construction. The reason the fundamental group construction associated to a continuous function a group homomorphism is that we started with a function. So, you can say let's generalize that too and instead of talking about continuous functions $f:X\to Y$ let's talk about continuous relations. Here the trouble start. To some degree it can be done, but the multi-valuedness really creates lots of difficulties (try to even define what it should mean for a relation to be continuous). And let's not even start trying to say what it would mean for a relation to be differentiable, holomorphic, and so on.

So, while you can always generalize for the sake of generalization, the question of how applicable/relevant is the resulting more general theory to real-life situations. In the sciences many phenomena are modelled by functions, not by relations. So, we need to have a good theory of functions. The more general theory of relations is (being a generalization) weaker, it can prove less theorems, so when applying it to functions it might not give strong enough results.

To conclude, you can generalize anything, including functions. But functions are everywhere, so we need to study them.

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Your proposed definition of homomorphisms as relations doesn't work when the relation $\phi$ isn't a function, at least if you make the reasonable assumption that $(e,e')\in\phi$, where $e$ and $e'$ are the identity elements of the two groups. Suppose, for example, that some $x$ is related by $\phi$ to two different values, say $(x,x')\in\phi$ and also $(x,x'')\in\phi$. Then, applying your definition with $y,y',z$,and $z'$ equal to $e,e',x$, and $x''$ respectively, we could infer from $x*e=x$ that $x'\times e'=x''$, which is wrong if $x'\neq x''$.

Instead of assuming that $(e,e')\in\phi$, you could deduce similar trouble from your definition plus the assumption that, for every $y$ in the domain group, there is at least one $y'$ in the codomain group with $(y,y')\in\phi$.

If you want to define homomorphic relations, you might reasonably define them to be subgroups of the direct product of the domain and codomain groups.

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Yes, good point; most algebraic structures prove that every left-total morphism is in fact a function. This doesn't apply to all structures, though - for instance, partially ordered sets. So maybe, the definition of a morphism as a function is appropriate in algebra, but less appropriate in, say, order theory, or graph theory. –  user18921 May 2 '13 at 1:37
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There's at least one example where homomorphisms aren't assumed to be functions.

This source defines "a linear relation" as a relation that is a linear subspace of the direct product of its domain and codomain. This generalizes the notion of a linear mapping. Furthermore, the composition of linear relations is itself a linear relation, and there is a category of vector spaces and linear relations.

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By definition groups are sets with additional structure. It makes sense to define homomorphisms as set functions which preserve this structure. We have the same situastion with topological spaces. If we wanted to, we could define continuous maps via the lattice of open sets of a topological space, and this is useful in some contexts, but not in others.

One of the benefits of using functions is that it is easy to "stack" structures. For example, we can put a group structure on a topological space, and demand that continuous maps are also group homomorphisms. This gives us the notion of a topological group, and it was trivially easy to define it because we use set functions. If we insisted on using the lattice of open sets for topological spaces and your relatio $\phi$, simply defining such a structure would probably be a tedious and awkward affair.

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I guess what I was trying to say is, we can define homomorphisms as relations that preserve this structure. But this is uncommon (indeed, I just found my first example, and posted it as an answer.) Hence the question. –  user18921 Jan 30 '13 at 18:07
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Your statement is not correct. People in the 1940s invented a very general framework for that which you asked. It is called category theory. Roughly speaking a category consists of objects and morphisms between them that behave like functions in a certain sense.

But there are lots of categories where the morphisms aren't functions. For example each partially ordered set is a category. The elements if the set are the objects of the category. And a morphism between $x$ and $y$ exists whenever $x\leq y$.

Monoids are categories with one object. In this case the elements of the monoid are identified with the morphisms.

There are lots of other examples of that kind.

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