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Speaking in the strictest sense, does a homomorphism map between the carriers of two algebras, or between the two algebraic structures? To explicate what I mean suppose we have two algebras (j, J) and (k, K), which have carriers (sets) j and k respectively, and have binary operations J and K. Does a homomorphism H come as a function H:j->k (maps the carriers) such that for all x, y in g, H (J (x, y))=K ( H(x), H(y)), or does it come as a function H:(j, J)->(k, K)? Or do the two definitions come as equivalent in some sense? Say we have the following two structures:

A  1  2
1  1  2
2  2  2

B  a  b  c
a  a  b  b
b  b  b  b
c  b  c  c

and the homomorphism L:{1, 2}->{a, b, c}, specified by L:1->a, 2->b. Also, suppose we have the same sets, but now with tables as follows:

A  1  2
1  1  2
2  2  2

B  a  b  c
a  a  b  c
b  b  b  b
c  b  c  c

And say we have function M:{1, 2}->{a, b, c}, M:1->a, 2->b. Does function M come as different than function L above, since they map between different structured sets (if they do such, of course), or does L basically come as the same function as M since the have the same domain, the same co-domain, and the same set of ordered pairs (1, a), (2, b)?

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1 Answer 1

Well, if you're implementing your algebraic structures concretely as sets, then every distinct homomorphism specifies a unique map of the underlying carrier sets — so long as you fix the domain and codomain. No additional data needs to be specified, so in that sense a homomorphism is nothing more than a map of sets satisfying certain properties. In the language of category theory we say that there's a faithful functor from the category of your algebraic structures to the category of sets. The only problem is that different algebraic structures may be mapped to the same underlying set. In that case it is possible that two homomorphisms between two different pairs of algebraic structures get mapped to the same map of carrier sets. To fix this we might say that a homomorphism $\varphi$ between two algebraic structures $\mathcal{A}$ and $\mathcal{B}$ is really a triple $(f, \mathcal{A}, \mathcal{B})$, where $f : A \to B$ is a map of sets and $A$ and $B$ are the carrier sets of $\mathcal{A}$ and $\mathcal{B}$, respectively.

Having said all that, I don't think there's any benefit to thinking this way. There is rarely any risk of confusion when we use the same symbol to refer to both an algebraic structure and its carrier set, nor when we use the same symbol to refer to both a homomorphism and its underlying map of sets. It is sometimes possible to dispense with carrier sets entirely while retaining the concept of homomorphisms — how then should we interpret them?

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If you spent time answering the question, what about spending a second voting up the said question? –  David Kohler May 30 '11 at 21:37
    
So, if I understand your definition correctly, a homomorphism between the first two structures goes ({(1, a), (2, b)}, ({1, 2}, {(1, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 2)}), ({a, b, c}, {(a, a, a), (a, b, b), (a, c, b), (b, a, b), (b, b, b), (b, c, b), (c, a, b), (c, b, c), (c, c, c)}) with {(1, a), (2, b)} as the map between sets, ({1, 2}, {(1, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 2)}) the first structure, and {(a, a, a), (a, b, b), (a, c, b), (b, a, b), (b, b, b), (b, c, b), (c, a, b), (c, b, c), (c, c, c)} the second structure, correct? The carrier set definition would go {(1, a), (2, b)}. –  Doug Spoonwood May 30 '11 at 21:57
1  
@Doug: I don't think it's a very interesting question. Details like this are not relevant in the practice of mathematics. Notice that the result of such formalism is highly unwieldy — you've just produced a very long string of symbols and I can't be bothered to go check that it is an example of what I said. As far as I can tell you've decided to identify your algebraic structures by the implementation of their binary operations — why? It's fine for your examples, but it breaks when the algebraic structure being studied has only partially-defined operations or has more than operation, etc. –  Zhen Lin May 31 '11 at 7:17
    
I don't see how such implementation breaks down when you have more than operation, you just have more to specify. If a homomorphism maps between structures of the same type, and the "carrier set" definition isn't equivalent to the definition where we have a map between structures, then it needs to map between not just between the carrier sets such that homomorphic equation gets satisfied, but also between their operations. So, of course, in such a case, there exists more to do for a complete verification. I think the two definitions equivalent, but I don't know for certain. –  Doug Spoonwood Jun 1 '11 at 3:32
    
@David, since you have so much time and so many votes, why not upvote it yourself? What a world... –  The Chaz 2.0 Feb 14 '12 at 3:51

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