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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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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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    @David, since you have so much time and so many votes, why not upvote it yourself? What a world...2012-02-14