I Terminology involving "homomorphic"

[Stephen Tashi]

I notice several current Wikipedia articles on "homomorphism" avoid using the adjective "homomorphic". (e.g. https://en.wikipedia.org/wiki/Group_homomorphism , https://en.wikipedia.org/wiki/Homomorphism).

Of course the problem with saying "A and B are homomorphic" is that there can be a homomorphism from A to B but no homomorphism from B to A may exist. So it would seem best to only to use the terminology "A is homomorphic to B", which suggests the possibility of an asymmetrical relation. But what do we want that terminology to mean? The natural definition (in my opinion) is that it should mean there is a homomorphism from B to A. However, I've heard people use it to mean that there is a homomorphism from A to B.

 

[martinbn]

Why would you want to use the word homomorphic?!

 

[member 587159]

Why would you want to use the word homomorphic?!

For example, if $A$ is a $p$-group, and $A$ is homomorphic to $B$, then there is a subgroup of $B$ such that $B$ is a $p$-group (this subgroup is the image of the homomorphism). I do agree that the terminology is ambiguous, and should therefore be avoided.

 

[Erland]

So if "A is homomorphic to B" means "There is a homomorpism from A to B", then this relation is not symmetric. For example (as fields), $\mathbb Q$ is homomorphic to $\mathbb R$ (take the inclusion map as the homomorphism), but not the other way round. This is a difference from the case with "isomorphic" and "isomorphism".

 

[Stephen Tashi]

For example (as fields), $\mathbb Q$ is homomorphic to $\mathbb R$ (take the inclusion map as the homomorphism), but not the other way round.

Couldn't we map the all the reals to the identity element? The identity element is a subset of $\mathbb Q$, so that would be a homomorphism from $\mathbb R$ to $\mathbb Q$.

 

[I like Serena]

Isn't any group homomorphic to any other group if we pick the trivial map?
That would make the concept meaningless.

 

[WWGD]

Maybe we should use the expression "there is a homomorphism of A _into_ B " , as opposed to a homomorphism of A _onto_ B.
Similarly, there may be homeomorphisms _into_ ( meaning embeddings) without homeomorphisms _onto_. I EDIT :AFAIK, a bijective homomorphism is an isomorphism , though the equivalent is not true for hemeomorphisms.

 

[Erland]

Couldn't we map the all the reals to the identity element? The identity element is a subset of $\mathbb Q$, so that would be a homomorphism from $\mathbb R$ to $\mathbb Q$.

Yes, you are right!

 

[Erland]

Isn't any group homomorphic to any other group if we pick the trivial map?
That would make the concept meaningless.

Exactly!

 

[Erland]

Maybe we should use the expression "there is a homomorphism of A _into_ B " , as opposed to a homomorphism of A _onto_ B.
Similarly, there may be homeomorphisms _into_ ( meaning embeddings) without homeomorphisms _onto_. I EDIT :AFAIK, a bijective homomorphism is an isomorphism , though the equivalent is not true for hemeomorphisms.

A homomorphism from A onto B is called an epimorphism.

 

[WWGD]

A homomorphism from A onto B is called an epimorphism.

Still, unfortunately many are not precise in this usage. Isn't this called an embedding in this " category" (Algebraic objects: rings, groups, etc.) so that A embeds in B?

 

[Erland]

Not in general. To call a map an embedding it must be one-to-one (injective). In algebra, an embedding is also called a monomorphism.
https://en.wikipedia.org/wiki/Embedding

 

[WWGD]

Not in general. To call a map an embedding it must be one-to-one (injective). In algebra, an embedding is also called a monomorphism.
https://en.wikipedia.org/wiki/Embedding

Yes, sorry, this is what I was assuming, and it is a stronger condition, meaning there is a "copy" of A living in B , much more stringent than just having a homomorphism..

 

[Stephen Tashi]

On second thought, it there is technicality in the definition of a "field homomorphism" that would disqualify mapping all of $Q$ to the multiplicative identity as a homomorphism between fields. http://planetmath.org/fieldhomomorphism

 

[I like Serena]

On second thought, it there is technicality in the definition of a "field homomorphism" that would disqualify mapping all of $Q$ to the multiplicative identity as a homomorphism between fields. http://planetmath.org/fieldhomomorphism

Ah yes, so a field homomorphism from $\mathbb Q \to \mathbb R$ has to be the canonical map, doesn't it?
And more generally, a field homomorphism has to be from the 'smaller' field to the 'larger' field, effectively eliminating the ambiguity of direction.

 

[WWGD]

Just curious, in the finite case, one obstruction to the existence of non-trivial ( non-identity) homomorphisms f
between finite rings A,B is that f(A) must be a subgroup/ring/etc. and then there are divisibility issues (right)?
Still, what are the obstructions for the existence of homomorphisms between infinite rings? I remember a similar
question in topology about the existence of continuous maps, and the answer was that of topological invariants.
(e.g., there cannot be a continuous , non-constant map between a connected space and a disconnected space).
What are the obstructions between infinite algebraic objects?

 

[I like Serena]

Just curious, in the finite case, one obstruction to the existence of non-trivial ( non-identity) homomorphisms f
between finite rings A,B is that f(A) must be a subgroup/ring/etc. and then there are divisibility issues (right)?
Still, what are the obstructions for the existence of homomorphisms between infinite rings? I remember a similar
question in topology about the existence of continuous maps, and the answer was that of topological invariants.
(e.g., there cannot be a continuous , non-constant map between a connected space and a disconnected space).
What are the obstructions between infinite algebraic objects?

Just like a field homomorphism, a ring homomorphism cannot be trivial - unless we have the zero ring with 1 element (0=1) as codomain.

 

[mathwonk]

all i can think of say for commutative rings with identity, and which take identity to identity, is that some quotient of the source must embed in the target. in particular for fields, since they have no non trivial quotients, i.e. no proper ideals, the source itself must embed in the target. the term is not much needed in general since we already have terms for the existence of surjective and injective homomorphisms, i.e. (isomorphism with) a quotient or subobject.

 

[WWGD]

Just like a field homomorphism, a ring homomorphism cannot be trivial - unless we have the zero ring with 1 element (0=1) as codomain.

Still, as an example of a non-trivial one between a ring without unity and one with unity, we have the inclusion of $2\mathbb Z \rightarrow \mathbb Z$.

 

[WWGD]

Just like a field homomorphism, a ring homomorphism cannot be trivial - unless we have the zero ring with 1 element (0=1) as codomain.

My bad, I am used to dealing with simpler objects like vector spaces , with a single structure ( addition of vectors) where we may map into ( additive) 0.