We say that $f: A \to B$ is a homomorphism iff it preserves the operations and relations of the structure: $f(o^A(\bar{a})) = o^B(f(\bar{a}))$ where $a ∈ A^{ar(o)}$, $o$ any function symbol from the signature, and similarly $R^A(\bar{a}) \implies R^B(f(\bar{a}))$.

Definition I know: We say that $f: A \to B$ is a strong homomorphism iff it's a homomorphism and it also reflects the relations: $R^B(f(\bar{a})) \implies R^A(\bar{a})$.

However there is another definition which can be found for example at http://en.wikipedia.org/wiki/Structure_%28mathematical_logic%29#Homomorphisms: $f: A \to B$ is a strong homomorphism iff it's a homomorphism and $R^B(\bar{b}) \implies ∃\bar{a}: R^A(\bar{a}) ∧ f(\bar{a}) = \bar{b}$.

If we call an element of a structure related iff there is some relation of the structure such that the element is related with some other elements via this relation, then the second definition is just the first one plus requirement that the image of $f$ contains all related elements of $B$ which seems to me a bit unnatural for general definition. Also with the second definition it doesn't hold that embeddings are precisely injective strong homomorphisms (which Wikipedia also states).

So my questions: What is the second definition good for? Is any of those definitions “more standard”? Am I missing something? Thank you.

Update: The second definition can be found also on https://www.encyclopediaofmath.org/index.php/Homomorphism. It seems that both occurrences of the definition come from Chang, Keisler: Model theory where it occurs only in two exercises (5.2.22, 5.2.23) and only in context when the homomorphisms are onto (through whole book they use only homomorphisms onto).


I can't answer your first question, but I can assure you that the definition you know (a strong homomorphism preserves and reflects relations) is more standard!

A small correction: You say "the second definition is just the first one plus requirement that the image of $f$ contains all related elements of $B$". It does require this, but it's also much weaker in its reflection requirement, only requiring the some preimage of related elements in $B$ be related in $A$, not all.

I've never seen the definition from Wikipedia before, and I agree that it seems very unnatural. I suspect it might be a typo, but I'm not a expert in general relational structures.

  • $\begingroup$ You're right about the weaker reflection in the second definition. Now it seems even more unnatural. $\endgroup$ – user87690 Aug 15 '13 at 6:51

I can tell you one thing (either definition of) strong homomorphisms are good for - showing that surjections are orthogonal (see http://ncatlab.org/nlab/show/orthogonality) to injections.

Just to be clear, let's call a homomorphism strongly reflective if

$R^B(f(\overline{a}))\Rightarrow R^A(\overline{a})$,

and weakly reflective if

$R^B(f(\overline{a}))\Rightarrow \exists a':R^A(\overline{a'})\wedge f(\overline{a'})=f(\overline{a})$

(so this is just like the Wikipedia definition, without requiring surjectivity, which I agree seems unnatural). Then you can show that if $e$ and $m$ are surjective and injective weakly reflective homomorphisms respectively, they are orthogonal in the following sense - whenever $ve=mu$, for any other weakly reflective homomorphisms, there exists a (unique) weakly reflective homomorphism $d$ with $de=u$ and $md=v$. One important consequence of this is that a weakly reflective homomorphism that is both injective and surjective must in fact be an isomorphism. This can fail for arbitrary homomorphisms - the identity map from $\mathbb{R}$ with the discrete order (i.e. $x\leq y\Leftrightarrow x=y$) to $\mathbb{R}$ with its usual linear order is an injective surjective homomorphism (of structures with a single relation $\leq$) but certainly not an (order) isomorphism.

Of course, the same orthogonality result holds also for strongly reflective homomorphisms, but you may not want to throw away any homomorphisms that you don't have to. For example, the (quite useful!) coordinate projections on a product $\mathbb{P}\times\mathbb{Q}$ of partially ordered sets $\mathbb{P}$ and $\mathbb{Q}$ are weakly reflective but not strongly reflective.


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