Why is the strong definition of L-homomorphism the preferred one?

Question

Some logic textbooks seem to define homomorphisms to have the clause for relation symbols be a biconditional:

A function $h:\mathfrak{A}\rightarrow\mathfrak{B}$ is homomorphism iff

$\vdots$

$$R^{\mathfrak{A}}(\overline{a})\iff R^{\mathfrak{B}}(h(\overline{a}))$$

The weaker version is: $$R^{\mathfrak{A}}(\overline{a})\implies R^{\mathfrak{B}}(h(\overline{a}))$$

What reasons to prefer the weak vs strong, and vice-versa?

Answer 1

In my correct opinion, the "weak" version is the correct definition of homomorphism.

To be clear on terminology:

• A homomorphism is a map $f\colon A\to B$ such that (0) $f(c^A) = c^B$ for all constant symbols, (1) $f(h^A(a_1,\dots,a_n)) = h^B(f(a_1),\dots,f(a_n))$ for all $n$-ary function symbols $h$, and (2) $R^A(a_1,\dots,a_n) \implies R^B(a_1,\dots,a_n)$ for all $n$-ary relation symbols $R$.
• A strong homomorphism is the same as a homomorphism, but replacing $\implies$ with $\iff$ in condition (2).
• An embedding is an injective strong homomorphism, but it is additionally required to be injective.

In the comments, you asked about $=$. The equals sign $=$ is not a relation symbol, but rather a logical symbol that is always interpreted as true equality in $L$-structures. Note that every function $f$ satisfies the analogue for $=$ of condition (2) in the definition of homomorphism: $a = b \implies f(a) = f(b)$. The analogue for $=$ of condition (2) in the definition of strong homomorphism is exactly injectivity: $a = b \iff f(a) = f(b)$.

Here are some reasons to prefer homomorphisms to strong homomorphisms:

1. Expanding on the discussion above, they're more natural. Under the weak definition, homomorphisms can be characterized as those maps which preserve all atomic formulas, in the sense that if $f\colon A\to B$ and $A\models \varphi(a)$, where $\varphi$ is atomic and $a\in A^x$, then $B\models \varphi(f(x))$. Analogously, embeddings (injective strong homomorphisms) can be characterized as those maps which preserve all atomic and negated atomic formulas - or equivalently, which preserve and reflect all atomic formulas, in the sense that if $f\colon A\to B$, then $A\models \varphi(a)$ if and only if $B\models \varphi(f(a))$ where $\varphi$ is atomic and $a\in A^x$. To characterize strong homomorphisms in this way, one needs to distinguish atomic formulas involving $=$ from those involving relation symbols: strong homomorphisms are those maps which preserve atomic formulas of the form $t = s$, where $t$ and $s$ are terms, and which preserve and reflect atomic formulas of the form $R(t_1,\dots,t_n)$ where the $t_i$ are terms.

2. As bof notes in the comments, homomorphisms are involved in the semantic characterization of the important positive existential fragment of first-order logic. A formula $\varphi$ is $T$-equivalent to a positive extistential formula (one built up from atomic formulas by $\land$, $\lor$, and $\exists$, without $\lnot$ or $\forall$) if and only if it is preserved by all homomorphisms between models of $T$. By contrast, strong homomorphisms preserve exactly those formulas which are equivalent to formulas built up from atomic formulas by $\land$, $\lor$, and $\exists$, plus the additional case that $\lnot$ can appear directly in front of a relation symbol, but nowhere else. This is a much less natural fragment.

3. With the proper definition of homomorphism, the category of $L$-structures has all small limits and colimits. If we use strong homomorphisms, the resulting category fails to even have an initial object, a terminal object, or binary products (when there are relation symbols in $L$).

I know of no reasons whatsoever to prefer the strong version. If anyone wants to make the case, I'd love to hear it!