You're right that these definitions are confusing. Is it possible that they've been translated from another language? It seems to me that Definition 1 is defining what I would call an embedding (though missing the condition of reflecting relations - unless that's contained in Definition 2). Definition 2 is defining homomorphism, though imprecisely, as I noted in my comment, and Definition 3 is defining isomorphism. Here are the proper/standard definitions:
Let $A$ and $B$ be $L$-structures, and let $h\colon A\to B$ be a function.
$h$ is a homomorphism if:
- $h$ preserves constants: For every constant symbol $c$ in $L$, $h(c^A) = c^B$.
- $h$ preserves functions: For every $n$-ary function symbol $f$ in $L$, and for all $a_1,\dots,a_n\in A$, $h(f^A(a_1,\dots,a_n)) = f^B(h(a_1),\dots,h(a_n))$.
- $h$ preserves relations: For every $n$-ary relation symbol $R$ in $L$, and for all $a_1,\dots,a_n\in A$, if $(a_1,\dots,a_n)\in R^A$, then $(h(a_1),\dots,h(a_n))\in R^B$.
$h$ is an embedding if:
- $h$ is a homomorphism.
- $h$ is injective.
- $h$ reflects relations: For every $n$-ary relation symbol $R$ in $L$, and for all $a_1,\dots,a_n\in A$, if $(h(a_1),\dots,h(a_n))\in R^B$, then $(a_1,\dots,a_n)\in R^A$.
[Note that reflecting relations is the dual of preserving relations. Note also that injectivity is equivalent to reflecting $=$; for all $a,b\in A$, if $h(a) = h(b)$, then $a = b$. Of course, all functions preserve $=$.]
$h$ is an isomorphism if:
- $h$ is an embedding.
- $h$ is surjective.
Equivalently, $h$ is an isomorphism if it is an invertible homomorphism (whose inverse is also a homomorphism).
$A$ and $B$ are isomorphic if there is an isomorphism $h\colon A\to B$.
Note that if $h$ is an embedding, then $A$ is isomorphic to a substructure of $B$ (namely the image of $h$).
In the comments you ask:
I have a complementary worry: in the same lecture notes the definitions above are followed by this theorem: If h is an isomorphism of M onto M', s is an assignment in M and s' is an assignment in M' such that for every variable x, [x]M',s'= h([x]M,s), then s satisfies a formula A iff s' satisfies a formula A. Would this theorem hold also for a homomorphism of M into M'?
No: Consider the language $L = \{\leq\}$, and let $h$ be the inclusion of the linear order $(\mathbb{N},\leq)$ into the linear order $(\mathbb{Z},\leq)$. Then $h$ is a homomorphism (even an embedding), but it does not preserve truth of the formula $\varphi(x) = \forall y\, x\leq y$. Indeed, $\mathbb{N}\models \varphi(0)$, but $\mathbb{Z}\not\models \varphi(h(0))$.
A map which preserves and reflects truth of formulas is called an elementary embedding, and the theorem in your notes says that every isomorphism is an elementary embedding (in the language of your notes, every isomorphism of $M$ onto $M'$ is an elementary embedding, but the example above shows that an "isomorphism of $M$ into $M'$", i.e. an embedding, need not be an elementary embedding).
However, there are elementary embeddings which are not isomorphisms; for example, the inclusion $(\mathbb{Q},\leq)\to (\mathbb{R},\leq)$ is an elementary embedding which is not surjective.
To sum up:
homomorphisms $\subseteq$ embeddings $\subseteq$ elementary embeddings $\subseteq$ isomorphisms
homomorphisms preserve atomic formulas, embeddings preserve and reflect atomic formulas, elementary embeddings preserve and reflect all formulas, and isomorphisms do all of the above and are bijective. In between homomorphisms and elementary embeddings, you can have fun defining many other classes of maps by picking a class of formulas and asking that the maps preserve/reflect/both the formulas in that class.
