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Like subalgebras of partial algebras, there are also three ways to define homomorphisms between partial algebras. Similar to the definition of homomorphisms between algebras, a homomorphism $\phi: \boldsymbol{A}\to \boldsymbol{B}$ between two partial algebras of type $\tau$ is a function from $A$ to
$B$ that satisfies the equation \begin{equation} \phi(f_{\boldsymbol{A}}(a_1,\ldots, a_n))= f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n)) \end{equation}for every $n$ -ary function symbol $f\in \tau$ . However, because $f_{\boldsymbol{A}}$ and $f_{\boldsymbol{B}}$ are not everywhere defined in their respective domains, care must be taken as to what the equation means.
- $\phi$ is a homomorphism if, given that $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is defined, so is $f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n))$ , and equation (1) is satisifed.
- $\phi$ is a full homomorphism if it is a homomorphism and, given that $f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ is defined and in $\phi(A)$ , for $b_i\in \phi(A)$ , there exist $a_i\in A$ with $b_i=\phi(a_i)$ , such that $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is defined.
- $\phi$ is a strong homomorphism if it is a homomorphism and, given that $f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n))$ is defined, so is $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ .
We have the following implications:
strong homomorphism $\rightarrow$ full homomorphism $\rightarrow$ homomorphism.
For example, field homomorphisms are strong homomorphisms.
Homomorphisms preserve constants: for each constant symbol $f$ in $\tau$ , $\phi(f_{\boldsymbol{A}}) = f_{\boldsymbol{B}}$ . In fact, when restricted to constants, $\phi$ is a bijection between constants of $\boldsymbol{A}$ and constants of $\boldsymbol{B}$ .
When $\boldsymbol{A}$ is an algebra (all partial operations are total), a homomorphism from $\boldsymbol{A}$ is always strong, so that all three notions of homomorphisms coincide.
An isomorphism is a bijective homomorphism $\phi: \boldsymbol{A}\to \boldsymbol{B}$ such that its inverse $\phi^{-1}: \boldsymbol{B}\to \boldsymbol{A}$ is also a homomorphism. An embedding is an injective homomorphism. Isomorphisms and full embeddings are strong.
The various types of homomorphisms and the various types of subalgebras are related. Suppose $\boldsymbol{A}$ and $\boldsymbol{B}$ are partial algebras of type $\tau$ . Let $\phi:A\to B$ be a function, and $C=\phi(A)$ . For each $n$ -ary function symbol $f\in \tau$ , define $n$ -ary partial operation $f_{\boldsymbol{C}}$ on $C$ as follows:
for $b_1,\ldots, b_n\in C$ , $f_{\boldsymbol{C}}(b_1,\ldots, b_n)$ is defined iff the set $$D:=\lbrace (a_1,\ldots, a_n)\in A^n \mid \phi(a_i)=b_i\rbrace \cap \operatorname{dom}(f_{\boldsymbol{A}})$$ is non-empty, where $\operatorname{dom}(f_{\boldsymbol{A}})$ is the domain of definition of $f_{\boldsymbol{A}}$ , and when this is the case, $f_{\boldsymbol{C}}(b_1,\ldots, b_n):=\phi(f_{\boldsymbol{A}}(a_1,\ldots, a_n))$ , for some $(a_1,\ldots, a_n)\in D$ .
If $\phi$ preserves constants (if any), and $f_C$ is non-empty for each $f\in \tau$ then $\boldsymbol{C}$ is a partial algebra of type $\tau$ .
Fix an arbitrary $n$ -ary symbol $f\in \tau$ . The following are the basic properties of $\boldsymbol{C}$ :
Proposition 1 $\phi$ is a homomorphism iff $\boldsymbol{C}$ is a weak subalgebra of $\boldsymbol{B}$ .
Proof. Suppose first that $\phi$ is a homomorphism. If $n=0$ , then $f_{\boldsymbol{A}} \in A$ , and $f_{\boldsymbol{B}} = \phi(f_{\boldsymbol{A}}) \in C$ . If $n>0$ , then for some $a_1,\ldots, a_n\in A$ , $f_{\boldsymbol{A}}(a_1, \ldots, a_n)$ is defined, and consequently $f_{\boldsymbol{B}}(\phi(a_1), \ldots, \phi(a_n))$ is defined, and is equal to $\phi(f_{\boldsymbol{A}}(a_1, \ldots, a_n)) \in C$ . By the definition for $f_{\boldsymbol{C}}$ above, $f_{\boldsymbol{C}}(\phi(a_1), \ldots, \phi(a_n)):=\phi(f_{\boldsymbol{A}}(a_1, \ldots, a_n))$ . This shows that $\boldsymbol{C}$ is a $\tau$ -algebra.
To furthermore show that $\boldsymbol{C}$ is a weak subalgebra of $\boldsymbol{B}$ , assume $f_{\boldsymbol{C}}(b_1,\ldots, b_n)$ is defined. Then there are $a_1,\ldots, a_n\in A$ with $b_i=\phi(a_i)$ such that $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is defined. Since $\phi$ is a homomorphism, $f_{\boldsymbol{B}}(\phi(a_1),\ldots,\phi(a_n))$ , and hence $f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ , is defined. Furthermore, $f_C(b_1,\ldots, b_n)=\phi(f_{\boldsymbol{A}}(a_1,\ldots, a_n))=f_{\boldsymbol{B}}(\phi(a_1),\ldots,\phi(a_n))=f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ . This shows that $\boldsymbol{C}$ is weak.
On the other hand, suppose now that $\boldsymbol{C}$ is a weak subalgebra of $\boldsymbol{B}$ . Suppose $a_1,\ldots, a_n\in A$ and $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is defined. Let $b_i=\phi(a_i)\in C$ . Then, by the definition of $f_{\boldsymbol{C}}$ , $f_{\boldsymbol{C}}(b_1,\ldots, b_n)$ is defined and is equal to $\phi(f_{\boldsymbol{A}}(a_1,\ldots, a_n))$ . Since $\boldsymbol{C}$ is weak, $f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ is defined and is equal to $f_{\boldsymbol{C}}(b_1,\ldots, b_n)$ . As a result, $\phi(f_{\boldsymbol{A}}(a_1,\ldots, a_n))=f_{\boldsymbol{C}}(b_1,\ldots, b_n)=f_{\boldsymbol{B}}(b_1,\ldots, b_n)= f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n))$
. Hence $\phi$ is a homomorphism. 
Proposition 2 $\phi$ is a full homomorphism iff $\boldsymbol{C}$ is a relative subalgebra of $\boldsymbol{B}$ .
Proof. Suppose first that $\phi$ is full. Since $\phi$ is a homomorphism, $\boldsymbol{C}$ is weak. Suppose $b_1,\ldots, b_n\in C$ such that $f_{\boldsymbol{A}}(b_1,\ldots, b_n)$ is defined and is in $C$ . Since $\phi$ is full, there are $a_i\in A$ such that $b_i = \phi(a_i)$ and $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is defined, and $\phi(f_{\boldsymbol{A}}(a_1,\ldots,a_n))=f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n))=f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ , so that $f_{\boldsymbol{B}}(b_1,\ldots,b_n)$ is defined and thus $\boldsymbol{C}$ is a relative subalgebra of $\boldsymbol{B}$ .
Conversely, suppose that $\boldsymbol{C}$ is a relative subalgebra of $\boldsymbol{B}$ . Then $\boldsymbol{C}$ is a weak subalgebra of $\boldsymbol{B}$ and $\phi$ is a homomorphism. To show that $\phi$ is full, suppose that $b_i\in C$ such that $f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ is defined in $C$ . Then $f_{\boldsymbol{C}}(b_1,\ldots, b_n)$ is defined in $C$ and is equal to $f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ . This means that there are $a_i\in A$ such that $b_i=\phi(a_i)$ , and $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is defined, showing that $f_{\boldsymbol{A}}$ is full. 
Proposition 3 $\phi$ is a strong homomorphism iff $\boldsymbol{C}$ is a subalgebra of $\boldsymbol{B}$ .
Proof. Suppose first that $\phi$ is strong. Since $\phi$ is full, $\boldsymbol{C}$ is a relative subalgebra of $\boldsymbol{B}$ . Suppose now that for $b_i\in C$ , $f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ is defined. Since $b_i=\phi(a_i)$ for some $a_i \in A$ , and since $\phi$ is strong, $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is defined. This means that $f_{\boldsymbol{B}}(b_1,\ldots, b_n)= f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n))=\phi(f_{\boldsymbol{A}}(a_1,\ldots, a_n))$ , which is in $C$ . So $\boldsymbol{C}$ is a subalgebra of $\boldsymbol{B}$ .
Going the other direction, suppose now that $\boldsymbol{C}$ is a subalgebra of $\boldsymbol{B}$ . Since $\boldsymbol{C}$ is a relative subalgebra of $\boldsymbol{B}$ , $\phi$ is full. To show that $\phi$ is strong, suppose $f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n))$ is defined. Then $f_{\boldsymbol{C}}(\phi(a_1),\ldots, \phi(a_n))$ is defined and is equal to $f_{\boldsymbol{B}}(\phi(a_1),\ldots, \phi(a_n))$ . By definition of $f_{\boldsymbol{C}}$ , $f_{\boldsymbol{A}}(a_1,\ldots, a_n)$ is therefore defined. So $\phi$ is strong. 
Definition. Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be partial algebras of type $\tau$ . If $\phi:\boldsymbol{A}\to \boldsymbol{B}$ is a homomorphism, then $\boldsymbol{C}$ , as defined above, is a partial algebra of type $\tau$ , and is called the homomorphic image of $A$ via $\phi$ , and is sometimes written $\phi(\boldsymbol{A})$ .
- 1
- G. Grätzer: Universal Algebra, 2nd Edition, Springer, New York (1978).
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