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homomorphism between algebraic systems
Let $(A,O),(B,O)$ be two algebraic systems with operator set $O$. Given operators $\omega_{A}$ on $A$ and $\omega_{B}$ on $B$, with $\omega\in O$ and $n=$ arity of $\omega$, a function $f:A\to B$ is said to be compatible with $\omega$ if
$f(\omega_{A}(a_{1},\ldots,a_{n}))=\omega_{B}(f(a_{1}),\ldots,f(a_{n})).$ 
Dropping the subscript, we now simply identify $\omega\in O$ as an operator for both algebras $A$ and $B$. If a function $f:A\to B$ is compatible with every operator $\omega\in O$, then we say that $f$ is a homomorphism from $A$ to $B$. If $O$ contains a constant operator $\omega$ such that $a\in A$ and $b\in B$ are two constants assigned by $\omega$, then any homomorphism $f$ from $A$ to $B$ maps $a$ to $b$.
Examples.
1. When $O$ is the empty set, any function from $A$ to $B$ is a homomorphism.
2. When $O$ is a singleton consisting of a constant operator, a homomorphism is then a function $f$ from one pointed set $(A,p)$ to another $(B,q)$, such that $f(p)=q$.
3. A homomorphism defined in any one of the well known algebraic systems, such as groups, modules, rings, and lattices is consistent with the more general definition given here. The essential thing to remember is that a homomorphism preserves constants, so that between two rings with 1, both the additive identity 0 and the multiplicative identity 1 are preserved by this homomorphism. Similarly, a homomorphism between two bounded lattices is called a $\{0,1\}$lattice homomorphism because it preserves both 0 and 1, the bottom and top elements of the lattices.
Remarks.

Like the familiar algebras, once a homomorphism is defined, special types of homomorphisms can now be named:

a homomorphism that is onetoone is a monomorphism;

an onto homomorphism is an epimorphism;

an isomorphism is both a monomorphism and an epimorphism;

a homomorphism such that its codomain is its domain is called an endomorphism;

finally, an automorphism is an endomorphism that is also an isomorphism.


All trivial algebraic systems (of the same type) are isomorphic.

If $f:A\to B$ is a homomorphism, then the image $f(A)$ is a subalgebra of $B$. If $\omega_{B}$ is an $n$ary operator on $B$, and $c_{1},\ldots,c_{n}\in f(A)$, then $\omega_{B}(c_{1},\ldots,c_{n})=\omega_{B}(f(a_{1}),\ldots,f(a_{n}))=f(\omega_{% A}(a_{1},\ldots,a_{n}))\in f(A)$. $f(A)$ is sometimes called the homomorphic image of $f$ in $B$ to emphasize the fact that $f$ is a homomorphism.
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