PlanetMath: first isomorphism theorem

Let $\Sigma$ be a fixed signature, and $\mathfrak{A}$ and $\mathfrak{B}$ structures for $\Sigma$ . If $f \colon \mathfrak{A}\to \mathfrak{B}$ is a homomorphism, then there is a unique bimorphism $\phi\colon\mathfrak{A}/\!\ker(f) \to {\mathrm{im}}(f)$ such that for all $a \in \mathfrak{A}$ , $\phi([\![a]\!]) = f(a)$ . Furthermore, if

has the additional property that for each $n \in \mathbb{N}$ and each

-ary relation symbol

of $\Sigma$ ,

Proof. Since the homomorphic image of a $\Sigma$ -structure is also a $\Sigma$ -structure, we may assume that ${\mathrm{im}}(f) = \mathfrak{B}$ .

Let $\sim\ = \ker(f)$ . Define a bimorphism $\phi \colon \mathfrak{A}/\!\!\sim \to \mathfrak{B}: [\![a]\!] \mapsto f(a)$ . To verify that $\phi$ is well defined, let $a \sim a'$ . Then $\phi([\![a]\!]) = f(a) = f(a') = \phi([\![a']\!])$ . To show that $\phi$ is injective, suppose $\phi([\![a]\!]) = \phi([\![a']\!])$ . Then , so $a \sim a'$ . Hence $[\![a]\!] = [\![a']\!]$ . To show that $\phi$ is a homomorphism, observe that for any constant symbol of $\Sigma$ we have $\phi([\![c^\mathfrak{A}]\!]) = f(c^\mathfrak{A}) = c^\mathfrak{B}$ . For each $n \in \mathbb{N}$ and each -ary function symbol of $\Sigma$ ,

$\displaystyle \phi(F^{\mathfrak{A}/\!\sim}([\![a_1]\!], \ldots, [\![a_n]\!]))$	$\displaystyle = \phi([\![F^\mathfrak{A}(a_1, \ldots, a_n)]\!])$
	$\displaystyle = f(F^\mathfrak{A}(a_1, \ldots, a_n))$
	$\displaystyle = F^\mathfrak{B}(f(a_1), \ldots, f(a_n))$
	$\displaystyle = F^\mathfrak{B}(\phi([\![a_1]\!], \ldots, \phi([\![a_n]\!])).$

For each $n \in \mathbb{N}$ and each

-ary relation symbol

of $\Sigma$ ,

$\displaystyle R^{\mathfrak{A}/\!\sim}([\![a_1]\!], \ldots, [\![a_n]\!])$	$\displaystyle \Implies R^\mathfrak{A}(a_1, \ldots, a_n)$
	$\displaystyle \Implies R^\mathfrak{B}(f(a_1), \ldots, f(a_n))$
	$\displaystyle \Implies R^\mathfrak{B}(\phi([\![a_1]\!], \ldots, \phi([\![a_n]\!])).$

Thus $\phi$ is a bimorphism.

Now suppose has the additional property mentioned in the statement of the theorem. Then

$\displaystyle R^\mathfrak{B}(\phi([\![a_1]\!]), \ldots, \phi([\![a_n]\!]))$	$\displaystyle \Implies R^\mathfrak{B}(f(a_1), \ldots, f(a_n))$
	$\displaystyle \Implies \exists a_i'[ a_i \sim a_i' \land R^\mathfrak{A}(a_1', \ldots, a_n')]$
	$\displaystyle \Implies R^{\mathfrak{A}/\!\sim}([\![a_1]\!], \ldots, [\![a_n]\!]) .$

Thus $\phi$ is an isomorphism. $\qedsymbol$