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 $\phi\colon\mathfrak{A}/\!\ker(f)\to\operatorname{im}(f)$ such that for all $a\in\mathfrak{A}$ , $\phi([\![a]\!])=f(a)$ . Furthermore, if $f$ has the additional property that for each $n\in\mathbb{N}$ and each $n$ -ary relation symbol $R$ of $\Sigma$ ,

$R^{\mathfrak{B}}(f(a_{1}),\ldots,f(a_{n}))\Rightarrow\exists a_{i}^{\prime}[f(% a_{i})=f(a_{i}^{\prime})\land R^{\mathfrak{A}}(a_{1}^{\prime},\ldots,a_{n}^{% \prime})],$

then $\phi$ is an isomorphism.

Proof.

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

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 $a\sim a^{\prime}$ . Then $\phi([\![a]\!])=f(a)=f(a^{\prime})=\phi([\![a^{\prime}]\!])$ . To show that $\phi$ is injective, suppose $\phi([\![a]\!])=\phi([\![a^{\prime}]\!])$ . Then $f(a)=f(a^{\prime})$ , so $a\sim a^{\prime}$ . Hence $[\![a]\!]=[\![a^{\prime}]\!]$ . To show that $\phi$ is a homomorphism, observe that for any constant symbol $c$ of $\Sigma$ we have $\phi([\![c^{\mathfrak{A}}]\!])=f(c^{\mathfrak{A}})=c^{\mathfrak{B}}$ . For each $n\in\mathbb{N}$ and each $n$ -ary function symbol $F$ 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 $n$ -ary relation symbol $R$ of $\Sigma$ ,

	$\displaystyle R^{\mathfrak{A}/\!\sim}([\![a_{1}]\!],\ldots,[\![a_{n}]\!])$	$\displaystyle\Rightarrow R^{\mathfrak{A}}(a_{1},\ldots,a_{n})$
		$\displaystyle\Rightarrow R^{\mathfrak{B}}(f(a_{1}),\ldots,f(a_{n}))$
		$\displaystyle\Rightarrow R^{\mathfrak{B}}(\phi([\![a_{1}]\!],\ldots,\phi([\![a% _{n}]\!])).$

Thus $\phi$ is a bimorphism.

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

	$\displaystyle R^{\mathfrak{B}}(\phi([\![a_{1}]\!]),\ldots,\phi([\![a_{n}]\!]))$	$\displaystyle\Rightarrow R^{\mathfrak{B}}(f(a_{1}),\ldots,f(a_{n}))$
		$\displaystyle\Rightarrow\exists a_{i}^{\prime}[a_{i}\sim a_{i}^{\prime}\land R% ^{\mathfrak{A}}(a_{1}^{\prime},\ldots,a_{n}^{\prime})]$
		$\displaystyle\Rightarrow R^{\mathfrak{A}/\!\sim}([\![a_{1}]\!],\ldots,[\![a_{n% }]\!]).$

Thus $\phi$ is an isomorphism. ∎

Title	first isomorphism theorem
Canonical name	FirstIsomorphismTheorem1
Date of creation	2013-03-22 13:50:42
Last modified on	2013-03-22 13:50:42
Owner	almann (2526)
Last modified by	almann (2526)
Numerical id	10
Author	almann (2526)
Entry type	Theorem
Classification	msc 03C07