# 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 $\Sigma$,

 $R^{\mathfrak{B}}(f(a_{1}),\ldots,f(a_{n}))\Rightarrow\exists a_{i}^{\prime}[f(% a_{i})=f(a_{i}^{\prime})\land R^{\mathfrak{A}}(a_{1}^{\prime},\ldots,a_{n}^{% \prime})],$
###### 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  FirstIsomorphismTheorem1 2013-03-22 13:50:42 2013-03-22 13:50:42 almann (2526) almann (2526) 10 almann (2526) Theorem msc 03C07