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.∎