# kernel of a homomorphism is a congruence

Let $\mathrm{\Sigma}$ be a fixed signature^{}, and $\U0001d504$ and $\U0001d505$ two structures^{} for $\mathrm{\Sigma}$. If $f:\U0001d504\to \U0001d505$ is a homomorphism^{}, then $\mathrm{ker}(f)$ is a congruence^{} on $\U0001d504$.

###### Proof.

If $F$ is an $n$-ary function symbol of $\mathrm{\Sigma}$, and $f({a}_{i})=f({a}_{i}^{\prime})$, then

$f({F}^{\U0001d504}({a}_{1},\mathrm{\dots},{a}_{n}))$ | $={F}^{\U0001d505}(f({a}_{1}),\mathrm{\dots},f({a}_{n}))$ | ||

$={F}^{\U0001d505}(f({a}_{1}^{\prime}),\mathrm{\dots},f({a}_{n}^{\prime}))$ | |||

$=f({F}^{\U0001d504}({a}_{1}^{\prime},\mathrm{\dots},{a}_{n}^{\prime})).\mathit{\u220e}$ |

Title | kernel of a homomorphism is a congruence |
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Canonical name | KernelOfAHomomorphismIsACongruence |

Date of creation | 2013-03-22 13:48:03 |

Last modified on | 2013-03-22 13:48:03 |

Owner | almann (2526) |

Last modified by | almann (2526) |

Numerical id | 8 |

Author | almann (2526) |

Entry type | Theorem |

Classification | msc 03C05 |

Classification | msc 03C07 |

Related topic | KernelOfAHomomorphismBetweenAlgebraicSystems |