# 準同型定理

$G$, $G'$ are Groups.
$f$ : $G \longrightarrow G'$ is a Homomorphism.
$\pi$ : $G \longrightarrow G / \ker{f}$ is a Natural Homomorphism.

Then
${}^{\exists 1} \phi$ such that $\phi : G / \ker{f} \longrightarrow G'$ and $f = \phi\circ\pi$ and $G / \ker{f} \simeq \text{Im}f$