Group homomorphism/Homomorphism theorem/Surjective and kernel/Fact/Proof

Proof

We show firstly the uniqueness. For every Element ${}u\in Q$, there exists some ${}g\in G$ with ${}\psi (g)=u$. The commutativity of the diagram ensures that

${}{\tilde {\varphi }}(u)=\varphi (g)\,$

holds. This means that there exists at most one ${}{\tilde {\varphi }}$.

We have to show that this condition yields a well-defined mapping. Hence, let ${}g,g'\in G$ be two preimages of ${}u$. Then

${}\psi {\left(g'g^{-1}\right)}=uu^{-1}=e_{Q}\,;$

therefore, we have ${}g'g^{-1}\in \operatorname {kern} \psi \subseteq \operatorname {kern} \varphi$. Hence, ${}\varphi (g)=\varphi (g')$, and the mapping is well-defined. Let ${}u,v\in Q$ be given, and let ${}g,h\in G$ be preimages. Then ${}gh$ is a preimage of ${}uv$. Therefore,

${}{\tilde {\varphi }}(uv)=\varphi (gh)=\varphi (g)\varphi (h)={\tilde {\varphi }}(u){\tilde {\varphi }}(v)\,.$

This means that ${}{\tilde {\varphi }}$ is a group homomorphism.

To fact