Vector spaces/Linear mapping/Homomorphism theorem/Surjective and kernel/Fact/Proof2

Proof

For every element $u\in Q$, there exists some $v\in V$ with $\psi (v)=u$. Due to the commutativity condition, we must have $\tilde{\phi }(u)=\phi (v)$. This means that there exists at most one $\tilde{\phi }$. We have to show that, by this condition, a well-defined mapping is given. So let $v,v^{\prime }\in V$ denote two preimages of $u$. Then

$v^{\prime }-v\in \mathrm{kern}\psi \subseteq \mathrm{kern}\phi ;$

therefore, $\phi (v)=\phi (v^{\prime })$. hence, the mapping is well-defined.
Let $u,u^{\prime }\in Q$ be gievn, with preimages $v,v^{\prime }\in V$. Then $v+v^{\prime }$ is a preimage of $u+u^{\prime }$; therefore, we have

$\tilde{\phi }(u+u^{\prime })=\phi (v+v^{\prime })=\phi (v)+\phi (v^{\prime })=\tilde{\phi }(u)+\tilde{\phi }(u^{\prime }).$

This means that $\tilde{\phi }$ is compatible with the addition.
Let $u\in Q$ be given with a preimage $v\in V$, and let $\lambda \in K$. Then $\lambda v$ is a preimage of $\lambda u$; therefore,

$\tilde{\phi }(\lambda u)=\phi (\lambda v)=\lambda \phi (v)=\lambda \tilde{\phi }(u),$

and $\tilde{\phi }$ is also compatible with the scalar multiplication.

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