Show that the following equation applies: $\dim(\phi^{-1}(U))=\dim(U\cap\phi(V))+\dim(\ker(\phi))$

Question

Let $V$ be a finite dimensional $\mathbb{K}$-vector space, $\phi : V \to W$ a linear mapping and $U$ a sub-vector space of $W$. Show that the following equation applies: $$\dim(\phi^{-1}(U))=\dim(U\cap\phi(V))+\dim(\ker(\phi))$$

EDIT

Let $\phi:V\to W$ and $v\mapsto \phi(v)$ and $\Phi :\phi^{-1}(u)\mapsto u$ with $v\in V,\;\color{red}{\phi(v)\in W},\;u\in U$ and $U\subseteq W$

We insert this into the Rank–nullity theorem:
$$\dim(\phi^{-1}(U))=\dim(\operatorname{im}(\Phi))+\dim(\ker(\Phi))$$

$\operatorname{im}\phi = \phi(v)$
$\operatorname{im} (\Phi)=\operatorname{im}\phi \cap U$
$\ker(\phi):=\{v\in \phi^{-1}(u):\phi(v)=0\}$
$0\in U\implies \ker \phi =\ker \phi^{-1} \implies \dim(\phi^{-1}(U))=\dim(U\cap\phi(V))+\dim(\ker(\phi))$

Any opinions on this "proof" (edited)?

Answer 1

You can use the nullity-rank theorem with the map $\Phi: \phi^{-1}(U)\to U$ :

$\dim(\phi^{-1}(U))=\dim(ker(\Phi))+\dim(im(\Phi))$

but

$ker(\Phi)=\ker(\phi)$

because if $v\in Ker(\Phi)$ then $\Phi(v)=\phi(v)=0$ so $v\in Ker(\phi) $while if $v\in Ker(\phi)$ then $\phi(v)=0\in U$ so $v\in \phi^{-1}(U)$ and you have that $\Phi(v)=\phi(v)=0$ then $v\in Ker(\Phi)$

and

$im(\Phi)=U\cap im(\phi)$

So

$\dim(\phi^{-1}(U))=\dim(ker(\phi))+\dim(U\cap im(\phi))$