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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)?

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  • $\begingroup$ It is not correct because \phi^{-1} could be not defined $\endgroup$ – Federico Fallucca Jun 24 at 22:03
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    $\begingroup$ The first line already fails. What is "$v$" in $v\mapsto\phi(V)$? $\phi^{-1}(U)$ is not an element of $V$, and $U$ is not an element of $W$, so "$\phi^{-1}(U)\mapsto U$" doesn't make sense. $\endgroup$ – user10354138 Jun 24 at 22:08
  • $\begingroup$ $v\in V,\; w\in \phi(v),\;u\in U;\; \phi^{-1}(u)= u$. Does that make more sense? $\endgroup$ – Doesbaddel Jun 24 at 22:11
  • $\begingroup$ Let $\phi:V\to W$ and $v\mapsto \phi(v)$ and $\Phi :\phi^{-1}(u)\mapsto u$ with $v\in V,\; w\in W,\;\phi(v)\in w,\;u\in U$ and $U\subseteq W$ Is this better? $\endgroup$ – Doesbaddel Jun 24 at 22:20
  • $\begingroup$ No it does not make sense. See $\phi(v)$ is an element, so how is $w$ an element of an element $\phi(v)$? An element is not a set. $\endgroup$ – TheLast Cipher Jun 25 at 7:33
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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

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

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  • $\begingroup$ @Frederico Falluca: Why do you think $\operatorname{ker}(\Phi)=\operatorname{ker}(\phi)$? I don't think this is true, since $\Phi$ is a restriction of $\phi$ to the subspace $\phi^{-1}(U)$ of $V$. So I can't see a reason why we can say that all elements of $V$ mapped to the identity of $U$ is necessarily an element of $\phi^{-1}(U)$ when we consider that $\phi^{-1}(U) \subseteq V$. $\endgroup$ – TheLast Cipher Jun 25 at 8:19
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    $\begingroup$ Have I convinced you now? $\endgroup$ – Federico Fallucca Jun 25 at 9:36
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    $\begingroup$ @Frederico: thanks! $\endgroup$ – TheLast Cipher Jun 25 at 9:56
  • $\begingroup$ Thank you for the additions! $\endgroup$ – Doesbaddel Jun 26 at 8:36

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