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

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)?

• It is not correct because \phi^{-1} could be not defined – Federico Fallucca Jun 24 at 22:03
• 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. – user10354138 Jun 24 at 22:08
• $v\in V,\; w\in \phi(v),\;u\in U;\; \phi^{-1}(u)= u$. Does that make more sense? – Doesbaddel Jun 24 at 22:11
• 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? – Doesbaddel Jun 24 at 22:20
• 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. – TheLast Cipher Jun 25 at 7:33

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

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

• @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 $\phi^{-1}(U) \subseteq V$. – TheLast Cipher Jun 25 at 8:19
• Have I convinced you now? – Federico Fallucca Jun 25 at 9:36
• @Frederico: thanks! – TheLast Cipher Jun 25 at 9:56
• Thank you for the additions! – Doesbaddel Jun 26 at 8:36