Maps between projective spaces

Question

Let $K$ be a field, let $V$ and $V'$ be two vector spaces on $K$ and let $f : V \to V'$ an isomorphism. If I consider the projective spaces $P(V)$ and $P(V')$, the map ${\sigma}_f : P(V) \to P(V')$, given by ${\sigma}_f(\langle u \rangle) = \langle f(u) \rangle$ for all $u \in V$ has some properties: ${\sigma}_{Id_V} = Id_{P(V)}$, ${\sigma}_{g \circ f} = {\sigma}_g \circ {\sigma}_f$, being $V''$ a vector space on $K$ and $g : V' \to V''$ an isomorphism, or ${\sigma}_f$ is inversible and ${\sigma}_f^{- 1} = {\sigma}_{f^{- 1}}$. If $S$ is a vector subspace in $V$, with $\dim P(S) = k$, how can I prove that ${\sigma}_f(P(S))$ is a vector subspace in $V'$, with $\dim {\sigma}_f(P(S)) = k$? Thank you very much.

Answer 1

$\dim P(S)=k$ means that $\dim S= k+1$. Since $f$ is an isomorphism $$\dim f(S) = \dim S= k+1$$ (you can prove this by choosing a basis of $S$ and proving that it is sent to a basis of $f(S)$). This implies by definition $$\dim P( f(S))=k$$

What it remains to prove is that $P(f(S))= \sigma_f(P(S))$, and I leave it to you: use double inclusion.