The is given vector space $V$ through a field $K$ and f: $V → V$ is an Endomorphism. Furthermore, $ U ⊂ V $ is a vector subspace and $ξ: V → V/U$ the canonical projection. Please prove that there exists an Endomorphism $ϕ : V/U → V/U$ with $ϕ ◦ ξ = ξ ◦ f$ if and only if $F(U)⊂ U.$
The official solution start like that:
First of all $F(U)⊂ U.$ is true. Then the mapping
$ϕ : V/U → V/U , v + U → f(v) + U $
is well defined.
I got informed that $V/U$ consists of equivalence classes with $v∼w$ iff $v−w∈U$. So in particular $[v]$ can be seen as $v+U$. What I don't understand is $U+f(v)$ How do we know that $v + U → f(v) + U $? If $ϕ : V/U → V/U , $, shouldn't it be just $v + U → w + U $ with $v,w ∈ V$?
Thanks for you help.
So we define $\phi : V/U \to V/U, v+U \mapsto f(v)+U$.
Formally we have $V/U = \{ [v] \mid v \in V \}$ with $[v] = \{ w \in V \mid v - w \in U \}$. However, we can show that $v +U := \{ v+ u \mid u \in U \} $ is equal to $[v]$.
So we define $\phi([v]) = f(v) +U$. However, this is only well-defined if $\phi([w]) = f(v) + U$ for all $w \in [v]$, since then $[v] = [w]$.
So let $w \in [v]$, then we have $w = v +u$ for some $u \in U$ and thus \begin{align} \phi([w]) = f(w) + U = f(v+u) + U = f(v) + f(u) + U = f(v) +U \end{align} where we used $f(v+u) = f(v) +f(u)$ as $f$ is an endomorphism and $f(U) \subset U$, thus $f(u) \in U$ as $u\in U$.
As $U$ is a subspace, so $f(u) + U = \tilde{u} + U = U$.
So regardless of the representative $w$ of $[v]$, we have $\phi([v]) = f(v) +U = [f(v)]$, and thus $\phi$ is well-defined.
And we choose $\phi([v])$ to be $[f(v)]$, as we then have \begin{align} \xi \circ f(v) = \xi(f(v)) = [f(v)] = \phi([v]) = \phi(\xi(v) = \phi \circ \xi (v), \end{align} i.e. $\xi \circ f = \phi \circ \xi$ as we wanted to show.