Linear Transformation Question $ \operatorname{Hom}(V,W)$ and $ \operatorname{Hom}(W,V)$ satisfying conditions Can I please get help on this questions. I'm not sure how to start it or end it.

Let $V$ and $W$ be vector spaces over a field $F$. Let $\alpha$ be an element of $ \operatorname{Hom}(V,W)$ and $\beta$ be an element of $ \operatorname{Hom}(W,V)$ satisfy the condition  $\alpha\cdot\beta\cdot\alpha=\alpha$. 
If $w \in  \operatorname{Im}(\alpha)$, show that $\alpha^{-1}(w) = \{ \beta(w)+v - \beta \alpha(v)  \mid v \in V \}$.

 A: If you  prove that
1) if  $k=b(w)+v-ba(v)$| $v \in V$
then $k \in a^{-1}(w)$ meaning $a(k)=w$.
2) if $v$ s.t $a(v)=w$ then $v=b(w)+v-ba(v)$ where 
a,b linear functions s.t $aba=a$
Can you use the linearity and $aba=a$ to prove those two? If you do, you proved that the set $\{b(w)+v-ba(v)|v \in V \}$ equals to the set $\{v \in V| a(v)=w\}$
A: It's generally true that, for $w$ in the image of $\alpha$, $\alpha^{-1}(w)=\{x+y:y\in\ker\alpha\}$, where $x\in V$ is a fixed vector such that $\alpha(x)=w$.
Since $w=\alpha(x)$, then $w=\alpha(x)=\alpha\beta\alpha(x)=\alpha\beta(w)$, so we can take $x=\beta(w)$.
Now it's generally true that $\alpha^{-1}(w)=\{\beta(w)+y:y\in\ker\alpha\}$, so we just need to determine $\ker\alpha$.
Let $v\in V$; then
$$
\alpha(v-\beta\alpha(v))=\alpha(v)-\alpha\beta\alpha(v)=0
$$
so $v-\beta\alpha(v)\in\ker\alpha$. Conversely, if $y\in\ker\alpha$, then $y=y-\beta\alpha(y)$. Thus
$$
\ker\alpha=\{v-\alpha\beta(v):v\in V\}
$$
A: Write $w=\alpha(u)$ with $u \in V$. Then $\alpha\beta(w)=\alpha\beta\alpha(u)=\alpha(u)=w$ and so $\beta(w) \in \alpha^{-1}(w)$.
Therefore, $\alpha^{-1}(w)=\beta(w) + \ker \alpha$.
So it remains to prove that $\ker \alpha = \{ v - \beta \alpha(v) : v \in V \}$.
Now $\alpha\beta\alpha=\alpha$ implies  $\ker \alpha \supseteq \{ v - \beta \alpha(v) : v \in V \}$.
Also, $v \in \ker \alpha$ implies $v=v - \beta \alpha(v)$ and so $\ker \alpha \subseteq \{ v - \beta \alpha(v) : v \in V \}$.
