Showing uniqueness property over a vector space that is the direct sum of two of its subspaces. Let $V$ be a vector space over a field $F$, and let $S$ and $T$ be subspaces of $V$ such that 
$V=S⊕T$.
Show that for every $x∈V$ ,there are unique $y_1∈S$ and $z_1∈T$ such that $x=y_1+z_1$. In other words, show that, if $y_2 ∈ S$ and $z_2 ∈ T$ also satisfy $y_1 +z_1 = x = y_2+z_2$, then
$y_1 =y_2$ and $z_1 =z_2$
$P(x):=$unique $y∈S$ such that $x=y+z$ for some $z∈T$
Prove that $P$ is a linear map, and also that we have $P^2(= P ◦ P) = P$. Show also
that $Range(P) = S$ and $Ker(P) = T$.
Attempt for the first part:
By definition of direct sum $S \cap T$={$0$}.
If $V=S⊕T$ then for every $x \in V$, $x=s+t$ for some $s \in S$ and $t\in T$.
Let $n \in S$ and $m \in T$.
Now assume $$n+m=s+t$$ Rearranging we have $0=(s-n)+(t-m)$
Then since $S,T$ are subspaces of $V$ and $s,n \in S$ and $t,m \in T$, then $s-n \in S$ and $t-m\in T$ Therefore both $s-n$ and $t-m$ are equal to $0$ which gives us our uniqueness property.
For the second part I'm stuck I know that this can be rearranged to get
 $P(x):=$ unique $y ̄ ∈ S$ such that $x − y ∈ T$. But I'm unsure how to continue from here. I know I need to prove the below but I'm confused where to begin to do this with this function and how the range and kernel would be S and T respectively.
$P()=P()$
$P(+)=P()+P()$

If $x = a+b, y=c+d$ are the unique decompositions for x and y respectively, then 


*

*$P(x+y) = P((a+c)+(b+d)) = a+c = P(x) + P(y).$

*$P(\alpha x) = P(\alpha a+ \alpha b) = \alpha a.$

*$P(P(a+b)) = P(a) = a.$

*$S \subseteq Range(P)$ since $a = a+0$ in $V$.

*$P(a+b) = 0 \Leftrightarrow$ $a=0$. Therefore the kernel is T.

