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.
$P()=P()$
$P(+)=P()+P()$
 A: First issue:

By definition of direct sum $S \cap T$ is the empty set.

No, $S \cap T = \{0\}$. It's not possible for the intersection of subspaces to be empty. This is an important distinction!
Second issue:

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.

I have no issue with the statements here, it's more about what isn't here. There is a small (but vitally important) observation skipped here. Specifically, since $t - m \in T$, then so is
$$s - n = -(t - m) \in T.$$
Therefore $s - n \in S \cap T = \{0\}$, so $s - n = 0$. You can conclude that $t - m = 0$ too.
This observation is really the heart of the proof, and where you get to use the assumption that $S$ and $T$ sum directly. If I were marking this proof, and I saw the above two errors, I would mark you down. I would assume that if the above steps were something you properly understood (even if you didn't write them down), then you wouldn't say that $S \cap T = \emptyset$.

For the second part, the question is tricky because the map $P$ is not defined explicitly, but implicitly. That is, there's no formula to take $x$ to $Px$. Instead, $Px$ satisfies an equation with a unique solution.
Let's start by taking arbitrary vectors $x_1, x_2 \in V$. Since $V = S \oplus T$, there exist unique $s_1, s_2, s_3 \in S$ and $t_1, t_2, t_3 \in T$ such that
\begin{align*}
x_1 &= s_1 + t_1 \\
x_2 &= s_2 + t_2.
\end{align*}
Note that $P(x_1) = s_1$ and $P(x_2) = s_2$.
In order to prove that $P(x_1 + x_2) = P(x_1) + P(x_2) = s_1 + s_2$, we need to find a decomposition of $x_1 + x_2$ into a component in $S$ (which hopefully will be $s_1 + s_2$ and a component in $T$. We can simply get this by adding the above equations:
$$x_1 + x_2 = (s_1 + s_2) + (t_1 + t_2).$$
By the first part, this decomposition is unique, and so $P(x_1 + x_2)$ is well-defined to be $s_1 + s_2$.
Now try with scalar multiplication!
A: Note that with
$V = S \oplus T \tag 1$
we have
$S \cap T = \{0\}, \tag 2$
and not
$S \cap T = \emptyset. \tag 3$
This distinction is of course consistent with the assertion that the decomposition of any $v \in V$ as
$v = s + t, \; s \in S, t \in T, \tag 4$
is unique, since if
$v = s_1 + t_1 = s_2 + t_2; \; s_1, s_2 \in S; t_1, t_2 \in T, \tag5$
then
$S \ni s_1 - s_2 = t_2 - t_1 \in T, \tag 6$
whence
$s_1 - s_2 = 0 = t_2 - t_1, \tag 7$
or
$s_1 = s_2, t_1 = t_2, \tag 8$
that is, the decompsition is unique.  This demonstration is, with minor variations, the same as that given by our OP Mathisoshardlmao in the body of the question itself.
Now for $v \in V$ with
$v = s + t \tag 9$
uniquely, we define
$P:V \to S \tag{10}$
via
$P(v) = s; \tag{11}$
we then have, from (9),
$\alpha v = \alpha(s + t) = \alpha s + \alpha t, \tag{12}$
whence, since 
$\alpha s \in S, \; \alpha t \in T, \tag{13}$
we see, again using uniqueness,
$P(\alpha v) = \alpha s = \alpha P(v); \tag{14}$
also, if
$v_1 = s_1 + t_1, \tag{15}$
and
$v_2 = s_2 + t_2 \tag{16}$
are the decompositions of $v_1$ and $v_2$ in accord with (1), then
$P(v_1 + v_2) = P((s_1 + t_1) + (s_2 + t_2))$
$= P((s_1 + s_2) + (t_1 + t_2)) = s_1 + s_2 = P(v_1) + P(v_2). \tag{17}$
(9)-(17) show that $P$ is a linear map.  
Now, yet again in accord with (9),
$P^2(v) = P^2(s + t) = P(P(s + t)) = P(s) = s = P(v), \tag{18}$
which shows 
$P^2 = P. \tag{19}$
Since with
$s \in S, \tag{20}$
$P(s) = s, \tag{21}$
it follows that
$\text{Range}(P) = S; \tag{22}$
with
$t \in T, \tag{23}$
$P(t) = 0, \tag{24}$
whence
$T \subset \ker P; \tag{25}$
and if
$t \in \ker P, \tag{26}$
$P(t) = 0, \tag{27}$
and if
$t = s + t \tag{28}$
is the unique decomposition of $t$, then
$P(t) = s = 0, \tag{29}$
so
$t \in T, \tag{30}$
and so
$\ker P \subset T; \tag{31}$
together (25) and (30) yield
$\ker P = T, \tag{32}$
and we are done.
A: 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.

