Show that a linear map $\phi:V\rightarrow W$ is uniquely determined by the values of a basis of $V$ To prepare this theorem we have concluded that every vector can be described as a finite linear combination of basisvectors:

Below is the proof of the theorem, I have highlighted the part that I don't understand:



Can someone explain me why $B_{u+v}\subseteq B_{u}\cup B_{v}$? I also don't understand how this implies $\sum_{b\in B_{u+v}}c_{u+v}(b)w_b=\sum_{b\in B_u}c_u(b)w_b+\sum_{b\in B_v}c_v(b)w_b$
 A: $B_{u + v}$ is the set of all the bases that appear with non-zero coefficient in unique linear combination for $ u + v$. Now, to find the linear combination of basis which gives $u + v$ is to sum up the linear combination of elements of B equal to $u$ to the one equal to $v$. This is exactly what the union does. 
( pay attention to $\subseteq$ and the reason why it is not equality) 
As for the equation in 6.6b, note that  $c_{u, v} = c_u + c_v$. This is because, $u + v  = \sum_{ b \in B_{u + v}} (c_u + c_v ) b$. And this is obtained by writing equation 5.28  for $u$ and $v$ and summing them up.  There is a only a unique linear combination of the elements of B that is equal to  $u + v$. Thus the coefficient of $b \in B_{u + v}$  in the linear combination for $u + v$, which we also denote by $c_{u + v}$, is equal to $c_u + c_v$. 
A: Because of the unicity of the coordinates in one basis. You can either directly decompose $u+v$, or decompose separately $u$ and $v$ and sum the result. However in both cases you get the same coordinates!
A: I am not sure if this is right but I want to show that 
if $b\in B_U\cap B_V\wedge (c_u+c_v)(b)\neq 0$ then $c_{u+v}(b)=(c_u+c_v)(b)$ and if $b\in B_V- B_W$ or $b\in B_W-B_V$ then $c_{u+v}=c_u$ or $c_{u+w}=c_v$ respectively. 
We define $\tilde{B}=B_V-B_W\cup B_W-B_V\cup B_V\cap B_W - \{b\in B_V\cap B_W|c_v(b)+c_u(b)=0\}$
$u+v=\sum_{b\in B_{u+v}}c_{u+v}(b)b=\sum_{b\in B_u}c_u(b)b+\sum_{b\in B_v}c_v(b)b$ then implies that the right sum can be splitted in three parts as suggested abbove. If we define an operation $\tilde{c}$ as suggested above (writing $\tilde{c}$ instead of $c_{u+v}$), then because the above unions are disjoint and because of Th. 5.19 (uniqueness) $\tilde{B}=B_{u+v}$ and $\tilde{c}=c_{u+v}$. 
Finally because $\sum_{b\in B_{u+v}}c_{u+v}(b)w_b=$
$\sum_{b\in\tilde{B}}\tilde{c}(b)w_b=$
$\sum_{b\in B_u-B_v}c_u(b)w_b+\sum_{b\in B_v-B_u}c_v(b)w_b+\sum_{b\in \{B_u\cap B_v|c_u+c_v\neq 0\}}(c_u+c_v)(b)w_b + 0 [=\sum_{b\in\{B_u\cap B_v|c_u+c_v=0\}}(c_u+c_v)(b)w_b]$
$=\sum_{b\in B_u}c_u(b)w_b+\sum_{b\in B_v}c_v(b)w_b$
the statement is shown. 
