Hey I want ot check if my solutions for this exercise are right. Can someone help me?
Let $V$ be a finite dimensional $K$-vector space and $U_1, . . . , U_n$ a family of $K$-subspaces in $V$ .
Show that the following statements are equivalent:
a)The natural mapping $U_1 ⊕ . . . ⊕ U_n → V$, $(u_1, . . . , u_n) → u_1 + . . . + u_n$ is bijective.
b) $V = U_1 + . . . + U_n$, and for every family $u_1, . . . , u_n$ in $V$ with $u_i ∈ U_i$ for all $1 ≤ i ≤ n$ and with $u_1 + . . . + u_n = 0$ then $u_1 = . . . = u_n = 0$.
c) $V = U_1 + . . . + U_n$ and $U_i ∩ \sum_{j=1, j\neq=i}^{n}U_j = 0$ for all $1 ≤ i ≤ n$.
d) $V = U_1 + . . . + U_n$ and $\dim(V) =\sum_{i=1}^{n}\dim(U_i)$.
so for $a)\Rightarrow b)$, since our function that I will call $f$ is bijective, is automatically injective and we have that $\ker(f)=0$.
So only with $u_1,=...=u_n=0$ follows $u_1+...+u_n=0$.
For $b)\Rightarrow c)$ let the implication be wrong so we have that there exists $U_i ∩ \sum_{j=1, j\neq=i}^{n}U_j \neq 0$ w.l.o.g. we have a vector $u_1\neq 0 \in U_1$ that can be generated from $U_2+...+U_n$.
But this is automatically wrong because so we could choose a combination of $u_2+...+u_n=-u_1$ and this would give $u_1+(u_2+...+u_n)= u_1-u_1=0$ which is a contradiction to b)
For $c) \Rightarrow d)$ we have the dimensions theorem (idk if this is the name of this formula in english) that says $\dim(V)=\sum_{i=0}^{n}\dim(U_i)-\dim(U_1 ∩...∩U_n)$ since we have that $U_i ∩ \sum_{j=1, j\neq=i}^{n}U_j = 0$ it means that the $U_1...U_n$ are independent and for that $\dim(U_1 ∩...∩U_n)=0$
$d)\Rightarrow a)$ we know that the $U_1,...,U_n$ are independent and for that every combination of vectors $u_1,...,u_n$ is linearly independent.
Have I done some errors? Thank you for any help.