Let $(\mathbb R,V,+)$ be a vector space, where $U_1,U_2\subset V$. We prove that $$ W=U_1\oplus U_2\iff W=U_1+U_2\text{ and }U_1\cap U_2=\{0\}. $$ "$\implies$": Assume $W=U_1\oplus U_2$. For any vector $w\in W$, there are unique vectors $u_1\in U_1$ and $u_2\in U_2$ such that $w=u_1+u_2$. Then for sure $W=U_1+U_2$. [...]
I'm confused by the way the theorem is worded. Shouldn't it go like this: $$ W=U_1\oplus U_2\implies W\subset U_1+U_2, $$ and $$ U_1\cap U_2=\{0\}\iff U_1\oplus U_2=U_1+U_2. $$
My motivation:
We know that $W=U_1\oplus U_2\implies W\subset U_1+U_2$. Assume $U_1\cap U_2\neq \{0\}$. That means that $\exists v: v\in U_1$ and $v\in U_2$. But then $v=v+0=0+v$, so we have that $v\notin U_1\oplus U_2$. So if $U_1\cap U_2\neq \{0\}$, we know that $W\neq U_1+U_2$.
Now assume $U_1\cap U_2=\{0\}$. Let $v=u_1+u_2=u_1'+u_2'$, where $u_1,u_1'\in U_1$ and $u_2,u_2'\in U_2$. Let's do some algebra: \begin{align} u_1+u_2&=u_1'+u_2'\\ u_1-u_1'&=u_2'-u_2. \end{align} We know that $u_1-u_1'\in U_1$ and $u_2'-u_2\in U_2$. Because they are equal, we know that $u_1-u_1'=u_2'-u_2\in U_1\cap U_2$. But that means that $u_1-u_1'=u_2'-u_2=0$. Thus $u_1=u_1'$ and $u_2=u_2'$. Therefore $W=U_1+U_2$.
Consider $U_1, U_2$ as vector subspaces of $W$. Take the following homomorphism:
$$\phi:U_1 \times U_2 \to W$$ $$(u_1,u_2) \mapsto u_1+u_2.$$
$u_1+u_2=0$ or $(u_1,u_2) \in \ker \phi$ if and only if $u_1=-u_2$, in which case $u_1 \in U_2$ since both are vector subspaces. Hence, $u_1 \in U_1 \cap U_2$. Similarly, we can show that $U_1 \cap U_2 \subseteq \ker \phi$.** Hence, we can conclude that $$\ker \phi= U_1 \cap U_2.$$
Hence, this function is noninjective, only when its kernel (the intersection of $U_1$ and $U_2$) is non-trivial.
**There is a slight abuse of notation here.