Assume $V=U\oplus W$ is a direct sum where $U$ and $W$ are subspaces of the vector space $V$. Then we can define a linear map $$E(v)=u,\\ with \qquad v=u+w\in V,\qquad u\in U,\ w\in W$$ Called the projection on $U$ along $W$.
Lemma: If $E$ is the projection on $U$ along $W$ then: $$(a)\qquad U=R(E)=N(I-E)\\ (b)\qquad W=N(E)=R(I-E)$$
In my lecture notes, the first part of $(a)$ is proved by stating that by definition $R(E)\subseteq U$ and that $\forall u\in U$ we have $u=Eu\in R(E)$, and therefore $R(E)=U$. However, this proof seems incomplete to me, since the definition of $E$ tells nothing about which $u$ $v$ is mapped to (the way I read it). Thus, I would imagine that we might as well have $Eu_1=u_2$, in which case we haven't shown that all $u\in U$ can be mapped to by $E$.
I guess I'm wrong but can someone tell me how?
If $u \in E$ then as $V=U\oplus W$ is a direct sum, we have $u= u + 0$ where $0 \in W$ and this is the unique way to decompose $u$ along $U,W$.
Therefore $E(u)=u$ and $E$ is onto $U$.