A bit of direct sum confusion So I'm reading through Serre's "Linear Representations of Finite Groups," and I'm a bit confused by what's probably a fairly minor point.  However, subsequent proofs are hinging on it, so I figure I'll turn to you guys for clarification.
So he defines the direct sum (I guess in this case the internal one), and then introduces the projection map.  So if $V=W+W'$, then  $p(x=w+w')=w$.  Then, for some converse implicating, he assumes we have some map $p$ from $V$ to $V$ whose image is $W$ and whose restriction to $W$ yields $p(x)=x$.  Then we can show that $V$ is the direct sum of $W$ and the kernel of $p$.  It's the next line that trips me up.  He says "A bijective correspondence is thus established between the projections of $V$ onto $W$ and the complements of $W$ in $V$."  My brain can't fill the gap in the logic, nor can I really understand what that statement is all about.
Apologies for the logorrhea. Any and all help is very much appreciated!
 A: Every projection $\,p:V\stackrel{\text{onto}}\rightarrow W\,$ determines a complement of $\,W\,$ , namely $\,\ker p\,$, since $\,V=W\oplus \ker p\,$, and the way around: for every decomposition $\,V=W\oplus U\,$ there exists a projection $\,p_U:V\to W\,$ , defined by $\,p_U(v=w\oplus u):= w\,$ . There you have your bijective correspondence. 
A: You just need to understand the following: Namely let $p : V \to V$ such that $p^2 = p.$ Then 
$$V = \textrm{Im} \hspace{1mm} p \oplus \textrm{Ker} \hspace{1mm} p. $$
Assuming this result, what Serre is saying is this. Suppose you have a subspace $W$ of $V$. Then you can always extend the basis of $W$ to a basis of $V$ just by adding in more basis vectors to the collection that is already a basis for $W$. In this way, those extra guys give you a complement of $W$ which you can call $W'$. I now claim that
$$V = W \oplus W'.$$
It is plain that every vector in $V$ can be written as $w + w'$ with $w \in W$, $w' \in W'$. It now suffices to show that $W \cap W' = \{0\}$. But this is trivial as well because by choice the set consisting of basis vectors of $W$ and $W'$ is linearly independent. This completes the proof of the claim.
So you can now get a projection $p$ from $V$ onto $W$ simply as follows. Write $\{w_1,\ldots w_n\}$ as a basis for $W$ and $\{u_1,\ldots,u_m\}$ a basis for $W'$. Then the matrix of $p$ with respect to the basis for $V$ obtained from concatenating the two is:
$$\left(\begin{array}{c|c} I & 0 \\ \hline 0& 0 \end{array}\right)$$
where the upper left block is the $n \times n$ identity matrix and the lower right block is a $m \times m$ matrix with all zeros as its entries.
Conversely if you have a projection, i.e. an operator that satifies $p^2 = p$ then by the result I told you above this gives rise to a subspace, namely $\textrm{im} p$ which is our $W$ now and its complement, the kernel of $p$.
A: Each direct decomposition of a module $M$ arises from an idempotent $e$ in the endomorphism ring $\mathrm{End}(M_R)$ in this way: $M=e(M)\oplus (1-e)(M)$. 
In this case, the module is a vector space and "projection" refers to the idempotent.
You can find some more discussion about this on the idempotence Wiki page.
(The other solutions are good by the way, I just wanted to give my own spin on the overview.)
