Connection between proof and algebraic factor theorem It is (no proof needed)
Let $f \in End(V)$ and $ f^2 = f$ Then $V = Ker(f) \oplus Im(f)$.
What does this mean for the following sentence??
"Factor Theorem. Let $F : V \to W $ be linear and $A = (u_1, ..., u_r, v_1, ..., v_k)$ be a Basis of V with ker(F) = span$(v_1, ..., v_k).$ 
By definition of $U = span(u_1, ..., u_r)$ we get:
1) $ V = U \oplus Ker(F)$
2) The Restriction $F|U: U \to Im(F)$ is an isomorphism
3) If $P: V = U \oplus Ker(F) \to U, v = u + v' \to u$ is the projection, so $F = (F|U) \circ P$.
Thank you very much for any kind of help, since I'm a bit confused :)
Well what I did see is that I proofed $V = Ker(f) \oplus Im(f)$. So according to 1) it is $Im(f) = U$....
 A: I'll expand here my comment, and try to answer to your second question.
Let us consider the general case where $F : V\rightarrow W$ is a homomorphism, and not an endomorphism of $V$.


*

*If $V = U\oplus \ker(F)$, then $f_U : U\rightarrow W$ defined by $F_U(x) = F(x)$ is an injective homomorphism. Indeed, its kernel is $\{x\in U : F_U(x) = 0\} = \{x\in U : F(x) = 0\} = \{x\in U : x\in \ker(F)\} = U\cap \ker(F)$ and this intersection is trivial (i.e, it is $\{0\}$) by definition of the direct sum. Hence $F_U$ is injective.

*Considering $F_U$ as a mapping between $U$ and $\mathrm{im}(F)$. This means that we forget about vectors that are in $W$ but that are not reached by $F$. This mapping is still injective, and now it is surjective: if $y\in \mathrm{im}(F)$ then by definition there is $x\in V$ such that $F(x) = y$. Consider the decomposition of $x$ as $x = u + n$, where $u\in U$ and $n\in\ker(F)$ (this exists by definition of direct sum, and is unique). Then $F(u) = F(x) - F(n) = F(x) - 0 = F(x)$ so that $F(u) = y$ and $u\in U$, i.e $y$ has a preimage by $F_U$ and $F_U$ is surjective. Hence $F_U : U\rightarrow \mathrm{im}(F)$ is an isomorphism.
Hence it always holds that if $V=U\oplus \ker(F)$, then $U\simeq \mathrm{im}(F)$, and you can derive from this fact the rank theorem: $\dim(V) = \dim(\ker(F)) + \mathrm{rk}(F)$.
In your special case where $F$ is an endomorphism, you have furthermore that $\mathrm{im}(F)\subseteq V$ and $F^2 = F$, and only with these hypotheses it is possible to prove that $V = \mathrm{im}(F) \oplus \ker(F)$.
Hoping this will help you see clearer.
