On proposition I.1.2 of "Quantum Groups" by Christian Kassel I am working through Christian Kassel's textbook on Quantum Groups. The Proposition states that 5 statemens are equivalent. The two I am having trouble with are as follows.
1.For any pair $V'\subset V$ of finite-dimensional $A$-modules, there exists an $A$-module $V''$ such that $V\cong V'\oplus V''$.
3.For any pair $V'\subset V$ of finite-dimensional $A$-modules, there exists an $A$-linear map $p:V\rightarrow V'$ with $p^2=p$.
In the proof it has the following for 3 implies 1: Let $V''=\operatorname{Ker}(p)$; it is a submodule of $V$. The relations $v=p(v)+\big(v-p(v)\big)$ and $p^2=p$ prove that $V$ is the direct sum $V''$ and $V'$.
I am completely lost as to how those relations prove the direct sum. Any help would be greatly appreciated.
 A: In this answer, $V'$ and $V$ are not necessarily finite dim.  In fact $A$ can be any ring (i.e., not necessarily an algebra over any field, and not necessarily unital).  That is the equivalence of the two statements is as follow.  

Theorem. Let $V$ be a module over a ring $A$ with a submodule $V'$.  Then, $V'$ is a direct summand of $V$ (that is, $V=V'\oplus V''$ for some submodule $V''$ of $V$) if and only if there exists a projection $p:V\to V$ such that $\operatorname{im}p=V'$.  (A projection on $V$ is an $A$-module endomorphism $p:V\to V$ such that $p^2=p$.)

Remark. The OP's version of this theorem is false.  If $p$ is not required to be surjective, we have a counterexample, e.g., when $V'$ is a non-zero submodule of $V$ with $p=0$.)
Proof. Suppose that $V'$ and $V$ are $A$-modules such that $V'\subseteq V$ and there exists a projection $p:V\to V$ with $V'=\operatorname{im}p$.  We will show that $V=V'\oplus V''$ where $V''=\ker p$.  
First, $V=V'+V''$.  To see this, we note that $v=p(v)+\big(v-p(v)\big)$.  Clearly, $p(v)\in V'$.  We must prove that $v-p(v)\in V''$, which is equivalent to $p\big(v-p(v)\big)=0$.  But $$p\big(v-p(v)\big)=p(v)-p^2(v)=p(v)-p(v)=0$$ by the hypothesis that $p^2=p$.  So $v-p(v)\in \ker p = V''$.
To finish the prove we must show that the sum $V=V'+V''$ is direct.  That is $V'\cap V''=\{0\}$.  Suppose that $w\in V'\cap V''$.  Since $w\in V'=\operatorname{im}p$, we must have $w=p(v)$ for some $v\in V$.  Since $w\in V''=\ker p$, $p(w)=0$.  Therefore,
$$0=p(w)=p\big(p(v)\big)=p^2(v)=p(v)=w,$$
where we use again the hypothesis that $p^2=p$.  This proves that $V'\cap V''=\{0\}$.  Therefore, $V$ is the internal direct sum $V\oplus V''$. $\square$

Corollary.  For a ring $A$, the following statements are equivalent.

(a) For any pair $(V,V')$ of $A$-modules such that $V'\subseteq V$, there exists an $A$-submodule $V''$ of $V$ s.t. $V$ is the internal direct sum $V'\oplus V''$.
(b) For any pair $(V,V')$ of $A$-modules such that $V'\subseteq V$, there exists a surjective $A$-linear map $p:V\to V$ with  $\operatorname{im} p =V'$ and $p^2=p$.

If $A$ is unital, then we have another equivalent statement.

(c) Any $A$-module is semisimple.



If you are curious about the whole statement in Kassel's book, I put it below.  However, I would like to note that the proposition is not true.  My correction will be in bold italic font.  In this setting, $A$ is an algebra over a field $k$, and an $A$-module means a left $A$ module.

Proposition I.1.2.   The following statements are equivalent.

(i) For any pair $V'\subseteq V$ of finite dim $A$-modules, there exists an $A$-module $V''$ s.t. $V\cong V'\oplus V''$.
(ii) For any pair $V'\subseteq V$ of finite dim $A$-modules where $V'$ is simple, there exists an $A$-module $V''$ such that $V\cong V'\oplus V''$.
(iii) For any pair $V'\subseteq V$ of finite dim $A$-modules, there exists a surjective $A$-linear map $p:V\to V'$ with $p^2=p$. 
(iv) For any pair $V'\subseteq V$ of finite dim $A$-modules where $V'$ is simple, there exists a surjective $A$-linear map $p:V\to V'$ with $p^2=p$.
(v) Any finite dim $A$-module is semisimple.


Note that (i) and (iii) are equivalent in the general setting as well.  See the corollary after the proof of the theorem above.
