Q:

Understanding proof that $\operatorname{Ext}^1(P, M)=0\; \forall M$ implies $P$ is projective

The claim is, that if $\operatorname{Ext}^1_R(P, M) =0$ for any R-module M, then P is a projective module. 
I found a proof here, but I don't understand what happens after the word "clearly". 

Assume $\operatorname{Ext}^1_R(P, M)=0$ for any $R$-module $M$. Pick a free resolution $F_{*} \to P$. Set $M = \operatorname{im}(F_1 \to F_0) = \ker (F_0 \to P)$. Consider an element $\xi \in \operatorname{Ext}^1_R(P, M)$ given by the class of the quotient map $\pi: F_1 \to M$. Since $\xi$ is zero, there exists a map $s: F_0 \to M$ such that $\pi = s \circ (F_1 \to F_0)$. Clearly, this means that 
  $$ F_0 = \ker(s) \oplus \ker(F_0 \to P) = P \oplus \ker(F_0 \to P) $$ and so $P$ is projective. 

So, I have two questions: firstly, why $F_0$ decomposes in a direct sum?  I can only see that $F_0 / \ker (s) = M = \ker(F_0 \to P)$ (since surjectivity of $\pi$ implies that of $s$)
And secondly, why is $\ker(s) = P$?  $P$ and $s$ seem to be unrelated. 

A:

You have that $\pi = s \circ (F_1 \rightarrow F_0)$. Since $F_1 \rightarrow F_0$ acts in the same way as $\pi$ we have that $s$ acts identically on $M$. That means that $F_0 = M \oplus \ker s$. Now we need to prove that $\ker s$ is canonically isomorphic to $P$. That is obvious since $M = \ker (F_0 \rightarrow P)$ and $\ker s \sim F_0 / M \sim P$.

