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

Question

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.

Answer 1

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 $\ker s \sim F_0 / M \sim P$.

Answer 2

We have the exact sequence $$ 0 \to M \xrightarrow{\iota_M} F_0 \to P \to 0, $$ and we need that $s: F_0\to M$ splits this exact sequence. But $M=\pi(F_1)$, and $F_1\to F_0 = \iota_M\circ \pi$. Thus the fact that $s\circ (F_1\to F_0) = \pi$ tells us that $s\circ \iota_M\circ \pi = \pi$. However $\pi$ is surjective, so we can cancel it from the right to get $s\circ \iota_M = \mathrm{id}$. In other words $s$ splits the exact sequence.

Since $s$ splits the exact sequence, it is the projection from $P\oplus M \to M$, so its kernel is $P$.

Answer 3

The key word is "splitting lemma." If you have a direct sum $A \oplus B$ then there is an exact sequence

$$ 0 \to A \to A \oplus B \to B \to 0$$

and there are also maps $A \oplus B \to A$ (projection) and $B \to A \oplus B$ (inclusion).

The splitting lemma says (roughly) that in an exact sequence $0 \to A \to C \to B \to 0$ we have $C \cong A \oplus B$ if and only if there is a projection map $C \to A$ if and only if there is an inclusion map $B \to C$.

In the proof, we have an exact sequence $$ 0 \to M \to F_0 \to P \to 0 $$ and a projection map $s : F_0 \to M$. "Projection," by the way, means that $s \circ (M \to F_0) = {\rm Id}_{M}$. It follows from the splitting lemma that $F_0 \cong P \oplus M$.

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What this proves is an important fact: $\DeclareMathOperator{\Ext}{Ext}\Ext^1_R(P, M) = 0$ if and only if every short exact sequence $0 \to M \to F \to P \to 0$ is split (where split means $F \cong M \oplus P$).

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The other way to look at is that for every short exact sequence $0 \to A \to B \to C \to 0$, the Ext functor turns it into a long-exact sequence

$$\DeclareMathOperator{\Hom}{Hom} 0 \to \Hom_R(P, A) \to \Hom_R(P, B) \to \Hom_R(P, C) \to \Ext^1_R(P, A) \to \Ext^1_R(P,B) \to \cdots. $$

If $\Ext^1_R(P,A)$ is always $0$ then $\Hom_R(P, -)$ is an exact functor:

$$ 0 \to \Hom_R(P, A) \to \Hom_R(P, B) \to \Hom_R(P, C) \to 0. $$