Why does the short exact sequence for projective module split?

I want to understand the definition of surjective module in terms of splitting sequence. The definition says for a projective $$R$$-module $$P$$, the following short exact sequence $$0 \to A \xrightarrow{f} B \xrightarrow{g} P \to 0$$ splits, where $$A,B$$ are also $$R$$-modules.

I want to see why $$B \cong A+P$$ ?

The Wikipedia, says, then there is a section map $$h:P \to B$$ such that $$gh=1_P$$.

$$(1)$$ Why is so ?

This then says $$B=\text{Im}(h)\oplus \text{ker}(g)$$.

$$(2)$$ why is so ?

I am trying in the following way:

Since the given sequence is short exact sequence, the map $$g$$ is surjective. This means that $$\text{Im}(g)=P$$.

Now as $$P$$ is surjective module any module homomorphism factors through an epimorphism to $$B$$ i.e., for any $$R$$-module $$C$$ there exists an epimorphism (surjective module homomorphism) $$i: C \twoheadrightarrow B$$ and a module homomorphism $$j: P \to C$$ such that $$ij=g.$$ I can't go further. What is the section map $$h$$ here ?

Any explanation of above two questions ?

• Which definition do you have for projective modules? Dec 25, 2020 at 18:07
• And what is a ‘surjective module’? Dec 25, 2020 at 18:14

Consider the diagram: \begin{align} &P \\ &\Big\|\operatorname{id_P} \\[-3ex] 0\longrightarrow A\xrightarrow{\enspace f\enspace} B \xrightarrow{\enspace g\enspace}&P\longrightarrow 0 \end{align} As $$P$$ is projective, and $$g$$ is onto, the map $$\operatorname{id}_P$$ factors through $$B$$, i.e. there exists a map $$s:P\longrightarrow B$$ such that $$g\circ s=\operatorname{id}_P$$.

Some more details: Any element $$b\in B$$ can be written as the sum of an element in $$f(A)$$ and an element in $$s(P)$$:

Indeed you easily check that $$g\bigl(b-s(g(b))\bigr)=g(b)-(g\circ s)\bigl(g(b)\bigr)=g(b)-g(b)=0$$, so there exists a (unique because $$f$$ is injective) $$a$$ such that $$b-s\bigl(g(b)\bigr)=f(a)\iff b= f(a)+s\bigl(g(b)\bigr).$$

• Can you answer my above two question $(2)$ ?
– MAS
Dec 25, 2020 at 18:44
• Oh!yes.I definitely should reread myself before posting… Thanks! Dec 25, 2020 at 18:44
• I think this $s$ is so-called $\text{section}$ map
– MAS
Dec 25, 2020 at 18:47
• The property of $s$ in my answer says it is a section of $g$ by definition. Also there results that $f$ has a retraction $r:B\longrightarrow A$, i.e. $r\circ f=\operatorname{id}_A$, and $B$ is isomorphic to the direct sum of (the image of) $P$ and the image of $A$. Dec 25, 2020 at 18:49
• @Masmath Once you have a section $s$ you can write any element $b\in B$ uniquely as $b=s(g(b)+(b-s(g(b))$. The two summands are in the image of $s$ and in the kernel of $g$ respectively. Dec 25, 2020 at 19:05

$$\DeclareMathOperator{\im}{im}\DeclareMathOperator{\id}{id}$$(2): Suppose $$y = s(x) \in \im(s) \cap \ker(g)$$. Then on the one hand $$g(y) = 0$$ and on the other hand $$g(y) = g(s(x)) = \id_P(x) = x$$ so $$x = 0$$. Therefore $$\im(s) \cap \ker(g) = \{0\}$$.

Next, suppose $$y \in B$$ and let $$x = g(y)$$, $$z = s(x)$$ and $$z' = y - z$$. Then

1. $$y = z + z'$$
2. $$z \in \im(s)$$
3. $$g(z') = g(y) - g(s(x)) = x - \id_P(x) = 0$$ so $$z' \in \ker(g)$$

Thus $$B = \im(s) \oplus \ker(g)$$.