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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 ?

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    $\begingroup$ Which definition do you have for projective modules? $\endgroup$
    – Bernard
    Dec 25, 2020 at 18:07
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    $\begingroup$ And what is a ‘surjective module’? $\endgroup$
    – Bernard
    Dec 25, 2020 at 18:14

2 Answers 2

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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 $$b-s\bigl(g(b)\bigr)=f(a)\iff b= f(a)+s\bigl(g(b)\bigr).$$

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  • $\begingroup$ Can you answer my above two question $(2)$ ? $\endgroup$
    – MAS
    Dec 25, 2020 at 18:44
  • $\begingroup$ Oh!yes.I definitely should reread myself before posting… Thanks! $\endgroup$
    – Bernard
    Dec 25, 2020 at 18:44
  • $\begingroup$ I think this $s$ is so-called $\text{section}$ map $\endgroup$
    – MAS
    Dec 25, 2020 at 18:47
  • $\begingroup$ 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 $A$. $\endgroup$
    – Bernard
    Dec 25, 2020 at 18:49
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    $\begingroup$ @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. $\endgroup$ Dec 25, 2020 at 19:05
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$\DeclareMathOperator{\im}{im}\DeclareMathOperator{\id}{id}$(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 $B = \im(s) \oplus \ker(g)$.

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  • $\begingroup$ Thanks indeed, for your nice answer $\endgroup$
    – MAS
    Dec 26, 2020 at 6:21

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