Solution: How does $s^{\prime}$ help us decompose $B$ into $A$ and $C$?

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In summary, $s^{\prime}$ acts as a bridge between $A$ and $C$ in decomposing $B$. It assists in the process by breaking down $B$ into smaller components and plays a crucial role in the overall solution. $s^{\prime}$ can be used to decompose other variables or equations, and its use contributes to scientific understanding and problem-solving by simplifying complex problems and providing a framework for analysis and solutions.
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Chris L T521
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Thanks to those who participated in last week's POTW! Here's this week's problem.

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Problem: Let $0\rightarrow A \xrightarrow{\phi}{} B \xrightarrow{\psi}{} C\rightarrow 0$ be an exact sequence of abelian groups. Let $s^{\prime}:B\rightarrow A$ be a homomorphism such that $s^{\prime}\circ \phi=1_A$. Show that $B\cong A\oplus C$.

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No one answered this question. Here's my solution.

Proof: Let $\phi:B\rightarrow A\oplus C$ be a homomorphism defined by $\phi(b)=(s^{\prime}(b),j(b))$.

Claim 1: $\phi$ is injective.

Proof of claim: Suppose we have
\[\phi(b_1)=(s^{\prime}(b_1),j(b_1))=(s^{\prime}(b_2),j(b_2))=\phi(b_2)\]
Thus $s^{\prime}(b_1)=s^{\prime}(b_2)$ and $j(b_1)=j(b_2)$. Now, note that
\[j(b_1)=j(b_2)\implies j(b_1)-j(b_2) = 0\implies j(b_1-b_2)=0\implies b_1-b_2\in\ker j = \text{Im}\,i\]
Thus, $\exists\,a\in A$ such that $i(a)=b_1-b_2$. Applying $s^{\prime}$ to both sides gives us
\[s^{\prime}(i(a))=s^{\prime}(b_1-b_2)\implies a=s^{\prime}(b_1)-s^{\prime}(b_2)=0\]
Thus, $a=0\implies i(a)=b_1-b_2=0 \implies b_1=b_2$. Therefore, $\phi$ is injective. Q.E.D.

Claim 2: $\phi$ is surjective.

Proof of claim: Suppose that $a\in A$ and $c\in C$. We seek to find $b\in B$ such that $\phi(b)=(a,c)$. Since $j$ is surjective, there exists a $b^{\prime}\in B$ (not necessarily unique) such that $j(b)=c$. Let us consider the element $i(a)+b^{\prime}-i(s^{\prime}(b^{\prime}))\in B$. Then it follows that
\[\begin{aligned}\phi(i(a)+b^{\prime}-i(s^{\prime}(b^{\prime}))) &= (s^{\prime}(b^{\prime}-i(s^{\prime}(b^{\prime}))),j(b^{\prime}-i(s^{\prime}(b^{\prime}))))\\ &= (s^{\prime}(i(a))+s^{\prime}(b^{\prime})-s^{\prime}(i(s^{\prime}(b^{\prime}))), j(i(a))+j(b^{\prime})- j(i(s^{\prime}(b^{\prime}))))\\ &= (a+s^{\prime}(b)-s^{\prime}(b), j(b^{\prime}))\\ &= (a,c)\end{aligned}\]
Thus, take $b=i(a)+b^{\prime}-i(s^{\prime}(b^{\prime}))$ and therefore $\phi$ is surjective. Q.E.D.

Claim 1 and Claim 2 implies that $\phi$ is an isomorphism and we have $B\cong A\oplus C$. Q.E.D.
 

FAQ: Solution: How does $s^{\prime}$ help us decompose $B$ into $A$ and $C$?

What is the purpose of $s^{\prime}$ in decomposing $B$ into $A$ and $C$?

The purpose of $s^{\prime}$ is to act as a bridge between $A$ and $C$ in the decomposition process. It allows for the transformation of $B$ into $A$ and $C$ by providing a common factor or component that connects them.

How does $s^{\prime}$ assist in the decomposition of $B$?

$s^{\prime}$ helps in the decomposition of $B$ by breaking it down into smaller, more manageable components of $A$ and $C$. By identifying the common factor or component, $s^{\prime}$ simplifies the process of decomposition and makes it easier to understand and analyze.

What role does $s^{\prime}$ play in the overall solution?

$s^{\prime}$ plays a crucial role in the overall solution as it enables the decomposition of $B$ into $A$ and $C$, which is essential for understanding and solving complex problems. Without $s^{\prime}$, it would be challenging to break down $B$ into its individual components and find a solution.

Can $s^{\prime}$ be used to decompose other variables or equations?

Yes, $s^{\prime}$ can be used to decompose other variables or equations, as long as there is a common factor or component that connects them. The concept of decomposition and using a common factor to break down a larger entity into smaller parts can be applied to various scientific and mathematical problems.

How does the use of $s^{\prime}$ contribute to scientific understanding and problem-solving?

The use of $s^{\prime}$ allows scientists to simplify complex problems and understand them better by breaking them down into smaller, more manageable components. It also aids in problem-solving by providing a framework for analyzing and approaching problems. By identifying and utilizing common factors, $s^{\prime}$ helps scientists find solutions more efficiently and effectively.

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