# Problem of the Week #4 - June 25th, 2012

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#### Chris L T521

##### Well-known member
Staff member
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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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
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.

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