hom and exact sequence

Question

Let $$ 0 \longrightarrow \operatorname{Hom}(M,Β_1) \stackrel{f^*}\longrightarrow \operatorname{Hom}(M,Β) \stackrel{g^*}\longrightarrow \operatorname{Hom}(M,Β_2) $$ be an exact sequence for any module $M$. I must to prove that $$ 0 \longrightarrow Β_1 \stackrel f\longrightarrow Β \stackrel g\longrightarrow Β_2 $$ is exact. I have difficulties to understand the point $\ker g \subseteq \operatorname{Im}f$: Let $ M=\ker g$, and $σ$ embedding of $\ker g$ to $B$. Then $g^*(σ)=gσ=0$ so $σ \in \ker g^*=\operatorname{im}f^* $, so for $ k \in \operatorname{Hom}(M,Β_1)$, $ f^*(k)=σ$ and $fk=σ$. Ι cant understand why $\ker g=\operatorname{im}σ \subseteq \operatorname{im}f$. Any help ...

Answer 1

With $M=\ker g$ and the canonical inclusion $\sigma\in\operatorname{Hom}(M,B)$, we have $g^*(\sigma)=g\circ \sigma=0$, hence $\sigma = f^*(k)$ for suitable $k\in\operatorname{Hom}(M,B_1)$. That is, $\sigma = f\circ k$. Thus for any $b\in B$ with $g(b)=0$ we have $b=\sigma(b)=f(k(b))\in\operatorname{im} f$.