I have a question about a step/ argument in the proof of Proposition 12 in Bosch's "Algebraic Geometry and Commutative Algebra" (page 352). Here the excerpt with red tagged argument:$\DeclareMathOperator{\Hom}{Hom}$
My question is:
Fix a ring $B$. If we have a sequence
(*)$$K \to L \to M \to 0$$
of $B$-modules why is (*) exact if and only if
the induced Hom-sequence
(**) $$0 \to \Hom_B(M, N) \to \Hom_B(L, N) \to \Hom_B(K, N) $$
for all $B$-modules $N$ is exact.
Indeed, the functor $\_B(-,N)$ is left exact so (*) $\Rightarrow$ (**) is clear.
What I don't understand is why if (**) is exact for all $N$ then (*) is also exact?
Best Answer
Write $K\xrightarrow{f}L\xrightarrow{g}M\to 0.$
Take $N=M,$ then $g^*(\text{id}_M)=g \in \text{im } g^*=\ker f^*,$ so $f^*g=gf=0.$
Take $N=L/\text{im} f.$ Let $\pi:L\to N$ be the projection. Then $f^*\pi=\pi f=0,$ so $\pi \in \text{im } g^*,$ i.e. $g^*h=hg=\pi$ for some $h.$ Then $\ker g\subset \ker \pi = \text{im } f.$
Take $N=M/\text{im } g.$ Let $\pi:M\to N$ be the projection. Then $g^*\pi=\pi g=0,$ so $\pi=0$ and therefore $M=\text{im }g.$