Is it true that the sequence $ M \to N \to P $ of $A$-modules is exact if the induced sequence $$\mathrm{Hom}_{A}(F, M) \to \mathrm{Hom}_{A}(F,N) \to \mathrm{Hom}_{A}(F,P) $$ and/or the sequence $$\mathrm{Hom}_{A}(P,F) \to \mathrm{Hom}_{A}(N, F) \to \mathrm{Hom}_{A}(M, F)$$ is exact for any $A$-module $F$?
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1Related: http://mathoverflow.net/questions/119680 – Martin Brandenburg Apr 12 '13 at 21:55
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1Related: http://math.stackexchange.com/questions/266241/hom-functor-and-left-exactness and http://math.stackexchange.com/questions/235372/homomorphisms-and-exact-sequences – Apr 13 '13 at 00:57
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Yes. In fact, it suffices to take $F = A$, since we have a natural isomorphism $\text{Hom}_A(A, M) \cong M$.
Qiaochu Yuan
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Consider the sequence (S) $M \xrightarrow{f} N \xrightarrow{g} P$. Let's see what the condition
(S2) For all $A$-modules $T$, the sequence $\hom(P,T) \xrightarrow{g^*} \hom(N,T) \xrightarrow{f^*} \hom(M,T)$ is exact.
tells us. For $T=P$ we see $0=(f^* g^*) (\mathrm{id}_P)=gf$. Thus, $\mathrm{im}(f) \subseteq \ker(g)$. Then $g$ induces a homomorphism $\overline{g} : Q:=N/\mathrm{im}(f) \to P$ with kernel $\ker(g)/\mathrm{im}(f)$. Then (S2) says that $\hom(P,T) \xrightarrow{\overline{g}^*} \hom(Q,T)$ is surjective for all $T$. By naturality (or the Yoneda Lemma), this reduces to the case $T=Q$ and that $\mathrm{id}_Q$ has a preimage. In other words, $\overline{g}$ has to be a split monomorphism. In particular, its kernel is $0$ and therefore $\ker(g)=\mathrm{im}(f)$. Besides, $\overline{g}$ identifies with $N/\mathrm{ker}(g) \cong \mathrm{im}(g) \hookrightarrow P$.
Summary: (S2) means that (S) is exact and $\mathrm{im}(g)$ is a direct summand of $P$.
Martin Brandenburg
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