What does it mean for the sequence $\text{Hom}(E,\text{Hom}(N,P))$ to be exact, where $E$ is an exact sequences of $A$-modules, and $N,P$ $A$-modules?

Question

Here $E$ is the exact sequence $M^{\prime} \stackrel{f}{\rightarrow} M \stackrel{g}{\rightarrow} M^{\prime \prime} \rightarrow 0$. I know what an exact sequence of $A$-modules is, but not what $\text{Hom}(E,\text{Hom}(N,P))$ means and what it means for it to be exact.

Answer 1

For any two $A$-modules $M, N$, the set $\operatorname{Hom}(M, N)$ of $A$-homomorphisms is an $A$-module again (via $(\phi + \psi)(m) = \phi(m) + \psi(m)$ and $(a\phi)(m) = a \phi(m)$ as usual).

Also, if you have an $A$-homomorphism $f : M' \to M$, you get an $A$-homomorphism $f^* : \operatorname{Hom}(M, N) \to \operatorname{Hom}(M', N)$ via $f^*(\phi) = \phi \circ f$. (Check!) This is also written $\operatorname{Hom}(f, N)$ occasionally. (Also note that the order in which $M$ and $M'$ appear is different for $f$ and $f^*$, one says that $\operatorname{Hom}(\bullet, N)$ is a contravariant functor.)

Given your modules $N, P$ and your sequence $E$, we can first apply the above construction to get a new $A$-module $\operatorname{Hom}(N, P)$ and then apply the construction again to this and each module in the sequence $E$ to get modules $\operatorname{Hom}(M', \operatorname{Hom}(N, P))$ and so on. We can also apply it to the homomorphisms $f, g$ to get $$ 0 \to \operatorname{Hom}(M'', \operatorname{Hom}(N, P)) \xrightarrow{g^*} \operatorname{Hom}(M, \operatorname{Hom}(N, P)) \xrightarrow{f^*} \operatorname{Hom}(M', \operatorname{Hom}(N, P)) $$ which is again a sequence of $A$-modules. This is what is meant by $\operatorname{Hom}(E, \operatorname{Hom}(N, P))$.

As it is a sequence of $A$-modules, the usual definition of “exactness” applies. (You will not need that the second argument is $\operatorname{Hom}(N, P)$; this form of exactness will hold with any $A$-module in the second position.)