Dual sequence and its exactness

Question

I am reading Lang's Algebra and trying to fill in the gaps in my mathematical background while I train for the quals. So I came across the following exercise (chapter 20, ex. 26 in the third edition).

$A$ is a commutative ring, $E$ an $A$-module and its dual is $E^*=\mathrm{Hom}_A(E; \mathbb Q/\mathbb Z)$.

It is asked to prove that a short sequence \begin{equation} 0 \rightarrow N \rightarrow M \rightarrow E \rightarrow 0 \end{equation} is exact iff the dual sequence \begin{equation} 0 \rightarrow E^* \rightarrow M^* \rightarrow N^* \rightarrow 0 \end{equation} is exact.

It is not hard to prove that the dual sequence is exact if the first one is, using the arguments Lang shows, in particular the injectivity of the module $\mathrm{Hom}_A(A; \mathbb Q/\mathbb Z)$. The vice versa turns out to be harder (at least, I can't find a clear way for that), besides the easy proof that $ N \rightarrow M$ is injective, thanks to the canonical embedding of a module $M$ in its bidual $M^{**}$. Any hint or reference? Also any interesting reading on the topics of injective and projective module, resolutions, etc, will be appreciated.

Thanks a lot.

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Edit (not by OP): This is also Exercise 3.2.4 from Weibel's Introduction to Homological Algebra.

Answer 1

$\DeclareMathOperator{\im}{im}\DeclareMathOperator{\Hom}{Hom}\newcommand{\C}[1]{\mathbf{#1}}$This is also Exercise 3.2.4 from Weibel's Introduction to Homological Algebra. The statement there is a bit stronger. Namely:

Exercise. Show that a sequence $A \xrightarrow{f} B \xrightarrow{g} C$ is exact iff its dual $C^{\ast} \to B^{\ast} \to A^{\ast}$ is exact.

There is one lemma that we use without proof (also a standard exercise).

Lemma. An abelian group $A$ is zero iff $\Hom_{\C{Ab}}(A, \mathbb{Q}/\mathbb{Z}) = 0$.

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Solution of Exercise. The direction $\Rightarrow$ is clear since $\Hom_{\C{Ab}}(-, \mathbb{Q}/\mathbb{Z})$ is exact because $\mathbb{Q}/\mathbb{Z}$ is injective.

For the converse, suppose that $C^{\ast} \to B^{\ast} \to A^{\ast}$ is exact.
Thus, we have $\psi \circ g \circ f = 0$ for all $\psi \in \Hom_{\C{Ab}}(C, \mathbb{Q}/\mathbb{Z})$. This implies that $g(f(A)) = 0$. Indeed, otherwise by the lemma there would be a nonzero map $g(f(A)) \to \mathbb{Q}/\mathbb{Z}$. Moreover, we could lift to a map $C \to \mathbb{Q}/\mathbb{Z}$ using injectivity. Thus, we have $\im(f) \subset \ker(g)$.

For the sake of contradiction, assume that the quotient $\ker(g)/\im(f)$ is nonzero. By the lemma, there exists a nonzero map $\ker(g)/\im(f) \to \mathbb{Q}/\mathbb{Z}$. Lifting this, we get a map $\ker(g) \to \mathbb{Q}/\mathbb{Z}$ which kills $\im(f)$. By injectivity, we can further lift this to a map $\varphi : B \to \mathbb{Q}/\mathbb{Z}$. Since this is a lift, we continue to have $f(A) \subset \ker(\varphi)$.
This means that $f^{\ast}(\varphi) = 0$. By exactness, there exists $\psi \in C^{\ast}$ such that $g^{\ast}(\psi) = \varphi$, i.e., $\psi \circ g = \varphi$. But this means that $\varphi$ must vanish on $\ker(g)$, contradicting our construction. $\qquad \square$