Consider two exact sequences $0\rightarrow N\rightarrow P\rightarrow A\rightarrow 0$ and $0\rightarrow M\rightarrow Q\rightarrow A\rightarrow 0$, where $P,Q$ are projective modules. The exercise(pg 26, 4.5) asked me to prove that $P\oplus M\cong Q\oplus N$. I do not know how to proceed.
Certainly the map $P\rightarrow A$ is surjective, and since we have the trivial map $A\rightarrow A$identify $A$ with itself, any map $Q\rightarrow A$ can be extended to the composition $Q\rightarrow P\rightarrow A$. Similarly any map $Q\rightarrow A$ can be extended to $P\rightarrow Q\rightarrow A$. I do not know how to proceed further from here; certainly the fact that any map $P\rightarrow A$ can be factored through $Q$ as $P\rightarrow Q\rightarrow P\rightarrow A$ is interesting, but this does not help me to prove the statement.
The author give the hint that I should look over Exercise 3.4(page 22), but I found it to be quite irrelevant as 3.4 was concerning a commutative diagram of two exact sequences, while this problem does not admit any arrow from $N$ to $M$ as only $P$ and $Q$ are supposed to be projective. So I got stuck. I think I need some help as the question is quite trivial.
Another trivial question is assume $A$ to be a finite abelian group, prove $\operatorname{Hom}_{\mathbb{Z}}(A,\mathbb{Q}/\mathbb{Z})$ is isomorphic to $A$. This would be trivial if I can prove it by $\mathbb{Z}$ and $\mathbb{Z}_{p^{n}}$ respectively. But plainly I do not see how $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Q}/\mathbb{Z})$ can be cyclic, since any such $\mathbb{Z}$ homomorphism must map $1$ to some element in $\mathbb{Q}/\mathbb{Z}$, and such choice can be totally arbitrarily among $\mathbb{Q}/\mathbb{Z}$. But we know that $\mathbb{Q}/\mathbb{Z}$ is not finitely generated, and it cannot be cyclic. I do not know where my reasoning got wrong, so I hope someone may help me point out my mistake and give me a simple proof on this.