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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.

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    The hint is not irrelevant at all: look up the proof of [Schanuel's lemma](http://en.wikipedia.org/wiki/Schanuel's_lemma) and find the short exact sequence they're talking about... A LaTeX thing: don't write `Hom$_{\mathbb{Z}}(...)$` rather use `$\operatorname{Hom}_{\mathbb{Z}}$`2011-12-22
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    The content of the exercise *is* Schanuel's lemma :)2011-12-22
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    By projectivity of $P$ you have a homomorphism $P\to Q$. Then you can always find a homomorphism $N\to M$ and thus extend the diagram while preserving the commutativity.2011-12-22
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    @Damian: Well, that's true but it won't help proving Schanuel's lemma.2011-12-22
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    Okay, this extension homomorphism $\alpha':N\to M$ can be found as follows. Let $x\in N$. Then $\mu(x)\in\text{ker}(\epsilon)$, where $\mu:N\to P,\epsilon:P\to A$. By commutativity of the diagram: $\alpha(\mu(x))\in\text{ker}(\epsilon')$, where $\alpha:P\to Q,\epsilon':Q\to A$. So there is a unique $y\in M: \mu'(y)=\alpha(\mu(x))$, where $\mu':M\to Q$. Now take $\alpha'(x):=y$. You should check that $\alpha'$ is a homomorphism.2011-12-22
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    @commenter: it will. There is a hint to use another exercise which joins these two exact sequences with another one, from which the Schanuel's lemma follows easily.2011-12-22
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    @Damian: no, that's not what the hint is about. You should take the pull-back of $P \to A$ and $Q \to A$. Then you get a commutative square with corner $X$. The two maps $X \to P $ and $X \to Q$ are both surjective, hence they split and their kernels are $M$ and $N$, respectively (that's what the exercise says). This then shows that $X$ is isomorphic to $P \oplus M$ and $Q \oplus N$.2011-12-22
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    @commenter: are you talking about hints and exercises from the book of Hilton & Stammbach? I am just right now looking into the book and the exercise is about obtaining some concrete exact seqence from these two beginning exact sequences (ex. 4.5 with the hint: ex. 3.4 following 3.3). But in general it is similar to your way of reasoning.2011-12-22
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    @Damian: I am, but I mistook exercise 3.3 for exercise 3.4, that's why I was so insistent. Now I understood how your argument works, nice! I haven't seen that way of putting it, I always did it the other way around. Apologies for the somewhat harshly formulated previous comments...2011-12-22
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    @commenter: it's ok, they weren't harsh. Greetz!2011-12-22
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    For the second problem: $\mathbb{Z}$ is not a finite abelian group, so the conclusion doesn't need to hold for it (in fact, $\operatorname{hom}_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Q/Z})$ is isomorphic to $\mathbb{Q/Z}$). As to how to prove the result, note that $\operatorname{hom}$ commutes with finite coproducts (*technically the result is a product, but products/coproducts coincide in the finite case...). Coupling this with the structure theorem for finite abelian groups reduces the problem to considering only cyclic groups (of prime power order).2011-12-22
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    @Riley E: It is embarassing that I ignored the "finite" condition in the statement..thank you.2011-12-23
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    @ChangweiZhou: A cool thing to note is that, although the finiteness condition is necessary for the exercise to be true, ignoring it _does_ lead to some interesting things. For example, $\operatorname{hom}_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Q/Z}) \simeq \mathbb{R}$. More generally, $\operatorname{hom}_{\mathbb{Z}}(-,\mathbb{Q/Z})$ tells a lot about $\operatorname{Ext}_{\mathbb{Z}}(-,\mathbb{Z})$ (cf. section III.6)2011-12-23
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    @Riley E: I can see the cardinality of Hom$_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})$ be $c$, but I could not tell why it is isomorphic to $\mathbb{R}$. I have not reached section III yet..:(2011-12-24
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    @ChangweiZhou: The (unfortunately unsatisfying) reason is that $\operatorname{hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q/Z})$ is a $\mathbb{Q}$-module (it's both divisible and torsionfree) that has cardinality $c$.2011-12-24

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