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suppose we are given a short exact sequence of $\mathbb{Z}G$-modules $$0\to K\to F\to A\to 0$$ where $F$ is free. and we form a diagram with that first row and with a second row $0\to L\to M\to N\to 0$ short exact but with no special condition and vertical arrows $\alpha\colon K\to L$, $\beta\colon F\to M$ and $\gamma\colon A\to N$. (Sorry about the horrible way of writing the diagram but I don't know how to write it without the xy package!) Suppose also that $\beta$ is an isomorphism. Could we say that $\gamma$ is an isomorphism as well? Could we use the five lemma to say so, or should we necessarily assume that $\alpha$ is an isomorphism too? Thank you in advance, bye!

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How about $K=F=M=N = \mathbb{Z}G$ and $A=L=0$ for the first question? – t.b. Jun 5 '11 at 10:29
oh that's right! How could I be so stupid!? Well... thanks a lot! And what about if I take first row $0\to F_0\to F_1\to A$ and second row $0\to F_0\to F_1\to N$ dropping in this way the surjectivity but requiring both $F_0$ and $F_1$ to be free? In this case I would have two isomorphisms $\alpha$ and $\beta$ and by diagram chasing I should be able to prove that $\gamma$ is an isomorphism. Is it right? Does any module $A$ admit an exact sequence $0\to F_0\to F_1\to A $? ìf I'm not wrong, this should mean that every module has a submodule isomorphic to the quotient of two free modules. – fatoddsun Jun 5 '11 at 11:08
up vote 1 down vote accepted

As I indicated in a comment, the first question has a negative answer, as shown by a silly example. The second one has a positive answer, though, as is easy to check (and it suffices to assume that $\alpha$ is epi and there's no need to assume that $F$ is free).

For the fun of it, let me mention an overkill way of doing it:

The sharp four lemma states:

If the rows of $$\begin{array}{ccccccc} A_1 & \to & A_2 & \to & A_3 & \to & A_4 \\\ \downarrow & & \downarrow & & \downarrow & & \downarrow \\\ B_1 & \to & B_2 & \to & B_3 & \to & B_4 \end{array}$$ are exact, $A_1 \to B_1$ is epi and $A_2 \to B_2$ and $A_4 \to B_4$ are isos then $A_3 \to B_3$ is mono. The proof is a simple diagram chase.

From this and its dual — or by a direct diagram chase — you can easily conclude the sharp five lemma: If the rows of $$\begin{array}{ccccccccc} A_1 & \to & A_2 & \to & A_3 & \to & A_4 & \to & A_5\\\ \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow\\\ B_1 & \to & B_2 & \to & B_3 & \to & B_4 & \to & B_5 \end{array}$$ are exact, $A_1 \to B_1$ is epi, $A_5 \to B_5$ is mono and $A_2 \to B_2$ and $A_4 \to B_4$ are isos then $A_3 \to B_3$ is an iso.

Now let $A_1 \to B_1$ be $\alpha$, $A_2 \to B_2$ be $\beta$ and let $A_4 = A_5 = B_4 = B_5 = 0$ to get the desired conclusion that $\gamma$ is an isomorphism.

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