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Show that if $0\rightarrow N\xrightarrow{\tau} P\xrightarrow{\epsilon} A\rightarrow 0$ and $0\rightarrow M\xrightarrow{\tau'} Q\xrightarrow{\epsilon'} A\rightarrow 0$ are exact sequences with $P,Q$ being projective modules then $P\oplus M\cong Q\oplus N$.

Consider surjective map $\epsilon$, as $Q$ is projective module and we have map $\epsilon'$ there exists $\eta:P\rightarrow Q$ such that $\epsilon'\circ \eta=\epsilon$

Moreover, $\epsilon'\circ (\eta\circ\tau)=(\epsilon'\circ \eta)\circ\tau=\epsilon\circ\tau=0$. Thus, $\eta\circ\tau$ is in the kernel of $\epsilon'$ and thus, has to be in the image of $\tau'$. So, we have a map $\eta':N\rightarrow M$ such that $\tau'\circ\eta'=\eta\circ\tau$

We then have a following diagram

$$\begin{array}{c} 0 & \rightarrow & N & \xrightarrow{\tau} & P & \xrightarrow{\epsilon} & A & \rightarrow & 0 \\ \downarrow{} & & \downarrow{\eta'} & & \downarrow{\eta} & & \downarrow{id} \\ 0 & \rightarrow & M & \xrightarrow{\tau'} & Q & \xrightarrow{\epsilon'} & A & \rightarrow & 0\\ \end{array}$$

From an exercise in Hilton Stammbach, Chapter I, exercise $3.4$ as map in last column is an isomorphism, we have a short exact sequence

$$0\rightarrow N\rightarrow P\oplus M\rightarrow Q\rightarrow 0$$

Now, as $Q$ is projective module, any surjective map onto $Q$ splits, so above short exat sequence is a split exact sequence, we then have $P\oplus M=Q\oplus N$.

Let me know if this is correct.

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  • $\begingroup$ The identity $\varepsilon \circ \gamma \circ \eta = \varepsilon$ does not imply $\gamma \circ \eta = \text{id}$, since $\varepsilon$ is not injective in general. In a correct proof, you will not be able to show that $P$ and $Q$ are isomorphic, as for example any exact sequence $0 \to N \to P \to A \to 0$ gives you an exact sequence $0 \to N \oplus P \to P \oplus P \to A \to 0$. $\endgroup$ – Matthias Klupsch Jul 28 '16 at 15:51
  • $\begingroup$ You can't cancel on the left in the equation $\epsilon \circ \gamma \circ \eta =\epsilon$ because $\epsilon$ is not surjective. When you have two projective resolutions, you generally don't have isomorphism between the resolutions, but you always have chain homotopy equivalence. $\endgroup$ – Noah Olander Jul 28 '16 at 15:57
  • $\begingroup$ Ok. Thanks to both comments, I was thinking i was using some thing more than that is given and i got it now. I will try to do in another way. Thanks @MatthiasKlupsch $\endgroup$ – user311526 Jul 28 '16 at 15:58
  • $\begingroup$ @NoahOlander : Chain homotopy equivalence is the right word as you said. I got some idea. I will use that. Thanks. $\endgroup$ – user311526 Jul 28 '16 at 15:59
  • $\begingroup$ @NoahOlander : Let me know if the edited content is correct. $\endgroup$ – user311526 Jul 28 '16 at 17:42

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