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Let $0{\rightarrow} K \stackrel{g}\rightarrow P\stackrel{f}\rightarrow Q \rightarrow 0$ and $0\rightarrow K' \stackrel{g'}\rightarrow P'\stackrel{f'}\rightarrow Q \rightarrow 0$

be exact sequences of left $R$-modules. Assume that $P$ and $P'$ are projective.

How to prove that $P\oplus K'\cong P'\oplus K$ ?

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This result is known as Schanuel's lemma.

A quick proof is given by introducing the pullback of $f$ and $f'$. It is the submodule of $P\oplus P'$ given by $$X = \{(p,p')\in P\oplus P'\mid f(p) = f(p')\}.$$ Then, the following sequences are exact : $$0 \longrightarrow \ker(f')\simeq K' \longrightarrow X \longrightarrow P \longrightarrow 0$$ and $$0 \longrightarrow \ker(f)\simeq K \longrightarrow X \longrightarrow P' \longrightarrow 0,$$ where $X \longrightarrow P$ and $X \longrightarrow P'$ are the natural projections. Since $P$ and $P'$ are projective, these sequences are split and one have : $$K'\oplus P \simeq X \simeq K\oplus P'$$

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