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Let $R$ be a ring. Let's say we have an exact sequence of $R$-modules $$0\rightarrow P\rightarrow R^2 \overset{f}\rightarrow R\rightarrow 0,$$

where $P\cong\ker(f)$.

Because of $R$ beeing a free $R$-module, the sequence splits and hence we have an $R$-homomorphism $g:R\rightarrow R^2$ such that $f\circ g=1_R$. Further we know that $P\oplus R\cong R^2$ and this means that $P$ is a stably free module of type $1$.

In the paper (page 779) I am reading it says that $P=\; $Im$(1_{R^2}-g\circ f)$.

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$Im(1 - gf) = \ker(f)$.

$f((1 - gf)(x)) = f(x) - fgf(x) = 0$ so $Im(1 - gf) ⊆ \ker(f)$.

On the other hand if $x ∈ \ker(f)$ then $(1 - gf)(x) = x$ so $\ker(f) ⊆ Im(1 - gf)$.

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  • $\begingroup$ Thanks! The first inclusion was already clear to me but I still did't get the second one. While reading this I also get the second one :) $\endgroup$ – Heffalump Aug 22 '13 at 23:33
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Assume $x \in R$, then $f( x -g \circ f(x)) = f(x) - f\circ g \circ f(x)$. Note $f \circ g$ is the identity on $R$.

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