# Exact sequence of modules

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$(1_{R^2}-g\circ f)$.

$Im(1 - gf) = \ker(f)$.
$f((1 - gf)(x)) = f(x) - fgf(x) = 0$ so $Im(1 - gf) ⊆ \ker(f)$.
On the other hand if $x ∈ \ker(f)$ then $(1 - gf)(x) = x$ so $\ker(f) ⊆ Im(1 - gf)$.
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 $R$.