Exact sequence of modules

Question

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 $R$ being 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 paper¹ (page 779) I am reading it says that $P=\mathrm{Im}(1_{R^2}-g\circ f)$.

¹Richard G. Swan. "Quaternionic Bundles on Algebraic Spheres." Rocky Mountain J. Math. 26 (2) 773 - 791, Spring 1996.

Answer 1

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

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

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

Answer 2

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 $R$.