Invertible endomorphisms of an $R$-module?

Question

Let $M$ be an $R$-module for some ring $R$, and $f,g\in E=\operatorname{End}M$ such that $fg=1_E$. Is it true that $gf=1_E$?
I also met a sililar problem where $f$ and $g$ are elements of a semi-simple ring. I can't see how either can be proved. I end up going round in circles proving useless relations between $f$ and $g$

Answer 1

No:

Take $M = \mathbb{Z}^\omega$ as a $\mathbb{Z}$ module.

• $f(x_0,x_1,x_2,\ldots) = (x_1,x_2,\ldots)$
• $g(x_0,x_1,x_2,\ldots) = (0,x_0,x_1,x_2,\ldots)$

Then $fg = 1$, but $gf(x_0,x_1,x_2,\ldots) = (0,x_1,x_2,x_3,\ldots)$.

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If you require $g$ be surjective, then the theorem is true:

$fg = 1$ implies $gfg = g$, so $(gf)(g x) = g(x)$ for every $x$.

Since $g$ is surjective, $gf(x) = gf(g(y)) = g(y) = x$ for some $y$, and the claim follows.

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I hope this helps ^_^