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Below is an excerpt from Berthelot's book on crystalline cohomology. I don't understand the last sentence, namely why it follows that $\sigma\circ \varepsilon$ is an isomorphism. For what it's worth, $P^1$ is the sheaf of principal parts and $E$ is an $\mathcal O _X$-module. I can elaborate on what $\sigma,\varepsilon,\tau$ are, but perhaps I'm just missing some basic algebra...

We have an endomorphism of a module which becomes the identity modulo a square zero ideal. Why, in this case, is it an isomorphism?

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Suppose you have a commutative ring $R$, a square zero ideal $I\subset R$, a $R$-module $M$ and an endomorphism $u$ of $M$ which is the identity modulo $IM$. Then $v:= 1_M-u$ maps $M$ to $IM$, hence $v^2=0$. Therefore $u=1_M-v$ is invertible — its inverse is $1_M+v$.

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  • $\begingroup$ Dear @abx, thanks for the lovely method! If I'm not mistaken, the calculation in the endomorphism ring should just be $(1-u)^2=0$ which gives $1=u(2-u)$, and not $u(1+u)=1$. Am I missing something? $\endgroup$
    – Arrow
    Commented May 6, 2021 at 20:30
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    $\begingroup$ Sorry, this is a typo — I meant that the inverse is $1+v=2-u$. I edit. $\endgroup$
    – abx
    Commented May 7, 2021 at 4:44

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