TITLE: Question about a proof in Berthelot's crystalline book
QUESTION [5 upvotes]: 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?

REPLY [10 votes]: 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 $1_M+v$.