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In Altman-Kleinman "Introduction to Grothendieck Duality Theory", page 105, there is the following Lemmaenter image description here:

I don't understand why by uniqueness, I can conclude $w \circ w'=id$. Any help?

Jotabeta
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    Can you describe the notations? E.g. what is $\Omega^1_{A/k}$? Anyways, typically, both $w\circ w'$ and ${\rm id}$ will satisfy a given universal property. – Berci Jan 07 '21 at 21:24

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As $d:A \rightarrow \Omega$ is a $K$-derivation there is a unique morphism $w‘: \Omega_{A\mid K}^1 \rightarrow \Omega$ satisfying $w’ \circ d_{A\mid K} = d‘$ by the universal property of the module of Kähler-differentials.

As $d_{A\mid K}$ is a derivation as well, there is a unique morphism $i: \Omega_{A\mid K}^1 \rightarrow \Omega_{A\mid K}^1$ satisfying $i \circ d_{A\mid K} = d_{A\mid K}$. Note that the identity is such a morphism, hence $i = id$.

Finally we are given a morphism $w: \Omega \rightarrow \Omega_{A\mid K}$ satisfying $w \circ d = d_{A\mid K}$. Hence we can calculate $$w \circ w‘ \circ d_{A\mid K} = w \circ d = d_{A\mid K}$$ which shows that $w \circ w‘$ satisfies the universal property of $\Omega_{A\mid K}$, so again by uniqueness we find $$w \circ w‘ = i = id.$$

Jonas Linssen
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