Lemma 86.3.1. Let $A$ be a Noetherian ring and let $I \subset A$ be a ideal. Let $B$ be an object of (86.2.0.2). The naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is well defined in $K(B)$.

**Proof.**
The lemma signifies that given a second presentation $B = A[y_1, \ldots , y_ s]^\wedge / K$ the complexes of $B$-modules

are homotopy equivalent. To see this, we can argue exactly as in the proof of Algebra, Lemma 10.134.2.

Step 1. If we choose $g_ i(y_1, \ldots , y_ s) \in A[y_1, \ldots , y_ s]^\wedge $ mapping to the image of $x_ i$ in $B$, then we obtain a (unique) continuous $A$-algebra homomorphism

compatible with the given surjections to $B$. Such a map is called a morphism of presentations. It induces a map from $J$ into $K$ and hence induces a $B$-module map $J/J^2 \to K/K^2$. Sending $\text{d}x_ i$ to $\sum (\partial g_ i/\partial y_ j)\text{d}y_ j$ we obtain a map of complexes

Of course we can do the same thing with the roles of the two presentations exchanged to get a map of complexes in the other direction.

Step 2. The construction above is compatible with compositions of morphsms of presentations. Hence to finish the proof it suffices to show: given $g_ i(x_1, \ldots , x_ r) \in A[x_1, \ldots , x_ n]^\wedge $ mapping to the image of $x_ i$ in $B$, the induced map of complexes

is homotopic to the identity map. To see this consider the map $h : \bigoplus B \text{d}x_ i \to J/J^2$ given by the rule $\text{d}x_ i \mapsto g_ i(x_1, \ldots , x_ n) - x_ i$ and compute. $\square$

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