The Stacks project

Proof. Since $P$ is a polynomial algebra over $R$ we can write $P = R[x_ a, a \in A]$ for some set $A$. As $\alpha '$ is surjective, we can choose for every $a \in A$ an element $f_ a \in P'$ such that $\alpha '(f_ a) = \phi (\alpha (x_ a))$. Let $\varphi : P = R[x_ a, a \in A] \to P'$ be the unique $R$-algebra map such that $\varphi (x_ a) = f_ a$. This gives the morphism in (1).

Let $\varphi $ and $\varphi '$ morphisms of presentations from $\alpha $ to $\alpha '$. Let $I = \mathop{\mathrm{Ker}}(\alpha )$ and $I' = \mathop{\mathrm{Ker}}(\alpha ')$. We have to construct the diagonal map $h$ in the diagram

where the vertical maps are induced by $\varphi $, $\varphi '$ such that

Consider the map $\varphi - \varphi ' : P \to P'$. Since both $\varphi $ and $\varphi '$ are compatible with $\alpha $ and $\alpha '$ we obtain $\varphi - \varphi ' : P \to I'$. This implies that $\varphi , \varphi ' : P \to P'$ induce the same $P$-module structure on $I'/(I')^2$, since $\varphi (p)i' - \varphi '(p)i' = (\varphi - \varphi ')(p)i' \in (I')^2$. Also $\varphi - \varphi '$ is $R$-linear and

Hence the induced map $D : P \to I'/(I')^2$ is a $R$-derivation. Thus we obtain a canonical map $h : \Omega _{P/R} \otimes _ P S \to I'/(I')^2$ such that $D = h \circ \text{d}$. A calculation (omitted) shows that $h$ is the desired homotopy.

Suppose that we have a commutative diagram

and that

1. $\alpha : P \to S$,

2. $\alpha ' : P' \to S'$, and

3. $\alpha '' : P'' \to S''$

are presentations. Suppose that

1. $\varphi : P \to P'$ is a morphism of presentations from $\alpha $ to $\alpha '$ and

2. $\varphi ' : P' \to P''$ is a morphism of presentations from $\alpha '$ to $\alpha ''$.

Then it is immediate that $\varphi ' \circ \varphi : P \to P''$ is a morphism of presentations from $\alpha $ to $\alpha ''$ and that the induced map $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha '')$ of naive cotangent complexes is the composition of the maps $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ and $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha '')$ induced by $\varphi $ and $\varphi '$.

In the simple case of complexes with 2 terms a quasi-isomorphism is just a map that induces an isomorphism on both the cokernel and the kernel of the maps between the terms. Note that homotopic maps of 2 term complexes (as explained above) define the same maps on kernel and cokernel. Hence if $\varphi $ is a map from a presentation $\alpha $ of $S$ over $R$ to itself, then the induced map $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha )$ is a quasi-isomorphism being homotopic to the identity by part (2). To prove (4) in full generality, consider a morphism $\varphi '$ from $\alpha '$ to $\alpha $ which exists by (1). The compositions $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha )$ and $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ are homotopic to the identity maps by (3), hence these maps are homotopy equivalences by definition. It follows formally that both maps $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ and $\mathop{N\! L}\nolimits (\alpha ') \to \mathop{N\! L}\nolimits (\alpha )$ are quasi-isomorphisms. Some details omitted. $\square$