Lemma 10.134.2. Suppose given a diagram (10.134.1.1). Let $\alpha : P \to S$ and $\alpha ' : P' \to S'$ be presentations.

1. There exists a morphism of presentations from $\alpha$ to $\alpha '$.

2. Any two morphisms of presentations induce homotopic morphisms of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$.

3. The construction is compatible with compositions of morphisms of presentations (see proof for exact statement).

4. If $R \to R'$ and $S \to S'$ are isomorphisms, then for any map $\varphi$ of presentations from $\alpha$ to $\alpha '$ the induced map $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\alpha ')$ is a homotopy equivalence and a quasi-isomorphism.

In particular, comparing $\alpha$ to the canonical presentation (10.134.0.1) we conclude there is a quasi-isomorphism $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits _{S/R}$ well defined up to homotopy and compatible with all functorialities (up to homotopy).

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

$\xymatrix{ I/I^2 \ar[r]^-{\text{d}} \ar@<1ex>[d]^{\varphi '_1} \ar@<-1ex>[d]_{\varphi _1} & \Omega _{P/R} \otimes _ P S \ar@<1ex>[d]^{\varphi '_0} \ar@<-1ex>[d]_{\varphi _0} \ar[ld]_ h \\ I'/(I')^2 \ar[r]^-{\text{d}} & \Omega _{P'/R'} \otimes _{P'} S' }$

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

$\varphi _1 - \varphi '_1 = h \circ \text{d} \quad \text{and}\quad \varphi _0 - \varphi '_0 = \text{d} \circ h$

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

$(\varphi - \varphi ')(fg) = \varphi (f)(\varphi - \varphi ')(g) + (\varphi - \varphi ')(f)\varphi '(g)$

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

$\xymatrix{ S \ar[r]_{\phi } & S' \ar[r]_{\phi '} & S'' \\ R \ar[r] \ar[u] & R' \ar[u] \ar[r] & R'' \ar[u] }$

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. $\square$

Comment #693 by Keenan Kidwell on

In the first part of the lemma, "there exist" should be "there exists."

Comment #1717 by Yogesh More on

Trivial remark but might be worth adding, after the line "we conclude that $\phi -\phi:P \to I'$.":

This implies that $\phi, \phi':P \to P'$ induce the same $P$-module structure on $I'/(I')^2$, since $\phi(p)i'-\phi'(p)i'=(\phi-\phi')(p)i' \in (I')^2$.

I suggest adding this (obvious) remark because you are implicitly using it in the equation showing that $(\phi-\phi')(fg)$ satisfies Leibniz rule (which I feel is the key to the result), and also in using the universal property for $\Omega_{P/R}$, i.e. to get the map $\Omega_{P/R} \to I'/(I')^2$, you are considering $I'/(I')^2$ as a $P$-module

Comment #2789 by Dario Weißmann on

Typo in the proof of (2): "Since both $\varphi$ and $\varphi$ are compatible with..." should read "Since both $\varphi$ and $\varphi'$ are compatible with..."

Comment #2791 by Dario Weißmann on

In the proof of (3) the last line should read: "...and NL$(\alpha') \to$NL$(\alpha'')$ induced by $\varphi$ and $\varphi'$."

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