# The Stacks Project

## Tag 00S1

Lemma 10.132.2. Suppose given a diagram (10.132.1.1). 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 (10.132.0.1) 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. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 33433–33452 (see updates for more information).

\begin{lemma}
\label{lemma-NL-homotopy}
Suppose given a diagram (\ref{equation-functoriality-NL}).
Let $\alpha : P \to S$ and $\alpha' : P' \to S'$ be presentations.
\begin{enumerate}
\item There exists a morphism of presentations from $\alpha$ to $\alpha'$.
\item Any two morphisms of presentations induce homotopic
morphisms of complexes $\NL(\alpha) \to \NL(\alpha')$.
\item The construction is compatible with compositions of morphisms
of presentations (see proof for exact statement).
\item If $R \to R'$ and $S \to S'$ are isomorphisms, then
for any map $\varphi$ of presentations from $\alpha$ to $\alpha'$
the induced map $\NL(\alpha) \to \NL(\alpha')$ is a homotopy equivalence
and a quasi-isomorphism.
\end{enumerate}
In particular, comparing $\alpha$ to the canonical presentation
(\ref{equation-canonical-presentation}) we conclude there is a
quasi-isomorphism $\NL(\alpha) \to \NL_{S/R}$ well defined
up to homotopy and compatible with all functorialities (up to homotopy).
\end{lemma}

\begin{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).

\medskip\noindent
Let $\varphi$ and $\varphi'$ morphisms of presentations from $\alpha$
to $\alpha'$. Let $I = \Ker(\alpha)$ and $I' = \Ker(\alpha')$.
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.

\medskip\noindent
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
\begin{enumerate}
\item $\alpha : P \to S$,
\item $\alpha' : P' \to S'$, and
\item $\alpha'' : P'' \to S''$
\end{enumerate}
are presentations. Suppose that
\begin{enumerate}
\item $\varphi : P \to P$ is a morphism of presentations from
$\alpha$ to $\alpha'$ and
\item $\varphi' : P' \to P''$
is a morphism of presentations from $\alpha'$ to $\alpha''$.
\end{enumerate}
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 $\NL(\alpha) \to \NL(\alpha'')$ of naive cotangent complexes
is the composition of the maps $\NL(\alpha) \to \NL(\alpha')$ and
$\NL(\alpha') \to \NL(\alpha'')$ induced by $\varphi$ and $\varphi'$.

\medskip\noindent
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
$\NL(\alpha) \to \NL(\alpha)$ 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 $\NL(\alpha) \to \NL(\alpha') \to \NL(\alpha)$ and
$\NL(\alpha') \to \NL(\alpha) \to \NL(\alpha')$ are homotopic to the identity
maps by (3), hence these maps are homotopy equivalences by definition.
It follows formally that both maps
$\NL(\alpha) \to \NL(\alpha')$ and $\NL(\alpha') \to \NL(\alpha)$ are
quasi-isomorphisms. Some details omitted.
\end{proof}

Comment #693 by Keenan Kidwell on June 16, 2014 a 7:30 pm UTC

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

Comment #1717 by Yogesh More on December 3, 2015 a 3:56 pm UTC

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 #1757 by Johan (site) on December 15, 2015 a 7:05 pm UTC

Thanks, fixed here.

Comment #2789 by Dario Weißmann on August 27, 2017 a 9:10 pm UTC

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 August 30, 2017 a 8:40 am UTC

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

Comment #2896 by Johan (site) on October 7, 2017 a 3:18 pm UTC

@#2789, #2791. THanks, fixed here.

There are also 7 comments on Section 10.132: Commutative Algebra.

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