Description of free Lie algebra in Weibel's book In Exercise 7.3.2 in Weibel's book An Introduction to Homological algebra the following description of the free Lie algebra over some $k$-module $M$ is given (where $k$ is any commutative ring):
First, consider the tensor algebra $T(M)$. Take the underlying Lie algebra $\mathrm{Lie}(T(M))$, i.e. $[x,y] = xy-yx$. Then, consider the Lie subalgebra $\mathfrak{f}(M) \subseteq \mathrm{Lie}(T(M))$ generated by $M$. The claim is that $\mathfrak{f}(M)$ is the free Lie algebra on $M$, i.e. that every $k$-linear map $f : M \to \mathfrak{g}|_k$ into the underlying $k$-module of some Lie algebra extends uniquely to a Lie algebra map $\overline{f} : \mathfrak{f}(M) \to \mathfrak{g}$.
Well, uniqueness is clear, since $M$ generates $\mathfrak{f}(M)$. Specifically, we have to map $[x_1,[x_2,[\dotsc[x_{n-1},x_n]\dotsc]]]$ to $[f(x_1),[f(x_2),[\dotsc[f(x_{n-1}),f(x_n)]\dotsc]]]$ and $\mathfrak{f}(M)|_k$ is generated by these elements. But I am not able to show existence, i.e. that this is well-defined.
By naturality, we may assume $M=\mathfrak{g}|_k$ and $f=\mathrm{id}$. The simplest special case for well-definedness is the following: If $[x,y]=0$ in $\mathfrak{f}(M)$, i.e. $x \otimes y = y \otimes x$ in $M^{\otimes 2}$, why do we have $[x,y]=0$ in $\mathfrak{g}$? This holds when $k$ is a field, because then $x \otimes y = y \otimes x$ implies $x \in \langle y \rangle$ and $[x,y]=0$ follows from $[y,y]=0$. If $k$ is arbitrary, I can only see $2 \cdot [x,y]=0$ (in fact, $[x,y]=[y,x]$ because the Lie bracket is bilinear, and $[y,x]=-[x,y]$ holds in general). So if $2 \in k^{\times}$, it would follow again that $[x,y]=0$, but otherwise it is not clear.
Shouldn't it be possible to construct a counterexample? It doesn't appear in the Errata, though. Is the description true, at least, when $k$ is a field? This should be connected to the PBW-theorem.
 A: 1. In the general case, Weibel's Exercise 7.3.2 is wrong: The Lie
subalgebra $\mathfrak{f}\left(  M\right)  $ of $\operatorname{Lie}\left(
T\left(  M\right)  \right)  $ is not always a free Lie algebra on $M$ (at
least not if the inclusion of $M$ into it is considered as the canonical map).
To see why not, we shall use the counterexample constructed in §5 of
P.M. Cohn, A remark on the Birkhoff-Witt theorem, J. London Math. Soc. 38
(1963), pp. 197--203.
The counterexample goes as follows (I have kept the notation but slightly
modified the definitions): Let $p$ be a prime, and let $K$ be a field of
characteristic $p$. Let $k$ be the commutative ring $k\left[  \alpha
,\beta,\gamma\ \mid\ \alpha^{p}=0,\ \beta^{p}=0,\ \gamma^{p}=0\right]  $. Let
$M$ be the $k$-module given by generators $x,y,z$ and relations $\alpha
x=\beta y+\gamma z$. Let $L$ be the free Lie algebra of $M$ (defined in the
usual way -- i.e., as a quotient of a free magmatic algebra, not using
$T\left(  M\right)  $). Then, the universal enveloping algebra $U\left(
L\right)  $ of $L$ is canonically isomorphic to the tensor algebra $T\left(
M\right)  $ (see, e.g., math.stackexchange #1030593). Cohn defines a
certain element $\Lambda_{p}\left(  \beta y,\gamma z\right)  $ of $L$ and
shows that $\Lambda_{p}\left(  \beta y,\gamma z\right)  =0$ in $U\left(
L\right)  $. Thus, $\Lambda_{p}\left(  \beta y,\gamma z\right)  =0$ in
$\mathfrak{f}\left(  M\right)  $ (since $\mathfrak{f}\left(  M\right)
\subseteq\operatorname*{Lie}\left(  T\left(  M\right)  \right)  \cong
\operatorname*{Lie}\left(  U\left(  L\right)  \right)  $).
On the other hand, Cohn claims that $\Lambda_{p}\left(  \beta y,\gamma
z\right)  \neq0$ in $L$. It is not clear if this latter claim has ever been
proven, but at least it can easily be checked by computation for $p=2$ and for
$p=3$, which means that at least for these primes $p$ we have a counterexample
(see mathoverflow #61981 and the comments beneath it for a discussion).
Anyway, we just want one counterexample, so $p=2$ is perfectly enough for us.
Combining the fact that $\Lambda_{p}\left(  \beta y,\gamma z\right)  \neq0$ in
$L$ with the fact that $\Lambda_{p}\left(  \beta y,\gamma z\right)  =0$ in
$\mathfrak{f}\left(  M\right)  $, we see that $\mathfrak{f}\left(  M\right)  $
cannot be a free Lie algebra on $M$ (because if it were, then there would be a
Lie algebra homomorphism $\mathfrak{f}\left(  M\right)  \rightarrow L$ sending
$x,y,z$ to $x,y,z$).
2. However, in many cases, Weibel's Exercise 7.3.2 is correct. One such
case is when $M$ is a free $k$-module. In this case, the Exercise follows from
Remark 6.42 in Darij Grinberg and Victor Reiner, Hopf Algebras in
Combinatorics, arXiv:1409.8356v4 (the linked file is the ancillary file
with solutions). Alternatively, it also follows from the Serre reference given
by user218931 in his comment (Jean-Pierre Serre, Lie Algebras and Lie
Groups, Corrected 5th printing, Springer 2006, Theorem I.IV.4.2). The
former proof is combinatorial, long and intricate (it is the standard argument
using the Hall property of Lyndon words). The latter is shorter and, while in
part still combinatorial (it does pass through Hall families), avoids much of
the work by tricking around with base change and linear algebra. Which of
these proofs you prefer is probably a matter of personal taste.
3. Another case where Exercise 7.3.2 is correct is the case when $k$ is a
$\mathbb{Q}$-algebra (that is, $1,2,3,\ldots$ are invertible in $k$). This
follows from Proposition 7.2.8 in Volume I of Benoit Fresse, Homotopy of
Operads and Grothendieck-Teichmüller Groups.
