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Let $k$ be a field of characteristic 0 and let $\varphi:\mathfrak{g}\rightarrow\mathfrak{f}$ and $\psi:\mathfrak{h}\rightarrow\mathfrak{f}$ be maps of Lie algebras. Is there a reference showing that the pullback (in the category of Lie $k$-algebras) of $\mathfrak{g}$ and $\mathfrak{h}$ along these maps is given by the following formula

$\mathfrak{g}\times_\mathfrak{f}\mathfrak{h}=\{(x,y)\in\mathfrak{g}\times\mathfrak{h}\mid \varphi(x)=\psi(y)\}$

EDIT: Just to be extra clear. I do know how to prove it, but I cannot find a place in the literature where this is proved or stated and I was wondering if you knew of one.

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    $\begingroup$ It is trivial and holds for every algebraic category: the forgetful functor creates limits. This statement can be found in texts on universal algebra or category theory. $\endgroup$
    – Martin Brandenburg
    Commented Jul 1, 2021 at 8:35
  • $\begingroup$ If you are looking for a citation for the purposes of off-loading the proof, then it might actually be better not to have the reader chase a reference just so that they end up discovering you directed them to something very simple that they could have done in the time it took to find the reference. $\endgroup$
    – Andrej Bauer
    Commented Jul 1, 2021 at 12:44
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    $\begingroup$ @AndrejBauer: I gave some reference in my answer, but I agree. This might be one of those cases where writing something like "it is well-known that..", "recall that..", or "it is easy to see that.." is better than giving any detail or a reference. $\endgroup$
    – spin
    Commented Jul 1, 2021 at 14:07

2 Answers 2

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A reference for pullbacks for modules over a ring: Proposition 5.11, p.222 in "An Introduction to Homological Algebra" by Rotman. For Lie $k$-algebras (or just $k$-algebras), you could refer to this and say the same construction/proof works.

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This follows from the existence of free Lie algebras. More precisely, the forgetful functor $U: \mathbf{Lie}_k \to \mathbf{Vect}_k$ has a left adjoint $F$ which takes $V$ to the free Lie algebra on $V$. Hence, $U$ preserves limits. This fact has many references, c.f. MacLane, Categories for the Working Mathematician, §V.5.

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    $\begingroup$ Though correct, this answer seems unnecessarily complicated. One can easily verify directly that the OP's proposed pullback has the required universal property. $\endgroup$
    – Andreas Blass
    Commented Jul 1, 2021 at 3:40
  • $\begingroup$ @AndreasBlass I am afraid my question was a bit misleading. I do know how to prove it. I was wondering if this fact can be found anywhere in the literature, maybe as an exercise in a book? I could not find a reference by googling it. I am specifically looking for a place in the literature where this is proved or at least stated. $\endgroup$
    – user15160811
    Commented Jul 1, 2021 at 4:02
  • $\begingroup$ @user15160811 I don't know offhand a reference for this fact specifically about Lie algebras, but I'd expect textbooks on universal algebra to prove this in much greater generality --- for arbitrary varieties of algebras. $\endgroup$
    – Andreas Blass
    Commented Jul 1, 2021 at 4:14
  • $\begingroup$ @user15160811 I have added a reference to the general statement about adjoint functors and limits. Hopefully this may do. $\endgroup$
    – Joshua Mundinger
    Commented Jul 1, 2021 at 4:20
  • $\begingroup$ @AndreasBlass does the general proof use the existence of free algebras? $\endgroup$
    – Joshua Mundinger
    Commented Jul 1, 2021 at 4:21

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