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Motivation: I'm trying to figure out whether the image of an injective Lie group homomorphism (between simply connected Lie groups) under the Lie functor is injective. Since the Lie functor is an equivalence of categories from the category of simply connected Lie groups to the category of finite-dimensional real Lie algebras, one knows that the image of an injection under the Lie functor is a monomorphism in the category of finite-dimensional real Lie algebras.

Question: Are monomorphisms in the category of finite-dimensional Lie algebras over a field injective?

My thoughts: I know that monomorphisms in the category of all Lie algebras are injective, since the forgetful functor from the category of all Lie algebras to the category of sets has a left adjoint. However, the forgetful functor from the category of finite-dimensional Lie algebras to the category of sets doesn't seem to have a left adjoint. I'm also aware of the results that the Lie functor preserves surjections, and that there exists an epimorphism in the category of finite-dimensional Lie algebras that is not surjective, but I don't think that is useful for this question.

(Edit: Using that taking the kernel commutes with applying the Lie functor, I have been able to prove the question that motivated me: the Lie functor preserves injections.)

kringelton4000
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    I don't know if you really need all the category theory for this. More directly, differentiating an injective Lie group homomorphism must give an injective Lie algebra homomorphism. – Callum Aug 01 '22 at 13:22
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    @Callum Yes, I have been able to prove that, see my edit. Nevertheless, I think it's interesting to know about the monomorphisms in this category, since their exponentials are exactly the Lie group monomorphisms. – kringelton4000 Aug 01 '22 at 13:38

2 Answers2

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I claim that a morphism $f:\mathfrak{g}\rightarrow\mathfrak{h}$ is monic iff it is injective.

Of course, in any concrete category, injective implies monic, so we'll focus on the more fun direction.

So, assume $f$ is monic and let $\mathfrak{k}$ denote the kernel of $f$. Consider the two functions $g_1,g_2:\mathfrak{k}\rightarrow \mathfrak{g}$ where $g_1$ is the inclusion and $g_2$ is the zero map. Then $f\circ g_1 = f\circ g_2$ since both are the $0$-map. As $f$ is monic, we conclude that $g_1 = g_2$, that is, the that identity map of $\mathfrak{k}$ is the zero map. Thus, $\mathfrak{k} = \{0\}$, so $f$ is injective.

Jason DeVito - on hiatus
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  • I don't know enough category theory to really understand to what extent this argument generalizes, but note that it works if the spaces are Lie algebras (finite dimensional or not), groups, rings, fields, vector spaces, etc. – Jason DeVito - on hiatus Aug 01 '22 at 14:06
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Some general categorical comments: let $U : C \to \text{Set}$ be a concrete category, by which I mean $U$ is a faithful functor with a left adjoint, which in particular preserves all limits. If $C$ has finite limits then a morphism $f : a \to b$ is a monomorphism iff its kernel pair is trivial; because this is a finite limit, any functor that preserves finite limits preserves monomorphisms. So $U$ preserves monomorphisms, meaning that monomorphisms are always injective. In the other direction, faithful functors always reflect monomorphisms, so injective morphisms are also always monomorphisms.

Now we just observe that the category of finite-dimensional Lie algebras has finite limits which agree with the corresponding limits in the category of all Lie algebras (from which it follows that, despite not having a left adjoint, the forgetful functor still preserves finite limits); to see this we just need to observe that a finite product of finite-dimensional Lie algebras remains finite-dimensional, as does an equalizer of a parallel pair of morphisms between finite-dimensional Lie algebras.

Qiaochu Yuan
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