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In classical finite-dimensional differential geometry, the Lie functor preserves surjections, sending a surjective Lie group homomorphism to a surjective Lie algebra homomorphism.

As pointed out below, the functor does not preserve all epimorphisms in the category of Lie groups, but only those which are also epimorphisms (surjective) in the ambient topos of sets.

Does this continue to hold in synthetic differential geometry? In this case, it seems that the appropriate definition of a "surjective" Lie group homomorphism is a Lie group homomorphism which is an epi within the ambient smooth topos.

To start, suppose we have a "surjective" Lie group homomorphism: $\phi: G \to H$. That is, for maps $\psi_1: H \to M$ and $\psi_2: H \to M$, if $\psi_1 \circ \phi = \psi_2 \circ \phi$ then $\psi_1 = \psi_2$.

Now if $f_1 : \mathfrak{h} \to N$ and $f_2 : \mathfrak{h} \to N$ are any maps, we suppose that $f_1 \circ \phi_* = f_2 \circ \phi_*$. Then for any Lie algebra element $(X: D \to G) \in \frak{g}$, we have $$f_1 (\phi \circ X) = f_2 (\phi \circ X).$$ But I am not sure how to conclude that $f_1 = f_2$. Indeed, as far as I know covariant representable functors do not in general preserve epimorphisms.

Sorry if this is obvious; I like to read up on SDG as a hobby, but I am not an expert in category theory or topos theory.

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    $\begingroup$ The condition that a covariant representable functor $\text{Hom}(P, -)$ preserves epimorphisms is one possible definition of $P$ being a projective object. So non-projective modules provide many counterexamples to this general statement. The abstract nonsense only tells you that covariant representable functors (regarded as functors to $\text{Set}$) preserve monomorphisms. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 20, 2016 at 22:09

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The inclusion $\mathbb Q\to \mathbb R$ is an epimorphism in the category of Lie groups.

Its image in the category of Lie algberas is not an epimorphism.

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  • $\begingroup$ Ah, I hadn't realized that epis in the category of Lie groups were not precisely surjective smooth homomorphisms. I believe the Lie functor does preserve surjections in classical differential geometry, so I have edited my question. $\endgroup$
    – ಠ_ಠ
    Commented Jan 20, 2016 at 22:09
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    $\begingroup$ Here, $\mathbb Q$ is endowed with the discrete topology (it is a zero-dimensional Lie group). Te map $\mathbb Q\to \mathbb R$ is an epimorphism because if $f:\mathbb R\to G$ is a homomorphism of Lie groups, then $f$ is completely determined by $f|_{\mathbb Q}$ (by continuity). $\endgroup$
    – André Henriques
    Commented Jan 21, 2016 at 1:24
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    $\begingroup$ Ah, right - sorry for the stupid comment. $\endgroup$
    – Peter Samuelson
    Commented Jan 21, 2016 at 1:34
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    $\begingroup$ By the way, the inclusion $\mathbb Q\to \mathbb R$ is not an epimorphism in the category of rings. It is the inclusion $\mathbb Z\to \mathbb Q$ which is an epimorphism in the category of rings. $\endgroup$
    – André Henriques
    Commented Jan 21, 2016 at 1:43
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    $\begingroup$ It's true that if you define a Lie group to be a second countable manifold equipped with a group structure, then a surjective homomorphism of Lie groups is mapped to a surjective homomorphism of Lie algebras. But this ceases to be true as soon as you allow manifolds with uncountably many connected components (think about the identity map from $\mathbb R$ with the discrete topology to $\mathbb R$ with its usual topology). So you see: it's fragile statement. And fragile statements often don't generalize very well. $\endgroup$
    – André Henriques
    Commented Jan 21, 2016 at 9:29

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