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Let $G$, $H$ be two compact Lie groups (possibly disconnected). Two short exact sequences of compact Lie groups $$ 0\rightarrow G\rightarrow M_1 \rightarrow H\rightarrow 0, $$ $$ 0\rightarrow G\rightarrow M_2\rightarrow H\rightarrow 0 $$ are considered equivalent is there is an isomorphism of compact Lie groups $M_1\rightarrow M_2$ making the rhomb-like diagram commute. Denote by $\mathrm{Ext}_{Lie}(G, H)$ the set of equivalence classes of short exact sequences of the above form.

We have a forgetful functor $F$ from the category of compact Lie groups to the category of groups. In its full image, one can also define the set of inequivalent extensions, $\mathrm{Ext}_{Grp}(\cdot, \cdot)$. Note that $F$ induces a well-defined map from $\mathrm{Ext}_{Lie}(G, H)$ to $\mathrm{Ext}_{Grp}(G, H)$. Is this map always an injection? What if we demand $G$ to be connected and $H$ to be zero-dimensional?

I do not think that this question can be answered by general nonsense, because analogous statement for the forgetful functor to closed manifolds does not hold (though I would be glad if I am proven wrong).

P.S.: Let $H$ be zero-dimensional. If we require the morphisms in short exact sequences to lie in the image of $F$ and fix a characteristic homomorphism $H\rightarrow \mathrm{OutAut}(G)$, then factor system argument described e.g. in chapter 18 of "Structure and geometry of Lie groups", should imply that the map is bijective. If we do not require the morphisms in the short exact sequence to lie in the image of $F$, then there may be more characteristic homomorphisms so the map is not surjective but it still should be injective.

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    $\begingroup$ For $T$ the circle group, $\mathrm{Ext}_{\mathrm{Grp}}(T,T)$ is huge (there are many non-abelian extensions), so clearly $\mathrm{Ext}_{\mathrm{TopGrp}}(T,T)\to\mathrm{Ext}_{\mathrm{Grp}}(T,T)$ is not surjective. (Here I write $\mathrm{Ext}_{\mathrm{TopGrp}}$ what you call $\mathrm{Ext}_{\mathrm{Lie}}$ since it's the same.) Also if $F$ is a nontrivial finite group $\mathrm{Ext}_{\mathrm{TopGrp}}(T,F)\to\mathrm{Ext}_{\mathrm{Grp}}(T,F)$ is not onto since $\mathrm{Ext}_{\mathrm{Grp}}(T,F)$ is huge too. I guess $\mathrm{Ext}_{\mathrm{TopGrp}}(G,H)\to\mathrm{Ext}_{\mathrm{Grp}}(G,H)$ is injective. $\endgroup$
    – YCor
    Commented Mar 7, 2019 at 15:59
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    $\begingroup$ Sure. My comment addressed the question when you asked about being bijective, before you edited. $\endgroup$
    – YCor
    Commented Mar 7, 2019 at 16:03
  • $\begingroup$ See section 15 (pages 169-190) of mat.univie.ac.at/~michor/dgbook.pdf for a quite comprehensive discussion of extensions of Lie algebras and Lie groups, including the discrete case. $\endgroup$ Commented Mar 7, 2019 at 17:40

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Here is a partial answer if $G$ is finite and Abelian, in this case we denote by $H^{\delta}$ the same group than $H$ with the discrete topology.The natural homomorphism $i:H^{\delta}\rightarrow H$ induces a continuous map $f:BH^{\delta}\rightarrow BH$.

In his paper entitled On the homology of Lie groups made discrete, Milnor conjectures the following fact: $f:BH^{\delta}\rightarrow BH$ induces an isomorphism of homology and cohomology with mod $p$ coefficients or more generally with any finite coefficient group.

Let $G$ be such finite coefficient group, the conjecture of Milnor says that $f$ induces an isomorphism $H^2(BH^{\delta},G)\rightarrow H^2(BH,G)$ and this is equivalent to saying that $Ext_{Lie}(G,H)$ is isomorphic to $Ext_{Gpr}G,H)$.

In that paper, Milnor shows its conjecture in different cases, for example if the identity component of the group $H$ is solvable. The paper of Milnor has been published a long time ago, perhaps there are other results in that direction.

Milnor, J. On the homology of Lie groups made discrete.

Comment. Math. Helvetici 58 (1983) 72-85. (eudml, authors' website, DOI: 10.1007/BF02564625)

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