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.
