Let $\mathit{cLieGrp}^{\mathrm{inj}}$ be the category of compact connected Lie groups, and injective continuous group homomorphisms. Is there a reasonable functor (some kind of degree $3$ differential cohomology?) $$ F:\mathit{cLieGrp}^{\mathrm{inj}} \to \mathit{Ab} $$ to the category of abelian groups with the following features:

(1) On the full subcategory of compact semisimple Lie groups, this functor agrees with $G\mapsto H^3(G,\mathbb Z)$ (integral 3rd cohomology of the underlying manifold of $G$).

(2) On the full subcategory of compact abelian Lie groups (tori), this functor agrees with $T\mapsto H^4(BT,\mathbb R)$.

Motivation for the question:
The functor $$G \mapsto \{ \text{full WZW models with gauge group } G\}$$ seems to behave like the functor $F$ above (except that the set of full WZW models with gauge group $G$ needs to be group-completed in order to make it into an abelian group).

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    $\begingroup$ Does "injective maps" mean "injective continuous group homomorphisms"? or "injective continuous maps" (or anything else)? $\endgroup$ – YCor Mar 11 at 14:42
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    $\begingroup$ Well, $H^3(T,Z)=Z$ and $H^4(BT,R)=R$, this looks impossible... $\endgroup$ – user43326 Mar 11 at 16:22
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    $\begingroup$ By the way I meant $T=S^1$ in the above. $\endgroup$ – user43326 Mar 11 at 18:08
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    $\begingroup$ @user43326. For $T:=S^1$, we have $H^3(T,\mathbb Z)=0$. It's just the cohomology of the underlying manifold (and as the manifold is 1-dimensional, there is no $H^3$). $\endgroup$ – André Henriques Mar 12 at 0:06
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    $\begingroup$ OK, I thought you meant "group cohomology". But in that case it still doesn't work since $H^4(BT,R)$ is still different from $H^3(T,Z)$. Or here does $B$ mean something other than the classifying space? $\endgroup$ – user43326 Mar 12 at 7:55

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