In light of point 1 of
Theorem 8.3.1, it suffices to check that
\(R\Hom_{\LCAb}(A,B) \cong R\Hom_{\CAb}(\underline{A}, \underline{B})\) when each of
\(A,B\) is either compact, discrete, or isomorphic to
\(\RR\text{.}\) We break down the argument as follows.
Suppose first that
\(A\) is discrete. Using resolutions and filtered colimits we may reduce to the case
\(A = \ZZ\text{.}\) In this case,
\(A\) and
\(\underline{A}\) are projective in their respective categories and the result is clear.
Suppose next that
\(A\) is compact. By taking a resolution of the dual group, we may reduce to the case where
\(A = \prod_I \TT\) with
\(I\) at most countable. At this point, when
\(B\) is discrete we may deduce the desired isomorphism by comparing
Lemma 8.3.2 with
[15], Example 4.11; when
\(B = \RR\) we may instead compare
Theorem 8.1.6 with
[15], Proposition 4.15 for the same effect. Combining these and using the exact sequence
(8.1), we deduce the claim also when
\(B = \TT\text{,}\) and consequently when
\(B = \prod_I \TT\text{;}\) again we may formally promote this to the general case of
\(B\) compact.
Finally, to resolve the case
\(A = \RR\) we may use
(8.1) to reduce to the previously treated cases
\(A = \ZZ\) (discrete) and
\(A = \TT\) (compact).