Proposition 5

The topology induced on $\mathsf{X}(A)$ by that of $\widehat{A}$ is coarser than the topology of $\mathsf{X}(A)$.

In fact, let T be a closed subset of $\widehat{A}$. Then T is the set of $\pi \in \widehat{A}$ whose kernel contains a subset M of A. Therefore $T\cap \mathsf{X}(A)$ is the set of $\chi \in \mathsf{X}(A)$ which vanish on M, that is, a closed subset of $\mathsf{X}(A)$. Hence the proposition.

In general, the topology of $\mathsf{X}(A)$ does not coincide with the topology induced by the topology of $\widehat{A}($cf. I, p. 193, Exercise 6, c)).