Let $\mathcal{A}$ be a commutative unital Banach algebra, $\mathcal{B} \subset \mathcal{A}$ a closed unital subalgebra, $\mathcal{I} \subset \mathcal{B}$ a closed ideal.
Is there in general a way to "identify" (soft question, perhaps) the maximal ideal spaces $\Sigma(\mathcal{B})$ and $\Sigma(\mathcal{A}/\mathcal{I})$, in terms of $\Sigma(\mathcal{A})$ and possibly some sort of other data? A couple of examples of the sort of thing I have in mind:
- If $\mathfrak{X}$ is a Banach space and $M \subset \mathfrak{X}$ a closed subspace, then $M^* \simeq \mathfrak{X}^*/M^\perp$ and $(\mathfrak{X}/M)^* \simeq M^\perp$, where $ M^\perp = \{f \in \mathfrak{X}^* \mid \forall m \in M: \, f(m) = 0\}. $
- In the special case where $\mathcal{A}$ is a $C^*$-algebra, the contravariant equivalence of categories with (compact Hausdorff spaces, continuous maps) implies that $C^*$-subalgebras of $\mathcal{A}$ correspond to quotients of $\Sigma(\mathcal{A})$, and quotients of $\mathcal{A}$ correspond to closed subspaces of $\Sigma(\mathcal{A})$.
Are there some sort of analogous relationships with commutative Banach algebras? An example: Viewing the disc algebra $A(\mathbb{D})$ as a closed subalgebra of $C(\mathbb{T})$, the above analogies might lead us to expect that $\Sigma(A(\mathbb{D}))$ is a quotient of $\Sigma(C(\mathbb{T}))$. But $\Sigma(A(\mathbb{D})) \simeq \overline{\mathbb{D}}$ while $\Sigma(C(\mathbb{T})) \simeq \mathbb{T}$, so it looks like in this case the relationship is a subspace rather than a quotient (and going the other direction).