Let $\mathcal{A}$ be a commutative unital Banach Algebra.
Suppose $a,b \in \mathcal{A}$, and $\mathcal{B}$ is a Banach subalgebra generated by $a,b$.
Let $\Delta_{\mathcal{B}}$ denote the maximal ideal space of $\mathcal{B}$. Show that there exists a homeomorphism $ \Gamma:\Delta_{\mathcal{B}} \rightarrow \sigma(a,b)=\{ (\phi(a), \phi(b) ) : \phi \in \Delta_{\mathcal{B}}\}$.
Attempt:
I thought of using this theorem:
Suppose $X$ is a compact space and $Y $ is a Hausdorff space. If $\Gamma: X \rightarrow Y$ is bijection and continuous. Then $\Gamma$ is a homeomorphism.
I know that $\hat{a}, \hat{ b} \in C(\Delta_{\mathcal{B}}).$ Also, $\sigma(a) = Range (\hat{a})$ and $\sigma(b) = Range (\hat{b})$
And $\Delta_{\mathcal{B}}$ and $\sigma(a,b)$ are compact and Hausdorff.
From here I'm not sure how to show it's injective.
Any help will be appreciated.
Thank you.
To see injectivity: If $\phi$ and $\phi'$ are two elements of $\Delta_{\mathcal B}$ such that $(\phi(a), \phi(b)) = (\phi'(a), \phi'(b))$, then we must have $\phi = \phi'$ on all of $\mathcal B$ because $a$ and $b$ generate $\mathcal B$. (In particular, the elements of $\Delta_{\mathcal B}$ are compatible with Banach algebra operations.)