Banach-algebra homeomorphism. Let $ A $ be a commutative unital Banach algebra that is generated by a set $ Y \subseteq A $. I want to show that $ \Phi(A) $ is homeomorphic to a closed subset of the Cartesian product $ \displaystyle \prod_{y \in Y} \sigma(y) $. Moreover, if $ Y = \{ a \} $ for some $ a \in A $, I want to show that the map is onto.
Notation: $ \Phi(A) $ is the set of characters on $ A $ and $ \sigma(y) $ is the spectrum of $ y $.
I tried to do this with the map
$$
f: \Phi(A) \longrightarrow \prod_{y \in Y} \sigma(y)
$$
defined by
$$
f(\phi) \stackrel{\text{def}}{=} (\phi(y))_{y \in Y}.
$$
I don’t know if $ f $ makes sense, and I can’t show that it is open or continuous. Need your help. Thank you!

Note that $\Phi  (A)$ is compact in the w$^*$-topology.  Also, $\prod \sigma(y)$ is compact Hausdorff in the product topology. For the map $f$ you defined, note that $Ker f = \{0\}$ since $Y$ generates $A$.
To prove continuity, take a net $\{\phi_\alpha\}_{\alpha \in I}$ in $\Phi(A)$, such that $\phi_\alpha \rightarrow \phi$ in w$^*$-topology. Then, $\phi_\alpha (y) \rightarrow \phi(y)$ in norm topology, for any $y \in Y$. ------------- (*)
Now consider a basic open set $V$ around the point $\prod_{y \in Y} \phi(y)$. Then, there exists $y_1, y_2, \ldots, y_k \in Y$ such that $V = \prod_{y \in Y} V_y $, where $V_y = \sigma (y)$ for any $y \in Y \setminus \{y_1, y_2, \ldots, y_k\}$ and $ V_{y_i} = V_i $ is an open ball in $\sigma (y_i)$ containing $\phi(y_i)$, for $i = 1, 2, \ldots, k$.
Using (*) we get that for each $i$, $\exists$ $\alpha_i$ such that $\phi_{\beta} (y_i) \in V_i$ for any $\beta \geq \alpha_i$. Since the index set $I$ is directed, $\exists$ $\alpha_0 \in I$ such that $\alpha_0 \geq \alpha_i$ for all $i$. Thus for any $\beta \geq \alpha_0$, we have that $\phi_{\beta} (y_i) \in V_i$ for each $i$, and hence $\prod_{y \in Y} \phi_{\beta} (y) \in V$. 
Thus, it follows that $\prod \phi_\alpha (y) \rightarrow \prod \phi(y)$ in $\prod \sigma(y)$, i.e, $f$ is continuous.
