$τ: Y \to Χ$ is a continuous map and $A: C(X)\to C(Y)$ is defined by $(Af)(y) = f(τ(y))$. How $||A||=1$? The following is from Conway's Functional Analysis :

If $X$ and $Y$ are compact spaces and $τ: Y \to Χ$ is a continuous map, define $A: C(X) \to C(Y)$ by $(Af)(y) = f(τ(y))$. Then $A \in \mathcal{B} (C(X), C(Y))$ and $||A||=1$.

Q.1 - $f$ is continuous because $f \in C(X)$ and $τ$ is continuous by hypothesis, then so is their composition with respect to y meaning that $Af$ is continuous so I couldn't reduce it to the continuity of $A$ and use equivalence of boundedness and continuity; so how $A \in \mathcal{B} (C(X), C(Y))$?
Q.2 - Why $||A||=1$? An attempt solution in here is unclear for both inequalities to reach $||A||=1$.
 A: $X$ and $Y$ are compact spaces and $τ: Y \to Χ$ is a continuous map, define $A: C(X) \to C(Y)$ by $(Af)(y) = f(τ(y))$.
Note that $Af = f \circ τ$.  It is easy to see that $A$ is linear. Now, let $E= \tau(Y)$. We have,
\begin{align*}  \| Af\|_\infty & = \|f \circ τ\|_\infty = \\ 
&= \sup\{|f \circ \tau|(y) : y \in Y\}= \\ 
&=\sup\{|f |(x) : x \in \tau(Y)\} \leq \\ 
&\leq  \sup\{|f |(x) : x \in X\} = \\ 
& =\|f\|_\infty
\end{align*}
So $A \in \mathcal{B} (C(X), C(Y))$ and $\|A\|\leq 1$.
To prove that $\|A\|= 1$, take $f$ to be the constant function defined on $X$ with value $1$. Then $f \circ \tau$ is the constant function defined on $Y$ with value $1$. So we have,
$$  \| Af\|_\infty = \|f \circ τ\|_\infty = 1 = \|f \|_\infty $$
It follows that  $\|A\|= 1$.
A: Showing linearity is not hard, simply check the axioms.
Showing boundedness is easy: since $\tau(Y) \subseteq X$ is a compact set
$$||Af-Ag|| = \max_{y \in Y} |f( \tau (y)) - g( \tau (y))| = \\ = \max_{x \in \tau (Y)} |f(x)-g(x)| \le \max_{x \in X} |f(x)-g(x)| = ||f-g||$$
Hence $||A|| \le 1$.
