Isometric isomorphism between $\mathscr{C}_0(X)/\mathscr{M}$ and $\mathscr{C}_0(F)$

Solution 1:

For what it's worth, a partial answer.

Question 2: How can we prove rigorously that $\|\hat{T}(f+\mathscr{M})\| = \|f+ \mathscr{M}\|$, where $\|\hat{T}(f+\mathscr{M})\| = sup_{x \in F} \{|f(x)|\}$ and $\|f+ \mathscr{M}\| = inf \{\|f+g\| \colon g \in \mathscr{M} \}$?

We are given an $f \in \mathscr C_0(X)$. Let $U$ be any neighborhood of $F$, i.e. open set with $F \subset U$. By Urysohn's lemma there exists a continuous $S_U:X \to [0,1]$ such that $S_U(x) = 0$ for all $x \in X \setminus U$ and $S_U(x) = 1$ for all $x \in F$. Define $$ f_U(x) = S_U(x)f(x). $$ We have $g = f_U - f \in \mathscr M$. Moreover, We note that $\|f + g\| = \|f_U\| \leq \sup_{x \in U} |f(x)|$. Thus, we have $$ \inf \{\|f+g\| \colon g \in \mathscr{M} \} \leq \inf\{\sup\{|f(x)| : x \in U, U\text{ is a nbhd of }F\}\}. $$ Because $f$ is in $\mathscr C_0(X)$, we can conclude that this infimum is at most $\sup\{|f(x)| : x \in F\}$, as was desired.