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

Question

I have some problems with the following exercise:

Let $X$ be a normal locally compact space and $F$ a closed subset of $X$. If $\mathscr{M} \equiv \{f \in \mathscr{C}_0(X) \colon f|_F \equiv 0\}$, then $\mathscr{C}_0(X)/\mathscr{M}$ is isometrically isomorphic to $\mathscr{C}_0(F)$. (Note that $\mathscr{C}_0(X) \equiv$ all continuous functions $f \colon X \to \mathbb{C}$ such that for all $\epsilon >0$ we have that $\{|f| \geq \epsilon\} = \{x \in X \colon |f(x)| \geq \epsilon \}$ is compact.)

It is clear that we are to work with the function $\hat{T} \colon \mathscr{C}_0(X)/\mathscr{M} \to \mathscr{C}_0(F)$ defined by $f+\mathscr{M} \mapsto f|_F$. Now it is easy to see that for $f+\mathscr{M} = g+\mathscr{M}$ we have $\hat{T}(f+\mathscr{M})=\hat{T}(g+\mathscr{M})$ and that $\hat{T}(f+\mathscr{M}) = f|_F \in \mathscr{C}_0(F)$, hence our function is well defined. (The second claim follows from the fact that for all $\epsilon$ we have $\{x \in F \colon |f(x)| \geq \epsilon\} = \{|f|\geq \epsilon\} \cap F = (|f|)^{-1}([\epsilon, \infty)) \cap F \subset \{|f|\geq \epsilon\}$; so we have, by continuity of $|f|$ and since $F$ is closed, a closed subset of a compact set, thus it is compact.) Again it is clear that $\hat{T}$ is injective. Now my first problem with this exercise comes with showing that $\hat{T}$ is surjective; Since $X$ is normal, one could make use of Tietze's extension Theorem to get a continuous extension of some function $g \in \mathscr{C}_0(F)$ to a continuous function $G$ on all of $X$, satisfying $G|_F = g$ and thus $\hat{T}(G + \mathscr{M}) = g$.

Question: Yet how can we ensure that $G \in \mathscr{C}_0(X)$?

Question: 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 $\|f+ \mathscr{M}\| = inf \{\|f+g\| \colon g \in \mathscr{M} \}$?

Thanks in advance!

Answer 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.