Show that this linear mapping on $\mathscr{C}_0(X)$ to $\mathscr{C}_0(F)$ is surjective Let $X$ be a normal locally compact space and $F$ a closed subset of $X$. We denote by $\mathscr{C}_0(X)$ the set of 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. 
If we define $T \colon \mathscr{C}_0(X) \to \mathscr{C}_0(F)$ by $f \mapsto f|_F$, then this map is well defined, since one can easily check that for all $f \in \mathscr{C}_0(X)$ we also have $f|_F \in \mathscr{C}_0(F)$. 

My question now is, does T necessarily have to be surjective?

 A: It is surjective, you can even drop the assumption that $X$ is normal. However, we shall assume that $X$ is Hausdorff.
Since $X$ is locally compact, it has a one-point compactification $X^+ = X \cup \{\infty\}$, where $\infty \notin X$. Open subsets of $X^+$ are all open subsets of $X$ plus all complements in $X^+$ of compact $C \subset X$. These complements are the open neigbborhoods of $\infty$. This construction works for both for non-compact and compact $X$. In the first case  $X$ is dense in $X^+$, in the second case $\infty$ is an isolated point of $X^+$. Note also that for compact $X$ we have $\mathscr{C}_0(X) = \mathscr{C}(X)$ and $\mathscr{C}_0(F) = \mathscr{C}(F)$.
Now a function $f : X \to \mathbb C$ is in $\mathscr{C}_0(X)$ iff the function $f^+ : X^+ \to \mathbb C, f^+(x) = f(x)$ for $x \in X$ and $f^+(\infty) = 0$, is continuous. To see this, note that if $f^+$ is continuous, then also $f$ is continuous and for each $\varepsilon > 0$ there exists a compact $C \subset X$ such that $\lvert f^+(\xi) \rvert < \varepsilon$ for $\xi \in X^+ \setminus C$. Thus $\{|f| \geq \varepsilon\} \subset C$. Since $\{|f| \geq \varepsilon\}$ is closed, it is compact. Conversely, if $f \in \mathscr{C}_0(X)$ it remains to show that $f^+$ is continuous in the point $\infty$ (in all other points it is continuous because $X$ is open in $X^+$). But $C = \{|f| \geq \varepsilon\}$ is compact and $\lvert f^+(\xi) \rvert < \varepsilon$ for $\xi \in X^+ \setminus C$.
Given $g \in \mathscr{C}_0(X)$, we can show as above that the function $g^+ : F^+ \to \mathbb C, g^+(x) = g(x)$ for $x \in F$ and $g^+(\infty) = 0$, is continuous. This amounts in showing that $g^+$ is continuous in the point $\infty$. But $D = \{|g| \geq \varepsilon\}$ is compact and $\lvert g^+(\xi) \rvert < \varepsilon$ for $\xi \in F^+ \setminus C$. The latter set is of course open in $F^+$.
$F^+ = F \cup \{\infty\}$ is closed in $X^+$ since $X^+ \setminus F^+ = X \setminus F$ is open in $X$, hence also in $X^+$. Compact Hausdorff spaces are normal and the Tietze extension theorem shows that $g^+ : F^+ \to \mathbb C$ has a continuous extension $\phi : X^+ \to \mathbb C$ (note that Tietze applies to real-valued functions, but we may extend to real and the imaginary part of $g^+$ to obtain $\phi$). Then $f = \phi \mid_X$ has the property $f^+ = \phi$, thus $f \in \mathscr{C}_0(X)$ and $T(f) = g$.
