Show that each $y\in Y_1$ has a unique $x_y\in X.$ Setting: Let $X,Y$ be compact Hausdorff spaces and $E,F$ be any Banach spaces. 
Let $C(X,E)$ be the collection of all $E$-valued continuous functions on $X.$
$C(Y,F)$ is defined similarly. 
Endow sup-norm to both $C(X,E)$ and $C(Y,F).$
Let $T:C(X,E)\to C(Y,F)$ be a continuous linear operator. 

In this paper, the authors defined the following sets in page $188:$
$$Y_3 = \{ y\in Y: (Tf)(y) = 0 \text{ for all } f\in C(X,E) \} \quad \text{and}$$
$$Y_1 = \{ y\in Y\setminus Y_3: \exists x_y\in X\text{ such that }(Tf)(y) = 0 \text{ if } f(x_y) = 0, f\in C(X,E) \}.$$
Then the author stated that it is easy to see that point $x_y\in X$ corresponding to each $y\in Y_1$ is uniquely determined. 

Question: Show the bolded statement. 

Suppose not, that is, there exist distinct $x_y, x_y'\in X$ that correspond to $y\in Y_1.$ 
Then choose a function $f\in C(X,E)$ such that $f(x_y) \neq 0$ and $f(x_y') =0.$
Since $y\in Y_1$ and $f(x_y') = 0,$ so $(Tf)(y) = 0.$
However, in order to obtain a contradiction, I would require if and only if in the definition of $Y_1,$ that is $Tf(y) = 0$ if and only if $f(x_y) = 0.$
With the new statement, since $f(x_y) \neq 0,$ so $Tf(y) \neq 0,$ a contradiction. 

As $y\notin Y_3$, we know that there exists $g\in C(X,E)$ such that $(Tg)(y)\neq 0$ and thus $g(x_y')\neq 0$ and $g(x_y')\neq 0$ (as both are in $Y_1$). Set $a:= g(x_y')$ and recall that in compact spaces the Urysohn lemma holds. Hence, there exists a continuous function $h: X \rightarrow \mathbb{R}$ such that $$h(x_y)=0 \qquad \text{and} \qquad h(x_y')=1$$ 
Now define $$f: X \rightarrow E, \ f(z)= -h(z) a$$
Clearly, $f\in C(X,E)$ and $f(x_y)= -h(x_y) a =0$ and $f(x_y')=-h(x_y') a = -a$. As $f(x_y)=0$ we get $(Tf)(y)=0$. Hence, we get 
$$T(f+g)(y)= (Tf)(y)+ (Tg)(y)= (Tg)(y) \neq 0$$
but 
$$(f+g)(x_y')= f(x_y') + g(x_y') = -a + a=0$$
 which is a contradiction.
