Elementary question about extension of continuous functions Let $X$ be a completely regular space and let $T$ a topological space such that $X \subseteq T \subseteq \beta X$. Then $\beta T = \beta X$, where $\beta$ denotes the Stone-Cech compactification.
Solution: Let $f: T \mapsto [0,1]$. Then it suffices to show that the restriction of f to $X$ can be extended to $\beta X$".. 
Why is this? 
 A: Here is my answer to the modified question: Because $X$ is dense in $\beta X$, it is also dense in $T$, and therefore every continuous function on $T$ is uniquely determined by its restriction to $X$.

Here is my answer to the original question, which asked when, in order to see that a continuous function $f:D\to [0,1]$ has a continuous extension to $Z$, it suffices to show that the restriction of $f$ to $C$ has a continuous extension to $Z$, where $C\subseteq D\subseteq Z$ and $Z$ is a topological space.:
This would suffice precisely when every continuous function from $D$ to $[0,1]$ is determined by its restriction to $C$.  This will always be true if $C$ is dense in $D$.  If $D$ is completely regular, then density of $C$ is also necessary.  To see this, suppose $D$ is completely regular and $C$ is not dense.  Then there is a continuous function $f:D\to [0,1]$ such that $f$ is zero on $\overline C$, but $f(x)=1$ for some $x\in D\setminus\overline C$.  Since the zero function on $D$ is also a continuous extension of the zero function on $C$, the zero function on $C$ has more than one continuous extension to $D$.
A: Recall that $\beta T$ is characterized by being (up to homeomorphism) the unique compact (Hausdorff) space of which $T$ is a dense subset and such that any continuous map $f:T\to[0,1]$ can be extended to a continuous map $g:\beta T\to[0,1]$.
Ok, so, given $f:T\to[0,1]$, continuous, we want to show that it extends to a continuous  $g:\beta X\to[0,1]$. 
First, $\beta T$ makes sense since $T$ is completely regular, being a subspace of $\beta X$.
Second, $X$ is dense in $T$, so any continuous $f:T\to[0,1]$ is completely determined by its restriction to $X$.
Third, by definition of $\beta X$, this restriction extends to a continuous $g:\beta X\to[0,1]$. 
Since $f\upharpoonright X$ uniquely determined $f$, we have shown that any continuous $f:T\to[0,1]$ extends to a continuous $g:\beta X\to[0,1]$.
This proves the theorem: Since $X\subseteq T$, then $\beta X\subseteq \beta T$. If $\beta T$ happened to be larger than $\beta X$, then we have that any continuous $h:X\to[0,1]$ can be extended to a continuous $j:\beta T\to[0,1]$, contradicting that $\beta X$ is largest with this property. The reason why this extension exists, is because if $i:\beta X\to[0,1]$ is the continuous extension of $h$, then its restriction to $T$ is continuous, and can therefore be extended.
A: The answer to this question (also about Stone-Cech compactification) is similar.
