Preference relations and the existence of extensions of functions representing them

Question

In a book I found the following question:

Let $\succsim$ be a complete preference relation on a nonempty set $X$, and let $\varnothing \neq B \subseteq A \subseteq X$. If $u \in [0,1]^A$ represents $\succsim$ on $A$ and $v \in [0,1]^B$ represents $\succsim$ on $B$, then there exists an extension of $v$ that represents $\succsim$ on $A$.

True or false?

The answer says that the statement is false, but I cannot really see a counterexample. Sure I am missing somebody, maybe about the very concept of extension of a function.

Is there somebody who can enlighten me on why it is the case that the answer is false?

Thank you for for your time.

Answer 1

Take $X=A=[0,1]$, and let $\succsim$ be simply $\ge$; clearly identity function on $A$ represents $\succsim$ on $A$. Let $B=\left[0,\frac12\right]$, and let $v:B\to[0,1]:x\mapsto 2x$; $v$ represents $\succsim$ on $B$, but clearly $v$ cannot be extended to a representation of $\succsim$ on $A$.