Exercise 18.13

Proof. Suppose that $g_1$ and $g_2$ are both continuous functions from $\overline{A}$ to $Y$ that extend $f$ so that $g_1(x)=g_2(x)=f(x)$ for all $x\in A$. Clearly $g_1=g_2$ if and only if $g_1(x)=g_2(x)$ for all $x\in \overline{A}$. So suppose that this is not the case so that there is an $x_0\in \overline{A}$ where $g_1(x_0)\ne g_2(x_0)$. Since $Y$ is a Hausdorff space and $g_1(x_0)$ and $g_2(x_0)$ are distinct, there are disjoint neighborhoods $V_1$ and $V_2$ of $g_1(x_0)$ and $g_2(x_0)$, respectively. Then there are also neighborhoods $U_1$ and $U_2$ of $x_0$ such that $g_1(U_1)\subset V_1$ and $g_2(U_2)\subset V_2$ by Theorem 18.1 since both $g_1$ and $g_2$ are continuous.

Now let $U=U_1\cap U_2$ so that $U$ is also a neighborhood of $x_0$. Since $x_0\in \overline{A}$, it follows that $U$ intersects $A$ so that there is a $y\in U$ where also $y\in A$ by Theorem 17.5. Since $y\in A$ we have that $g_1(y)=g_2(y)=f(y)$. We also have that $y\in U_1$ and $y\in U_2$ since $U=U_1\cap U_2$. Thus $g_1(y)\in g_1(U_1)$ so that $f(y)=g_1(y)\in V_1$ since $g_1(U_1)\subset V_1$. Similarly $f(y)=g_2(y)\in V_2$, but then we have that $f(y)\in V_1\cap V_2$, which contradicts the fact that $V_1$ and $V_2$ are disjoint! Hence it must be that $g_1=g_2$, which shows uniqueness. □