Exercise 13, Section 18 of Munkres’ Topology

Question

Let $A \subset X$; let $f \colon A \to Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g \colon \overline{A} \to Y$, then $g$ is uniquely determined by $f$.

My attempt: let $g_1:\overline{A}\to Y$ and $g_2:\overline{A}\to Y$ are continuous maps, and $g_1|_A=g_2|_A=f$. Clearly $A\subseteq \{x\in \overline{A}| g_1(x)=g_2(x)\}=G\subseteq \overline{A}$. By Exercise 13, Section 17 of Munkres’ Topology And Using This Result to Prove $\{x|f(x)=g(x)\}$ is Closed in $X$ result, $G$ is closed in $\overline{A}$. So $\overline{A}\subseteq G$. Thus $\overline{A}=\{ x\in \overline{A}|g_1(x)=g_2(x)\}$. In other word, $g_1(x)=g_2(x),\forall x\in \overline{A}$. Hence $g_1=g_2$. Is this proof correct?

Answer 1

It is natural to use nets to show that $g_1 = g_2$ on $\overline{A}$. Let $x \in \overline{A}$ be arbitrary. Since $x \in \overline{A}$, there is a net $(x_{a})_{a \in A}$ in $A$ converging to $x$. Since $g_1$ and $g_2$ are continuous on $\overline{A}$, it follows that $g_1(x_a) \to g_1(x)$ and $g_2(x_a) \to g_2(x)$. Since limits of nets in a Hausdorff space are unique, it follows that $g_1(x) = g_2(x)$.

Answer 2

The proof is correct as written.

To see that the hypothesis is necessary, let $Y=[0,1]\cup \{1'\}$ be the interval with double endpoint. Then $Y$ is a compact $T_1$ space (and a manifold) and for $X=[0,1]$ and $A=[0,1)$, you have two distinct continuous extensions of $f\colon A\to Y$ given by $f(x)=x$.

(I believe that with a bit more work, you can show that this property actually characterises Hausdorff spaces.)