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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?

user264745
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2 Answers2

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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 $g_1(x) = g_2(x)$.

Mason
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  • I known this approach in metric space. I haven’t prove $x\in \overline{A} \Rightarrow$ there is a net $(x_a)_{a\in A}$ and you used limit of sequence of continuity, which I know exist in metric space but I don’t about topological space. – user264745 Jan 17 '22 at 18:38
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    @user264745 I didn't use sequences anywhere. Nets are an adaptation of sequences that allow metric space sequence arguments to work in an arbitrary topological space. One such result that adapts is that $f$ is continuous at $x$ if and only if for every net $x_a \to x$ we have $f(x_a) \to f(x)$. – Mason Jan 17 '22 at 19:00
  • I accepted your answer because this is the third way to solve exercise $18.13$. – user264745 Jan 18 '22 at 05:35
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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.)

tomasz
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