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Problem. Let $A \subseteq X$. Let $f:A \to Y$ be continuous. Let $Y$ be Hausdorff. Prove that if $f$ is extended to a continuous function $g: \overline{A} \to Y$, then $g$ is uniquely determined by $f$.

My questions are,

  • Is the condition $Y$ is Hausdorff necessary? More specifically, is the there any counterexample to the following proposition?

    Let $A \subseteq X$. Let $f:A \to Y$ be continuous. Let $Y$ be $T_1$-space. Prove that if $f$ is extended to a continuous function $g: \overline{A} \to Y$, then $g$ is uniquely determined by $f$.

  • If the answer to the above question is negative then is the converse of the problem true? More specifically, is the following proposition true?

    Let $A \subseteq X$. Let $f:A \to Y$ be continuous. Suppose that $f$ is extended to a unique continuous function $g: \overline{A} \to Y$. Then $Y$ is Hausdorff.

Community
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    Suppose $Y$ has the indiscrete topology. Then any function $g: \bar{A} \to Y$ is continuous. I am not sure about your $T_1$ condition. – J126 Nov 11 '16 at 13:39
  • I haven't thought of an explicit counterexample using it yet, but I feel like the line with two origins could also be a counterexample http://math.stackexchange.com/questions/1038320/line-with-two-origins-is-a-manifold-but-not-hausdorff – Chill2Macht Nov 11 '16 at 13:40
  • Also, your converse should be: If every extension is uniquely determined by $f$, then $Y$ is Hausdorff. – J126 Nov 11 '16 at 13:41
  • The point of my first comment is that you need some condition on $Y$ to make it true. But, I don't know if $T_1$ is good enough. Since you have edited your question, I'm guessing you don't need clarification for my second comment. – J126 Nov 11 '16 at 13:45
  • @JoeJohnson126: Regarding your first comment, I understand that function $g:\overline{A}\to Y$ is continuous but is the function unique? –  Nov 11 '16 at 13:54
  • Since all functions into $Y$ are continuous, it doesn't have to be unique. Let $X = {a,b}$ with the indiscrete topology, and $Y = {1,2}$ with the indiscrete topology. Then we may extend $f: {a} \to Y$, $f(a) = 1$, by either $g(b) = 1$ or $g(b) = 2$. – J126 Nov 11 '16 at 14:02

2 Answers2

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The condition is necessary. Let $X$ be the space $\{a,b\}$ with the topology whose open subsets are the empty subset and $X$. The adherence of $\{a\}$ is $X$. And you can extend a the continuous function $f:X\rightarrow X$ on $\{a\}$ $f(a)=a$ by any element of $X$.

Tsemo Aristide
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Here is an answer to your first question. If we let $X = \mathbb{R}$ with the standard topology and let $Y$ be $\mathbb{R}$ with the finite complement topology, then any function $f: X \to Y$ with $f^{-1}(y)$ a finite set of points for all $y \in Y$ is continuous. This is because the closed sets of $Y$ are the finite sets. If $C \subseteq Y$ is closed, then $f^{-1}(C)$ is a finite collection of points, which is also closed in $X$. Let $A = [0,1) \subseteq X$ and $f = \operatorname{Id}$. Then we can extend $f$ to $\bar{A}$ by either $g(1) = 1$ or $g(1) = 10$. And $Y$ is $T_1$.

J126
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  • I still don't know the answer to your second question. My gut feeling is that it is false. There are a lot of really weird spaces out there. – J126 Nov 11 '16 at 14:36