Uniqueness of a continuous extension of a continuous map from a set to its closure

Question

Suppose $f$ is a continuous map from a space $A$ to a Hausdorff space Y. Then I know that $f$ can be extended uniquely to a continuous map from closure of $A$ to Y. What is a counterexample to the fact that this need not be true if Y is not Hausdorff? I think two point sets will work (and this has been hinted here Uniqueness of continuous extension from $A$ to $\overline{A}$ for maps into a Hausdorff space), but how?

Answer 1

This isn't a small example, but it might be intuitive.

Add a "second zero", $0^*$, to the reals to get a set set $Y = \mathbb{R} \cup \{ 0^* \}$. Topologize $Y$ so that

• the basic open neighbourhoods of each $x \in \mathbb R \subseteq Y$ are the usual basic open neighborhoods of $x$ in the real line,
• the basic open neighborhoods of $0^*$ are of the form $( - r , 0 ) \cup \{ 0^* \} \cup ( 0 , r )$ (so just the open interval $(-r,r)$ but replacing $0$ with $0^*$) for $r > 0$.

Note that $0$ and $0^*$ cannot be separated by disjoint open neighborhoods.

Consider $A = \mathbb R \setminus \{ 0 \}$ as a subset of the real line, and let $f : A \to Y$ be the identity function. This is continuous, but can be extended to $\overline{A} = \mathbb R$ in two ways: either $f(0) = 0$ or $f(0) = 0^*$. (Basically, the two subspaces $Y \setminus \{ 0^* \}$ and $Y \setminus \{ 0 \}$ of $Y$ are both homeomorphic to the real line.)

Answer 2

Let $Y = \{a, b\}$ with the indiscrete topology. Let $f : \mathbb Q \to Y$ be any function. Since $Y$ is indiscrete, $f$ is continuous. Any extension of $f$ to $\mathbb R$ is also continuous for the same reason. Thus, the extension is not unique.

Answer 3

Let $A=(0,1) \subset [0,1]$ with the usual topology, and $Y= \{a, b\}$ where the open sets of $Y$ are $\emptyset, Y$ and $\{ a \}$.

Consider $f_a: A \longrightarrow Y$ the constant function $x \mapsto a$.

Then $f_a$ can be extended by (at least) two distinct continuous functions $F_1, F_2: [0,1] \longrightarrow Y$ defined as

$$F_1(x) = a$$

and

$$F_2(x) = \begin{matrix} a & \mbox{ if $x \in A$} \\ b & \mbox{ if $x \notin A$} \end{matrix} $$