extending a continuous map to quotient spaces.

Question

Let $A\subset B$ and $C\subset D$ metric spaces. Suppose that there is $f: B-A\rightarrow D-C$ a homeomorphism such that for any $x\in A$, and any converging sequence $x_{n}\rightarrow x$ such that $\{x_{n}\}\in B-A$ for any $n\in \mathbb{N}$, we have $\lim_{n\rightarrow \infty} f(x_{n})\in C$.

Does $f$ extend to $B/A\rightarrow D/C$ where $B/A, D/C$ are the quotient space? Can we generalize this result to non metrizable hausdorff spaces ?

Answer 1

The extension to $B/A\to D/C$ is not continuous in general. For instance, let $B=[0,1)$, $A=(0,1)$, $D=(0,1)\cup\{2\}$, and $C=(0,1)$. Then $B-A$ and $D-C$ are both singletons and so are homeomorphic, and the condition on sequences in $B-A$ is trivial. However, the extension to a map $B/A\to D/C$ that sends $A$ to $C$ is not continuous, since $\{C\}$ is closed in $D/C$ but $\{A\}$ is not closed in $B/A$.

Answer 2

It is entirely possibile that $f$ does not "extend" to a continuous function - i.e. that there is no continuous function $g:B/A\to D/C$ such that $g\circ \left.\pi_A\right\rvert_{B\setminus A}=\pi_C\circ f$.

Consider \begin{align}B&=\left\{x\in\Bbb R^2\,:\, (x_2=0\land x_1\ge 0)\lor \left(x_1=1\land \frac1{x_2}\in\Bbb N\right)\right\}\\ A&=\{x\in\Bbb R^2\,:\,x_2=0\land x_1>0\}\\ D&=\left\{x\in\Bbb R\,:\, \frac1x\in\Bbb N\right\}\cup\{0,-1\}\\ C&=\{0\}\\ f(x)&=\begin{cases}x_2&\text{if }x_1= 1\\ -1&\text{if } x=0\end{cases}\end{align}

Now, $f:B\setminus A\to D\setminus C$ is a homeomorphism. Moreover, if $x^n\to y\in A$ and $x^n\notin A$, then eventually $x^n_1=1$ and therefore $f(x^n)=x^n_2\to 0$. However, there is no continuous function $g:B/A\to D/C$ such that $g([x])=f(x)$ for all $x\in B\setminus A$. Notice preliminarily that the quotient ma $D\to D/C$ is actually a homeomorphism, so we can identify points in $D$ with their classes in $D/C$. Every neighbourhood of $A$ contains a tail of the sequence $y^n=[(1,1/n)]$, on which $g$ takes value $\frac1n$. Therefore, $g(A)$ must be $0$. However, $\overline{\{A\}}=\{[0],A\}$, and therefore $0=g(A)=g([0])$, since $0$ is closed in $D$. However, by hypothesis $g([0])=-1$: absurd.