glueing lemma on connected subset.

Question

The glueing lemma is as followed: if $X = A \bigcup B$, $A$ and $B$ both open or closed and $f:A \rightarrow Y$, $g:B \rightarrow Y$ is continuous, $f$ and $g$ agrees on $A \bigcap B$, then you can have a continuous function $f \bigcup g: X \rightarrow Y$ such that $f \bigcup g(x) = f(x)$ for $x \in A$ and $f \bigcup g(x) = g(x)$ for $x \in B$. Every version of this lemma I found online is of this form. I wonder whether this will hold true if I remove the condition $A$,$B$ be opened or closed and replaced it with $A$,$B$ connected and their intersection is non-empty.

Answer 1

Let $A=\Bbb R\times\{0\}$, and let

$$B=\big([0,1]\times[0,1]\big)\cup\big([1,2]\times(0,1]\big)\,;$$

$A\cap B=[0,1]\times\{0\}$, so $A,B$, and $A\cap B$ are all connected.

Let

$$f:A\to\Bbb R:\langle x,0\rangle\mapsto 0$$

and

$$g:B\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases} 0,&\text{if }x\le 1\\ (x-1)^2,&\text{if }x>2\,; \end{cases}$$

then $f$ and $g$ are continuous and agree on $A\cap B$, but they have no continuous extension to $A\cup B$: if $1<a\le 2$,

$$\lim_{y\to 0}g(\langle a,y\rangle)=(a-1)^2\ne 0=\lim_{x\to a}f(\langle x,0\rangle)\,.$$