Let me write definition of Pasting Lemma: Let $X=A \cup B$ where $A$ and $B$ are closed subsets of $X$. Let $g: A\to Y$ and $h:B\to Y$ be continuous function such that $g|A\cap B = h|A\cap B$. Define $f:X\to Y$ by $f\left( x\right) =\begin{cases}g\left( x\right) ,x\in A\\
h\left( x\right) ,x\in B\end{cases}$
then $f$ is continuous.
My request is can you summary the pasting lemma for me?
I looked some applications of pasting lemma and author says that $f$ is continuous since $g$ and $h$ are continuous. Does theorem say that like this? Is it enough to check $g$ and $h$ are continuous? Thank all of you for now.
Best Answer
I think the best way to understand the pasting lemma is to understand why a "naive" version of it is false:
To see why this is false, consider for example $A=[0,1]$, $B=(1,2]$, $f:[0,1]\rightarrow\mathbb{R}: x\mapsto 0$, and $g: B\rightarrow\mathbb{R}: x\mapsto 1$ (where all spaces involved have the usual topologies). The union of $f$ and $g$ is the function $$[0,2]\rightarrow\mathbb{R}: x\mapsto\begin{cases} 0 & \mbox{ if } x\le 1\\ 1 & \mbox{ if } x>1\\ \end{cases}$$ which is not continuous.
So we have a naive idea that gluing together continuous functions should result in a continuous function, but a simple counterexample to the most obvious way of stating that. The correct pasting lemma is what we get when we realize what additional hypotheses we were missing in order to rule out counterexamples like the above - namely, that both $A$ and $B$ must be closed (this can actually be weakened somewhat, but that's not worth focusing on at first).