Pasting Lemma

 

Thm 1

Let $X$ be a topological space  and  $Y$ be subspace of $X$.

Let $U_{\alpha}$ be open set in $X$ for $\alpha \in \Lambda$  such that

$X = \displaystyle \bigcup_{\alpha \in \Lambda} U_{\alpha}$

Let $f_{\alpha} : U_{\alpha} \to Y$ be continuous  such that

$x \in U_{\alpha} \cap U_{\beta}$  $\Rightarrow$  $f_{\alpha}(x) = f_{\beta}(x)$

( Consider $U_{\alpha}$ as the subspace of $X$ )

 

Define $f : X \to Y$ by

$f(x) = f_{\alpha}(x)$

 

Then

$f$ is continuous

 

  Let $U$ be an open set in $Y$

 

  By definition of $f$,

  $f^{-1}(U) = \displaystyle \bigcup_{\alpha \in \Lambda} f_{\alpha}^{-1}(U)$

 

  Since $f_{\alpha}$ is continuous,

  $f_{\alpha}^{-1}(U)$ is open set in $U_{\alpha}$

  Since $U_{\alpha}$ is open set in $X$, by Thm 2 - 2

  $f_{\alpha}^{-1}(U)$ is open set in $X$

 

  $\therefore$  $f^{-1}(U)$ is open set in $X$

  $\therefore$  $f$ is continuous

 

 

 

Thm 2 Pasting Lemma

$X$ : topological space, $Y \subseteq X$

Let $A$ and $B$ be closed set in $X$  such that

$X = A \cup B$

Let $f : A \to Y$, $g : B \to Y$ be continuous  such that

$x \in A \cap B$  $\Rightarrow$  $f(x) = g(x)$

( Consider $A$, $B$ as the subspace of $X$ )

 

Define $h : X \to Y$ by

$h(x) = \begin{cases}
f(x) & x \in A \\
g(x) & x \in B
\end{cases}$

 

Then

$h$ is continuous

 

  Let $C$ be a closed set in $Y$

 

  By definition of $h$,

  $h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C)$

 

  Since $f$ is continuous,

  $f^{-1}(C)$ is closed in $A$

  Since $A$ is closed, by Thm 2 - 2

  $f^{-1}(C)$ is closed in $X$

 

  Similarly we can show

  $g^{-1}(C)$ is closed in $X$

 

  $\therefore$  $h^{-1}(C)$ is closed in $X$

 

  By Thm 2 -5,

  $h$ is continuous