Tietze Extension Theorem and Pasting Lemma

Tietze Extension Theorem

If $X$ is a normal topological space and $\displaystyle f:A\to\mathbb{R}$ is a continuous map from a closed subset $A\subseteq X$ , then there exists a continuous map $\displaystyle F:X\to\mathbb{R}$ with $F(a)=f(a)$ for all $a$ in $A$ .

Moreover, $F$ may be chosen such that $\sup\{|f(a)|:a\in A\}=\sup\{|F(x)|:x\in X\}$ , i.e., if $f$ is bounded, $F$ may be chosen to be bounded (with the same bound as $f$ ). $F$ is called a continuous extension of $f$ .

Pasting Lemma

Let $X$ , $Y$ be both closed (or both open) subsets of a topological space $A$ such that $A=X\cup Y$ , and let $B$ also be a topological space. If both $f|_X: X\to B$ and $f|_Y: Y\to B$ are continuous, then $f$ is continuous.

Proof:

Let $U$ be a closed subset of $B$ . Then $f^{-1}(U)\cap X$ is closed since it is the preimage of $U$ under the function $f|_X:X\to B$ , which is continuous. Similarly, $f^{-1}(U)\cap Y$ is closed. Then, their union $f^{-1}(U)$ is also closed, being a finite union of closed sets.