## 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.

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