## § Pasting lemma

Let $f: X \rightarrow Y$ be a function. Let $A, B$ are closed subets of $X$ such that $X = A \cup V$. Then $f$ is continuous iff $f|_A$ and $f|_B$ are continous.

#### § Forward: $f$ continuous implies restriction continuous

This is clear from how restrictions work. Pick $i: A \rightarrow X$ to be the function that embeds $A$ with the subspace topology into $X$. This is continuous by the definition of the subspace topology. Now define $f|_A : A \rightarrow Y \equiv f \circ i$. which is continous since it is the composition of continuous functions.

#### § Backward: restrictions continuous implies $f$ continuous.

Let $V \subseteq Y$ be closed. Then $f_A^{-1}(V)$ is closed in $A$ by the continuity of $f_A$ Now see that $f_A^{-1}(V)$ is closed in the subspace topology of $A$ means that there is some closed $P \subseteq X$ such that $f|_A^{-1}(V) = A \cap P$. Since both $A$ and $P$ are closed in $X$, this means that $f^{-1}(V)$ is closed in $X$ (see that we have filted "closed in $A$" to "closed in $X$). Similarly, we will have that $f|_B^{-1}(V) = B \cap Q$ for some closed $B$ and $Q$. Then we can write: \begin{aligned} &f^{-1}(V) \\ &= f|A^{-1}(V) \cup f|B^{-1}(B) \\ &= (A \cap P) \cup (B \cap Q) \\ &= \texttt{finite union of closed sets} \\ &= \texttt{closed} \\ \end{aligned}