----- > [!proposition] Proposition. ([[the pasting lemma]]) > Let $X=A \cup B$, where $A$ and $B$ are both [[closed set|closed]] (or both open) in $X$. Let $f:A \to Y$ and $g:B \to Y$ be [[continuous]]. If $f(x)=g(x)$ for every $x \in A \cap B$, then $f$ and $g$ combine to give a [[continuous]] function $h: X \to Y$, defined by setting $h(x):=\begin{cases} f(x) & x \in A ;\\ g(x) & x \in B. \end{cases}$ Said differently, a function $h:X \to Y$ is [[continuous]] provided that $h |_{A}$ and $h |_{B}$ are [[continuous]]. \ [[domain restriction preserves continuity|The converse clearly holds]]: if a function $h:X \to Y$ is [[continuous]], then so must be $h |_{A}$ and $h |_{B}$. > [!proof]- Proof. ([[the pasting lemma]]) > Let $A,B$ [[closed set|closed]] in $X$ and suppose $h |_{A}$ and $h |_{B}$ are [[continuous]], [[subspace topology|in the sense that]] for any closed $V \subset Y$ we have that $h |_{A}^{-1}(V)=h_{}^{-1}(V) \cap A$ and $h_{B}^{-1}(V)=h ^{-1}(V)\cap B$ are closed in $X$. [[closed sets behave complementarily to open sets|The union of these two sets is closed]] too and equals $h ^{-1}(V)$, from which the result follows. > The converse is just [[domain restriction preserves continuity]]. > Note that if we replace 'closed' with 'open' above the same argument goes through. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```