# Pasting Lemma It has been suggested that this page or section be merged into Continuity from Union of Restrictions. (Discuss) It has been suggested that this page or section be merged into Continuous Mapping on Finite Union of Closed Sets. (Discuss)

## Theorem

Let $X$ and $Y$ be topological spaces.

Let $A$ and $B$ be open in $X$.

Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.

Let $f \cup g$ be the union of the mappings $f$ and $g$:

$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$

Then the mapping $f \cup g : A \cup B \to Y$ is continuous.

### Corollary 1

Let $A$ and $B$ be closed in $X$.

Let $f : A \to Y$ and $g : B \to Y$ be continuous mappings that agree on $A \cap B$.

Then the mapping $f \cup g : A \cup B \to Y$ is continuous.

### Corollary 2

Let $\mathcal A = \left\{ {A_i: i \in I} \right\}$ be a set of sets that are open in $X$.

Let $f: \bigcup \mathcal A \to Y$ be a mapping such that:

$\forall i \in I : f \restriction A_i$ is continuous

Then $f$ is continuous on $\bigcup \mathcal A$.

## Proof

First it is shown that $f \cup g$ is well-defined.

It is sufficient to check that $f \cup g$ maps any $x \in A \cap B$ to a single value in $Y$.

Let $x \in A \cap B$.

Because $x \in A$:

$\map {f \cup g} x = \map f x \in Y$

Because $x \in B$:

$\map {f \cup g} x = \map g x \in Y$

But by definition of agreement of mappings:

$\map f x = \map g x$

so $f \cup g$ maps $x$ to a unique value in $Y$, as was to be shown.

$\Box$

Next it is shown that $f \cup g$ is continuous:

We show that $\map {\paren {f \cup g}^{-1} } U$ is open in $A \cup B$ for every open $U$ in $Y$.

Observe first that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$.

For if $x \in (f \cup g)^{-1}(U) \subseteq A \cup B$, then $(f \cup g)(x) \in U$; but if $x \in A$, then $f(x) = (f \cup g)(x) \in U$ i.e. $x \in f^{-1}(U)$ or if $x \in B$, then $g(x) = (f \cup g)(x) \in U$ i.e. $x \in g^{-1}(U)$. In both cases, $x \in f^{-1}(U) \cup g^{-1}(U)$. Conversely if $x \in f^{-1}(U) \cup g^{-1}(U)$, then without loss of generality consider the case of $x \in f^{-1}(U) \subseteq A$; in this case $(f \cup g)(x) = f(x) \in U$ so that $x \in (f \cup g)^{-1}(U)$.

Having established that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$, note that by continuity of $f$ and $g$, we have that $f^{-1}(U)$ is open in $A$ and $g^{-1}(U)$ is open in $B$. So there are an open sets $P$ and $Q$ of $X$ such that $f^{-1}(U) = P \cap A$ and $g^{-1}(U) = Q \cap B$. Since $A$ and $B$ are open in $X$ and intersections of open sets are open, this implies that $f^{-1}(U)$ and $g^{-1}(U)$ are open in $X$. Furthermore, note that $P \cap A$ and $Q \cap B$ are both subsets of $A \cup B$. Hence, $f^{-1}(U) = f^{-1}(U) \cap (A \cup B)$ and $g^{-1}(U) = g^{-1}(U) \cap (A \cup B)$.

Thus:

$(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U) = (f^{-1}(U) \cup g^{-1}(U)) \cap (A \cup B)$

demonstrating that $(f \cup g)^{-1}(U)$ is open in the subspace topology of $A \cup B$.

$\blacksquare$