# Pasting Lemma/Pair of Continuous Mappings on Open Sets/Proof 2

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## 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.

## Proof

From Union of Mappings which Agree is Mapping

- $f \cup g$ is well-defined.

By Definition of Continuous Mapping:

- $f \cup g$ is continuous if and only if $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$ for every open $U$ in $Y$.

Let $U$ be an arbitrary open subset in $Y$.

From Preimage of Union Mapping is Union of Preimages:

- $\paren {f \cup g}^{-1} \sqbrk U = f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$

By Definition of Continuous Mapping:

From Open Set in Open Subspace:

- $f^{-1} \sqbrk U$ and $g^{-1} \sqbrk U$ are open in $X$.

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:

From Open Set in Open Subspace:

- $f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$ is open in $A \cup B$

Hence:

- $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$

Since $U$ was an arbitrary open subset of $Y$:

- for all open subsets $U$ of $Y$, $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$

By Definition of Continuous Mapping:

- $f \cup g$ is continuous.

$\blacksquare$