Pasting Lemma. Let $A$ and $B$ be both open or closed subsets of a topological space $X$ such that $A \cup B = X$. Let $f: A \to Y$ and $g: B \to Y$ be continuous such that $f = g$ for all $x \in A \cap B$. Prove that $h: X \to Y$ is continuous such that $h\restriction_A = f$ and $h\restriction_B = g$.
Proof: Assume $A$ and $B$ are both closed. Take a subset $C \subset Y$ to be closed. We must prove that $h^{-1}(C)$ is closed, and then we are done. (cont'd)
Question 1: Why does it suffice to prove that $h^{-1}(C)$ is closed? Don't we also have to prove that $h\restriction_A = f$ and $h\restriction_B = g$? It says in the problem to prove two things: that $h: X \to Y$ is continuous and that $h\restriction_A = f$ and $h\restriction_B = g$. I understand that if we prove that $h^{-1}(C)$ is closed, then we have successfully proved the continuity of $h: X \to Y$. But it doesn't address the second statement f the proof that $h\restriction_A = f$ and $h\restriction_B = g$.
Proof (cont'd): We know that $h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C)$. (cont'd)
Question 2: How do we know this? There's nothing in our given information that says this. All the information we know is that "Let $A$ and $B$ be both open or closed subsets of a topological space $X$ such that $A \cup B = X$. Let $f: A \to Y$ and $g: B \to Y$ be continuous such that $f = g$ for all $x \in A \cap B$." How can we construct $h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C)$ from just this information?
Proof (cont'd): (Omitted because I understand the rest of the proof)
Question 3: Can $A$ and $B$ both be clopen? Or can $A$ and $B$ be disjoint subsets of $X$?
Best Answer
The function $h$ is defined by $h\restriction_A=f$ and $h\restriction_B=g$. There's a unique function with those properties: since $X=A\cup B$, an element $x\in X$ either is in $A$ or in $B$; in the former case define $h(x)=f(x)$, in the latter define $h(x)=g(x)$. This is a good definition, because $f(x)=g(x)$ if $x\in A\cap B$.
Now a function is continuous if and only if the inverse image of closed sets is closed.
Let $x\in h^{-1}(C)$; then either $x\in A$ or $x\in B$; in the former case $h(x)=f(x)$, so $x\in f^{-1}(C)$, in the latter $x\in g^{-1}(C)$. Therefore $h^{-1}(C)\subseteq f^{-1}(C)\cup g^{-1}(C)$. Can you prove the converse inclusion?
Finally there's nothing in the question that suggests $A$ and $B$ be clopen or disjoint.