Suppose that $\{A_i\}$ is a finite collection of closed sets in $X$ such that $X = \bigcup A_i$, and $f : X \to Y$ restricted to each $A_i$ is continuous. Then $f$ is continuous.
Okay. The base case ($n=2$) is just the pasting lemma. I am having a little trouble with the inductive case, specifically with what the hypothesis is and how to deal with continuity with respect to subspaces. I believe the inductive hypothesis is to assume that any function over a topological space that is the union of $n-1$ closed sets such that the function restricted to each of these $n-1$ closed set is continuous must be continuous over the whole space. Then I assume I have a topological space $X = \bigcup_{i=1}^n A_i$, where each $A_i$ is closed, and a function $f : X \to Y$ such that $f|_{A_i}$ is continuous for each $i$. Thus $X = \bigcup_{i=1}^{n-1} A_i \cup A_n$, where $C = \bigcup_{i=1}^{n-1} A_i$ is closed since it is a finite union of closed sets. If I can show $f|_C$ is continuous, then I simply apply the pasting lemma to finish the problem. Give $C$ the subspace topology, and notice that each $A_i$ is closed in $C$. Thus we have a topological space $C$ that is the union of $n-1$ closed. But do I still have continuity $f$ restricted to $A_i$ since I am no longer considering them subspaces of $X$ but of $C$? When one says $f|_{A_i}$ is continuous, is one saying that $f$ is continuous when restricted to $A_i$ and $A_i$ is given the subspace topology?
EDIT:
I think I figured out the lemma I need. Claim: Let $A$ and $C$ be subspaces of $X$ such that $A \subseteq C$. Topologize $C$ with subspace topology with respect to $X$. Then the subspace topology on $A$ as a subspace of $X$ (call it $\tau_X$) is the same as the subspace topology of $A$ as a subspace of $C$ (call it $\tau_C$).
Proof: Let $U$ be open in $X$. Then $A \cap U$ is some open set in $A$ as a subspace of $X$. Since $C$ is a subspace of $X$, $C \cap U$ is open in $C$ as a subspace of $X$. Then $A \cap (C \cap U) = (A \cap C) \cap U = A \cap U$ is open in $A$ as a subspace of $C$, thereby proving $\tau_X \subseteq \tau_C$. Now let $O$ be some open set in $C$ as a subspace of $X$. Then $A \cap O$ is some open subset of $A$ as a subspace of $C$. But $O = C \cap U$ for some open set in $X$. Therefore $A \cap O = A \cap (C \cap U)= A \cap U$, which is also open in $A$ as a subspace of $X$. Hence $\tau_C \subseteq \tau_X$
Kind of obvious. But it was hard to spot the need for this lemma when I was considering subspace of subspaces and continuous functions, etc. Am I right in thinking this is the lemma I need?
Best Answer
You don't really need such a lemma (which is true and in general is known as the transitive law for initial topologies, in a much more abstract setting, see my long answer here).
What you need is the following observation: for any subset $A$ of $X$ and any function $f: X \to Y$: $$(f|_A)^{-1}[C] = f^{-1}[C] \cap A$$ for all subsets $C$ of $Y$. This holds as $x \in (f|_A)^{-1}[C]$ iff $x \in A$ and $f(x) \in C$.
Now if $f|_{A_i}$ is continuous for each $i$ and $C$ is closed in $Y$, we know that $(f|_{A_i})^{-1}[C] = f^{-1}[C] \cap A_i$ is closed in $A_i$ and hence closed in $X$ as well (closed in a closed subspace is closed in the whole space). And then $$f^{-1}[C] = f^{-1}[C]\cap X = f^{-1}[C] \cap (\bigcup_i A_i) = \bigcup_i (f^{-1}[C] \cap A_i) $$ is a finite union of closed subsets of $X$ hence closed.
So $f$ is continuous, as inverse images of closed sets are closed.