Proof verification: Restriction of f to two closed subsets of X covering X continuous implies f on X is continuous

Question

Let $A,B \subseteq X$ be closed in $X$ with $A \cup B = X$.

Suppose there exists a function, $f: X \rightarrow Y$, with both $f {\restriction_{A}}: A \rightarrow Y$ and $ f {\restriction_{B}}: B \rightarrow Y$ continuous.

Prove that $f$ is continuous.

Proposed proof:

It is sufficient to show that if $F \subseteq Y$ is closed in $Y$, then $f^{-1}(F)$ is closed in $X$.

Let $F \subseteq Y$ be closed in $Y$. By continuity of $f{\restriction_{A}}$ and $f{\restriction_{B}}$, one has it that $f^{-1}{\restriction_{A}}(F)$ and $f^{-1}{\restriction_{B}}(F)$ are closed in their subspace topologies.

I.e. $f^{-1}{\restriction_{A}}(F) = A \backslash (\bigcup_\limits{{i \in I}}U_i \cap A) $ and similarly, $f^{-1}{\restriction_{B}}(F) = B \backslash (\bigcup_\limits{{j \in J}}U_j \cap B) $

Also being restrictions of $f$ we have that: $$f^{-1}{\restriction_{A}}(F) = f^{-1}(F) \cap A$$ $$f^{-1}{\restriction_{B}}(F) = f^{-1}(F) \cap B$$ $\Rightarrow f^{-1}{\restriction_{A}}(F) \cup f^{-1}{\restriction_{B}}(F) = (f^{-1}(F) \cap B) \cup (f^{-1}(F) \cap A) = f^{-1}(F) \cap (A \cup B) = f^{-1}(F)$

Since $A \cup B = X$.

It now suffices to show that $f^{-1}{\restriction_{A}}(F) = A \backslash (\bigcup_\limits{{i \in I}}U_i \cap A) $ is closed in $X$ and similarly for $f^{-1}{\restriction_{B}}(F)$, as the union of two closed sets is closed.

$$X \backslash f^{-1}{\restriction_{A}}(F) = X \backslash (A \backslash (\bigcup_\limits{{i \in I}}U_i \cap A)) $$

$$= X \backslash ((A \backslash \bigcup_\limits{{i \in I}}U_i) \cup (A \backslash A) $$ $$= (X \backslash A) \cup \bigcup_\limits{{i \in I}}U_i $$

Which is just the union of open sets in X (since A is closed by assumption).

Hence $X \backslash f^{-1}{\restriction_{A}}(F)$ is open $\Rightarrow f^{-1}{\restriction_{A}}(F)$ is closed.

One can show $f^{-1}{\restriction_{B}}(F)$ is closed similarly, thereby completing the proof.

Answer 1

What is the meaning of the family $(U_i)_{i \in J}$ ???

And why so complicated ??

If $f^{-1}{\restriction_{A}}(F)$ is closed in the subspace topology of $A$, then there is a closed subset $Z$ of $X$ such that $f^{-1}{\restriction_{A}}(F)=A \cap Z$ .

Since $A$ is closed (in $X$), we have that $f^{-1}{\restriction_{A}}(F)=A \cap Z$ is closed in $X$.

Answer 2

If $R$ is a subset of (the closed set) $A$ that is closed in $A$ then also it will be closed in $X$.

This because $R$ is closed in $A$ iff it can be written as $A\cap S$ where $S$ is a closed subset of $X$.

So $R$ appears to be the intersection of two sets that are both closed in $X$, hence $R$ is closed in $X$.

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This can be applied here.

If $F$ is a closed subset of $Y$ then: $$f^{-1}(F)=(f{\restriction_{A}})^{-1}(F)\cup(f{\restriction_{B}})^{-1}(F)$$where $(f{\restriction_{A}})^{-1}(F)$ is closed in $A$ hence closed in $X$ and $(f{\restriction_{B}})^{-1}(F)$ is closed in $B$ hence closed in $X$.

Then as a union of two sets both closed in $X$ the preimage $f^{-1}(F)$ is closed in $X$.