$f|_{A_i} $is continuous, $A_i$ closed $\implies$ $f:X\to Y$ is continous

Question

Take topological spaces $X$ and $Y$, a function $f:X \to Y$ and a family $(A_i)_{i\in I}$ os subsets of $X$ and $X= \cup A_i$. Assume that $f|_{A_i} $is continuous for each $i \in I$. Show that if I is finite and each $A_i$ is closed in $X$, then $f$ is continous.

$f$ is continous if preimage of every closed set in $Y$ is closed in $X$. I know that a union of finitely many closed sets is closed but I am not sure how to mix it.

For each $i \in I$ $f_i : A_i \to Y$ is continous, that means for each closed $C \subset Y$, $f^{-1}(C)$ is closed in $A_i$. Can I write something like this $f^{-1}(C_i)=A_i $?

Answer 1

You ought to be precice. First off, you need to say explicitly that $f_i$ means $f|_{A_i}$. Also, what $f_i$ being continuous really means is that $f_i^{-1}(C)$ is closed in $A_i$, not that $f^{-1}(C)$ is closed in $A_i$. What would that even mean if $f^{-1}(C)\not\subseteq A_i$?

From here, note that we have $f_i^{-1}(C) = A_i\cap f^{-1}(C)$. Can you express $F^{-1}(C)$ in terms of $A_i\cap f^{-1}(C)$? Finally, can you use this to show that $f^{-1}(C)$ is closed in $X$?

Answer 2

In general, $f^{-1}(C)$ will not be a subset of $A_i$. Therefore, in makes no sense to assert that it is a closed subset of $A_i$.

However, ${f|_{A_i}}^{-1}(C)=f^{-1}(C)\cap A_i$, wich is a closed subset of $A_i$ and therefore a closed set. Since $f^{-1}(C)=\bigcup_if^{-1}(C)\cap A_i$, it follows that $C$ is closed.