$f$ is continuous if and only $f|x_1$ and $f|x_2 $ are continuous

Question

Exercise:

Let X and Y be topological spaces and $f: X → Y$ be a map.

Assume that $X =X_1\bigcup X_2$, where $X_1$ and $X_2$ are two subsets of X.

Assume that $X_1$ and $X_2$ are closed/open subsets.

Show that $f$ is continuous $ \iff f|x_1$ and $f|x_2 $ are continuous

Definition: A map $f: X → Y$ between two topological spaces is continuous if $f^{-1}$ is open/closed in X for every open/closed set in Y.

Solution:

$ \Rightarrow$:

Assume that $f$ is continuous. Then from the definition we have that for every open V in Y is $f^{-1}(V)$ open i X.

But $f^{-1}(V) =(f^{-1}(V) \bigcap X_1) \bigcup(f^{-1}(V) \bigcap X_2) $, so $f|x_1$ and $f|x_2 $ are continuous because $f$ is continuous?

$\Leftarrow$:

Want to show that for every V open in Y then $f^{-1}(V)$ is open in X, which is equivalent to the continuity of $ f $.

Assume that $f|x_1$ and $f|x_2 $ are continuous. Let $V \subset Y$ be open.

Then $f^{-1}(V) =(f^{-1}(V) \bigcap X_1) \bigcup(f^{-1}(V) \bigcap X_2) $ is open, since $f|x_1$ and $f|x_2 $ are continuous. Hence $f$ is continuous.

I tried to prove the statement when $X_1$ and $X_2$ are both open, since a similar argument applies when X and Y are both closed. But I'm not completely sure that my method is right. Any help would be greatly appreciated.

Answer 1

Your proof is flawless.

You are using the fact that $$f^{-1}(V) =(f^{-1}(V) \bigcap X_1) \bigcup(f^{-1}(U) \bigcap X_2)$$

This is OK if you have already proved it, which I assume you did.

Also, in your last statement, you need to change $X$, $Y$ to $X_1 $, $X_2$ are both closed.

Answer 2

Rephrasing:

Assume: $X_1,X_2$ open, and $X_1 \cup X_2=X.$

$V \subset Y$, $V$ open .

$\Rightarrow:$

$f^{-1}(V)$ is open since f continuous.

Hence :

$f^{-1}(V) \cap X_i$ is open in $X_i,$ $i=1,2$,

i.e. $f_{x_i},$ $i=1,2$ is open.

$\Leftarrow:$

$f_{x_i},$ $i=1,2$ is continuous.

Then $f_{x_i}^{-1}(V) = f^{-1}(V) \cap X_i$ is open in $X_i$, $i=1,2$.

$\rightarrow:$

$f_{x_i}^{-1}(V) = X_i \cap O_i$, $i=1,2$, for some open subset $O_i$ of $X$.

Since $X_i$, $i=1,2$ is open in $X$ :

$f^{-1}(V) \cap X_i$ is open in $X$, and

finally:

$f^{-1}(V)=$

$ (f^{-1}(V) \cap X_1) \cup (f^{-1}(V) \cap X_2)$ is open in $X$.