1. Continuous and connectivity

Suppose that $\{A,B\}$ is a partition of a topological space $X$ and that $f: X \rightarrow Y$ is a map to another space $Y$. Prove that if the restrictions $f|A$ and $f|B$ are both continuous then $f$ is continuous.
Let $E \subset X$ s.t $E \cap B \neq \phi$ and $E \cap A \neq \phi$.

I need to show that $A$ and $B$ are connected (since if $f$ is continuous on $A$ and $B$, then the only place it might be discontinuous is at the boundary, where $A$ meets $B$).

$A$ is connected (as is $B$) since $f$ is a continuous map between any two points.

This is what I know, but I can't see what I can do from here! Can someone help?

2. What's the definition of a partition? I'm going to assume it's not just "sets whose union is the space", else a counterexample is $X=[0,2],\:A=[0,1),\:B=[1,2]$ and

$f(x)=\begin{cases}0&x\in A\\1&x\in B.\end{cases}$

3. If $\{A,B\}$ is a partition of a set $X$, then $A \cup B=X$ and $A \cap B= \phi$.

I didn't think the second condition helped, so I left it out.

4. Are you sure there's not another condition, such as A and B are both open or both closed? Else it can fail, as I demonstrated above.

5. Ah yes! Both $A,B$ have to be open.

Sorry! I forgot the most important condition.

6. Originally Posted by Showcase_22
Ah yes! Both $A,B$ have to be open.

Sorry! I forgot the most important condition.
Well, there's a "lemma" that talks about "pasting" continuous functions together. ;-) I don't know if it would be kosher just to cite it, but the proof isn't hard.

7. Okay.

I know that A and B are open. Since f is continuous then $f(A), \ f(B)$ are also open.

$Im(f)=f(A) \cup f(B)$. This is the union of two open sets so $Im(f)$ is open.

Therefore for any $E$ as defined before, $f(E)$ is also going to be open so $f$ maps an open set to an open set. Therefore f is continuous.

I think that's about right. The odd thing is that this is in the connectedness chapter, that's why I was trying to approach it from a connectedness angle.

8. Originally Posted by Showcase_22
I know that A and B are open. Since f is continuous then $f(A), \ f(B)$ are also open.
This is false--it only works in the inverse direction. For example, constant functions are continuous, but their images are single points. And the image of $f(x)=x^2$ is $[0,\infty)$, also not open.

A function with the property you gave is called an "open map".

9. Sorry I keep getting this wrong! I think i'm getting either flustered or tired.

let $V \in Y$ be open.

Then $f^{-1}(V) \subset A \cup B$.

If $f^{-1}(V) \subset A$ or $f^{-1}(V) \subset B$ then $f^{-1}(V)$ is open since we know that $f$ is continuous on these intervals.

If $f^{-1}(V)$ has points in both $A$ and $B$ then partition $V$ into two sets $I,J$ where $f^{-1}(I) \subset A$ and $f^{-1}(J) \subset B$.

$f^{-1}(I),\ f^{-1}(J)$ are both open since $f$ is continuous on both $A$ and $B$.

Hence $f^{-1}(V)$ is always open. The preimage of an open set is also an open set so $f$ is continuous.

10. Originally Posted by Showcase_22
Sorry I keep getting this wrong! I think i'm getting either flustered or tired.

let $V \in Y$ be open.

Then $f^{-1}(V) \subset A \cup B$.

If $f^{-1}(V) \subset A$ or $f^{-1}(V) \subset B$ then $f^{-1}(V)$ is open since we know that $f$ is continuous on these intervals.

If $f^{-1}(V)$ has points in both $A$ and $B$ then partition $V$ into two sets $I,J$ where $f^{-1}(I) \subset A$ and $f^{-1}(J) \subset B$.

$f^{-1}(I),\ f^{-1}(J)$ are both open since $f$ is continuous on both $A$ and $B$.

Hence $f^{-1}(V)$ is always open. The preimage of an open set is also an open set so $f$ is continuous.
How do you know that both I and J are open?

Here is a simple way to do this, take U open in Y, then $f^{-1}|_A(U)=f^{-1}(U) \cap A,f^{-1}|_B(U)=f^{-1}(U) \cap B$ are both open in X. Then $f^{-1}(U)=f^{-1}(U) \cap (A\cup B)=(f^{-1}(U) \cap A)\cup(f^{-1}(U) \cap B)$ a union of two open sets, which is open.