1. ## Topological Spaces

Let X and Y be a topological spaces.
Suppose X is the union of two closed subsets A and B of X .
Let f: A -> Y and g: B-> Y be continuous.
If f(x)=g(x) for every x in the intersection of A and B, then let h: X -> Y be a function defined by setting h(x)=f(x) for x in A, and h(x)=g(x) for x in B.

Then could we say that h is continuous ? Why?

2. Originally Posted by lttlbbygurl
Let X and Y be a topological spaces.
Suppose X is the union of two closed subsets A and B of X .
Let f: A -> Y and g: B-> Y be continuous.
If f(x)=g(x) for every x in the intersection of A and B, then let h: X -> Y be a function defined by setting h(x)=f(x) for x in A, and h(x)=g(x) for x in B.

Then could we say that h is continuous ? Why?
This is called the gluing lemma. It applies to finitely many closed sets and arbitrarily many open.

So, suppose that $\displaystyle X=A\cup B$ where $\displaystyle A,B$ are closed subspaces of $\displaystyle X$ and $\displaystyle f:A\to Y$ and $\displaystyle g:B\to Y$ are continuous functions such that $\displaystyle f\mid_{A\cap B}=g\mid_{A\cap B}$ then $\displaystyle f\sqcup g:X\to Y:x\mapsto\begin{cases}f(x)\quad\text{if}\quad x\in A\\g(x)\quad\text{if}\quad x\in B\end{cases}$ is well-defined since $\displaystyle f,g$ agree on the intersection of their domains. Also, it's continuous since if $\displaystyle E\subseteq Y$ is closed it is easy to see that $\displaystyle \left(f\sqcup g\right)^{-1}(E)=f^{-1}(E)\cup g^{-1}(E)$. But, since $\displaystyle f,g$ are continuous $\displaystyle f^{-1}(E),g^{-1}(E)$ are closed subspaces of $\displaystyle A,B$ respectively but since they themselves are closed subspaces of $\displaystyle X$ it follows that so are $\displaystyle f^{-1}(E),g^{-1}(E)$. Thus, $\displaystyle \left(f\sqcup g\right)^{-1}(E)$ is the finite union of closed sets and thus closed. The conclusion follows.

This can be generalized, as I said, to a finite partition of closed sets or an arbitary partition of open ones.