# Topological Spaces

#### 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?

#### Drexel28

MHF Hall of Honor
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.