Let and be topological spaces and a function. The graph of is the set .
Prove that is continuous if and only if the function given by is a homeomorphism where has the topology relative to the product topology .
If both the function h and the inverse function are continuous, then h is called a homeomorphism.
Lemma 1. Let be given by the equation
Then g is continuous if and only if the functions
Suppose f is a continuous function.
Since f is a well-defined function, it is clear that defined by is a bijection. Let be an open set of relative to the product topology containing an arbitrary point . It follows that
Since both an identity map and f are continuous functions, and are open, so is their intersection. It follows that is open. Thus, is continuous.
Now, we need to show that defined by is a continuous function. Let U be an open set containing an arbitrary point of . Then, , where is an intersection of a neighborhood of with . Specifically,
where M is a set and V is an open set in Y containing M.
Since has the topology relative to the product topology , is open in . Thus, is continuous.
By the defintion given the above, is a homeomorphism.
Suppose is a homeomorphism.
Since an identity map is continuous and given by is continuous by hypothesis, is continuous by lemma 1.
Note: This is my attemp to this problem. Any feedback or correction is welcomed.