Let be defined as above. The following are equivalent:
is the identification topology on .
is open under iff is open in .
() If is open under , then is open in as is continuous under . Now, suppose is not open under and is open in . Let be a subbase of . Define . Then the topology generated by is a strictly finer topology than making continuous, a contradiction.
() Let be the topology defined by 2. Then is continuous. Suppose is another topology on making continuous. Let be -open. Then is open in , which implies is -open. Thus and is finer than . ∎
is a subbasis for , using the subspace topology on of the identification topology on .
More generally, let be a family of topological spaces and be a family of functions from into . The identification topology on with respect to the family is the finest topology on making each a continuous function. In literature, this topology is also called the final topology.
Let be defined as above. Define binary relation on so that iff . Clearly is an equivalence relation. Let be the quotient . Then induces an injective map given by . Let be given the identification topology and the quotient topology (induced by ), then is continuous. Indeed, for if is open, then is open in . But then , which implies is open in . Furthermore, the argument is reversible, so that if is open in , then so is open in . Finally, if is surjective, so is , so that is a homeomorphism.
|Date of creation||2013-03-22 14:41:26|
|Last modified on||2013-03-22 14:41:26|
|Last modified by||rspuzio (6075)|