Let be a function from a topological space to a set . The identification topology on with respect to is defined to be the finest topology on such that the function is continuous.

Theorem 1   Let be defined as above. The following are equivalent:

1. is the identification topology on .
2. is open under iff is open in .

Proof. ( ) 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 .

Remarks.

• is open in 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.
• The dual concept of this is the initial 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.