Is the Identity Function between Topologies Continuous?

[ak416]

Let X and X' denote a single set in the two topologies T and T', respectively. Let i: X' -> X be the identity function.
a) Show that i is continuous <=> T' is finer than T.

Ok I am able to show that for any set in T|X this set is in T'. This is done as follows: Assume i is continuous. For any open set X^U (in X), i^-1(X^U) = X^U is open in X'. Since X' is open in the space with T', X^U is open in that space (i.e. X^U is an element of T').
However, i don't see how this necessarily applies to every open set defined by T. Maybe i interpreted the question wrong but i don't think so, so please help if you can..

 

[matt grime]

You're probably just missing some quantifiers in your statements. Please try to write everyting clearly and in full, grammatically correct English. There are several issues with what you wrote. How can i be the identity map unless X and X' are the same set, for instance?

It is trivial that if T and T' are two topologies on a space X that the identity from (X,T') to (X,T) is continuous if and only if T' is finer (has more open sets). But this is not what you wrote (though it couldbe what you meant to write).

 

[ak416]

Thats exactly how it is written in Munkres-Topology 2nd edition p.111 and you i am a little curious as to how that could be an identity function.