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(It goes on to define the term "quotient map.")Definition:

If X is a toplogical space, Y is a set, and \(\displaystyle f: ~ X \rightarrow Y\) is an onto mapping, then the collection \(\displaystyle \tau _f\) of subsets of Y defined by \(\displaystyle \tau _f = \{ F \subset Y | f^{-1}(F) \text{ is open in } X \}\) is a topology on Y, called the quotient topology on Y by f.

I have a difficulty with the use of \(\displaystyle f^{-1}()\) which I think is best demonstrated by the following theorem.

My difficulty is probably simple. The definition talks of an onto map, f, which means that \(\displaystyle f^{-1}(U)\) may not be well defined as an inverse function. In fact (if I'm doing this right) for an onto map f we have that \(\displaystyle f ( f^{-1}(U) ) \subseteq U\) rather than strict equality.Theorem:

If X and Y are topological spaces and \(\displaystyle f: X \rightarrow Y\) is continuous and open, then the topology \(\displaystyle \tau\) on Y is the quotient topology \(\displaystyle \tau _f\).

Proof:

Suppose f is continuous and open. Since \(\displaystyle \tau _f\) is the largest topology making f continuous, \(\displaystyle \tau \subset \tau _f\). But if \(\displaystyle U \in \tau _f\), then by definition of \(\displaystyle \tau _f\), \(\displaystyle f^{-1}(U)\) is open in X. Now f is open as a map to \(\displaystyle \{ Y, \tau \}\) so \(\displaystyle f ( f^{-1} (U) ) = U\) belongs to \(\displaystyle \tau\). Thus \(\displaystyle \tau _f \subset \tau\) and this establishes equality.

Where am I going wrong with this?

Thanks!

-Dan