# [SOLVED]Quotient map question

#### topsquark

##### Well-known member
MHB Math Helper
Here's the deal:

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

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

Where am I going wrong with this?

Thanks!

-Dan

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey Dan,

The function $f$ is indeed not necessarily an invertible function.
We might for instance have $X=\{u,v\}$ and $f(u)=f(v)=y$.
Now $f^{-1}(y)$ is undefined, since it is not a single element in $X$.

However, $f^{-1}(\{y\})$ is defined. It is the subset of $X$ that maps to $\{y\}$.
It is called the preimage of $U$ with respect to $f$.
So $f^{-1}(\{y\})=\{u,v\}$.

And since $f$ is 'onto', every element in $U$ must have at least 1 element in $X$ that maps to it.
That element is in the preimage of $U$.
If we then map it again with $f$, we get the original element in $U$.
So $f(f^{-1}(U))=U$.

#### topsquark

##### Well-known member
MHB Math Helper
That makes all sorts of sense. I was stuck on thinking of $$\displaystyle f^{-1}(y)$$ in my "example" concept as opposed to $$\displaystyle f^{-1}( \{ y \} )$$. (I was thinking of , as a model, $$\displaystyle f^{-1}$$ as being a projection map using individual points.)

Thanks again!

-Dan