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[SOLVED] Quotient map question


Well-known member
MHB Math Helper
Aug 30, 2012
The Astral plane
Here's the deal:


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.

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 \(\displaystyle \tau _f\).


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?



Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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 $f(f^{-1}(U))=U$.


Well-known member
MHB Math Helper
Aug 30, 2012
The Astral plane
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!