1. Quotient Topology

Let $h$ be a (real-valued) continuous function on some closed interval $[c,d].$ Let, $\lambda =\min_{[c,d]}h$ and $\gamma = \max_{[c,d]}h$ . Also, $h$ can be considered as a surjective map (onto) from $[c,d] \rightarrow [\lambda,\gamma]$.

**Question** How can one show that if $[c,d]$ has the usual topology, then the quotient topology on $[\lambda,\gamma]$ is also the usual topology ?

I am genuinely frustrated because this is an example from the quotient topology chapter, but the solution to this the author gave I cannot understand at all. I would really appreciate some help.

Note the quotient topology I am working with here is given by: a quotient topology on $Y$ is defined to be $T_Y=\{ V\subset Y : f^{-1}(V) \in T_X\}$,where $f:X\rightarrow Y$, and a topological space $X$ with topology $T_X.$

2. Re: Quotient Topology

Under the usual topology, the preimage of open sets are open. To prove two topologies are equal, show that every open set in one is open in both.

For an interval under the usual topology, you can use the set of open intervals as a basis. For the quotient topology, you can use the set of sets whose preimage is an open interval as a basis for the quotient topology. Show that any arbitrary open interval in the Image has a preimage that is open. Then show that any set with a preimage that is an open set is a union of open intervals.

3. Re: Quotient Topology

Thank you for the reply, I may need some help with writing that in terms of clear set notation. Could you show me how to do that, I am sorry my topology is quite bad as I just started to learn topology this month.

4. Re: Quotient Topology

What have you already proven? Do you already have that the continuous image of an interval is an interval? Have you studied compact sets? What sort of tools are you using?

5. Re: Quotient Topology

I think the general idea I have is to use draw the graph h, then try to use the epsilon delta to construct the usual topology sets on [c,d] on some image h. Then see that the only possible sets that are formed are sets on the usual topology. But its easier said than done and have trouble putting it in notations.

6. Re: Quotient Topology

ok so after some playing around with the notations, if we want to find the $T_{[\max_{[c,d]} h ,\min_{[c,d]} h}$, then we just have to find some sets $U \subset [\max_{[c,d]} h ,\min_{[c,d]}h]$ such that $h^{-1}(U)$ is open in $[c,d]$. But after drawing the graph of $h$, it seems to me that any open $(a,b) \subset U$ its inverse is indeed an open set in $[c,d].$ However I have trouble showing this part. Is this because h is given to be continuous, then for any open set in h, its inverse must also be open by the definition of continuity? If so how can we treat the last part where the preimage of the open interval is the union of open intervals?

7. Re: Quotient Topology

By definition of a continuous function, the preimage of any open set is open. So, it is trivial to show that the usual topology is a sub-topology of the quotient topology. That is true by definition of continuous. You do not need any set notation to prove it. It is already proven by definition.

All that you need to show is that for any set $U \subseteq [\lambda, \gamma]$ such that $f^{-1}(U)$ is open in $[c,d]$, you need to show that $U$ is open in the usual topology (to show that the quotient topology is a subtopology of the usual topology).

8. Re: Quotient Topology

To prove the final step consider the complement of the preimage of $U$. It is closed. A closed subset of a compact set is compact. The continuous image of a compact set is compact. The function is a surjection, so this image is the complement of $U$. The complement of a compact set is open. Therefore $U$ is open in the usual topology.

9. Re: Quotient Topology

Originally Posted by gaussrelatz
Let $h$ be a (real-valued) continuous function on some closed interval $[c,d].$ Let, $\lambda =\min_{[c,d]}h$ and $\gamma = \max_{[c,d]}h$ . Also, $h$ can be considered as a surjective map (onto) from $[c,d] \rightarrow [\lambda,\gamma]$.
**Question** How can one show that if $[c,d]$ has the usual topology, then the quotient topology on $[\lambda,\gamma]$ is also the usual topology ?
Croom prove a general case, by noting: $f^{-1}(\emptyset)=\emptyset~\&~f^{-1}(Y)=X$ so both $\emptyset~\&~Y$ are open sets. Then notes that ${f^{ - 1}}\left( {\bigcup\limits_{n = 1}^\infty {{O_n}} } \right) = \bigcup\limits_{n = 1}^\infty {{f^{^{ - 1}}}\left( {{O_n}} \right)}~\&~ {f^{ - 1}}\left( {\bigcap\limits_{n = 1}^J {{O_n}} } \right) = {f^{ - 1}}\left( {\bigcap\limits_{n = 1}^J {{O_n}} } \right)$
This means that the quotient topology is a topology on $Y$.