Results 1 to 9 of 9
Like Tree4Thanks
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal
  • 1 Post By SlipEternal
  • 1 Post By Plato

Thread: Quotient Topology

  1. #1
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    45

    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 $T_X.$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,562
    Thanks
    1436

    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.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    45

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,562
    Thanks
    1436

    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?
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    45

    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    45

    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?
    Last edited by gaussrelatz; Mar 21st 2018 at 06:33 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,562
    Thanks
    1436

    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).
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,562
    Thanks
    1436

    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,880
    Thanks
    2890
    Awards
    1

    Re: Quotient Topology

    Quote Originally Posted by gaussrelatz View Post
    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 made no comment on this thread unit now. I just did not follow the notation.
    But text PRINCIPLES OF TOPOLOGY, by Fred Croom has a good treatment. In my copy begins on page 217.

    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 $Y$.

    Please ignore this if it is only confusing you.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mobius Band as a Quotient Topology
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Mar 29th 2014, 07:45 PM
  2. Algebraic Topology - Quotient Spaces
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Jun 8th 2012, 06:58 PM
  3. Topology homeomorphisms and quotient map question
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Apr 22nd 2011, 11:24 AM
  4. Show quotient topology on [0,1] = usual topology on [0,1]
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Nov 5th 2010, 04:44 PM
  5. Topology Quotient Space
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: May 17th 2008, 01:42 PM

/mathhelpforum @mathhelpforum