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Glueing together normal topological spaces at a closed subset

  1. Mar 8, 2012 #1
    Hi all!

    My question is the following. Suppose we have two normal topological spaces X and Y and we have a continuous map from a closed subset A of X to Y. Then we can construct another topological space by "glueing together" X and Y at A and f(A). By taking the quotient space of the disjoint union of X and Y by the equivalence relation that

    x is equivalent to y if:

    1) x=y
    2) x,y are elements of A and f(x)=f(y)
    or
    3) x is an element of A and y is an element of Y and f(x)=y
    or x is an element of Y and y is an element of A and x=f(y).

    My question is how can you prove that this constructed space is again normal?
     
  2. jcsd
  3. Mar 8, 2012 #2

    morphism

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    The space obtained from this construction is called an adjunction space and is typically denoted by ##X \cup_f Y##. It forms a pushout in the category of topological spaces. With this and the normality of X and Y in mind, it's not too difficult to show that you can separate closed sets in ##X \cup_f Y## by a continuous function.
     
  4. Mar 9, 2012 #3
    Okay I believe this, but I would rather find a way to prove it without using any abstract non-sense so that I have an idea of where it is coming from.

    So basically the question is if I have non-intersecting closed sets in this adjunction I look at the pre-image in the disjoint union of X and Y (where they are again closed and non-intersecting). Since I know X and Y are normal I can construct open sets now in both X and Y that don't intersect and contain the earlier closed sets (i.e. definition of normal).
    but how then do I make sure that when I project them onto the adjunction they are again open and non-intersecting?
     
  5. Mar 9, 2012 #4

    morphism

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    The abstract nonsense here is just a reformulation of what it means to give a space the quotient topology, so it's really not so abstract.

    It's probably best not to try to separate by open sets, but to separate with continuous functions, like I mentioned in my post. This is where the abstract nonsense (which again is not so abstract) will be helpful.
     
  6. Mar 11, 2012 #5
    aaah thank you!

    Suddenly it all makes sense thanks!
     
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