Glueing together normal topological spaces at a closed subset

  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)
    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. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    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. 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. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    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. aaah thank you!

    Suddenly it all makes sense thanks!
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