glueing together normal topological spaces at a closed subset


by conquest
Tags: glueing, normal, spaces, subset, topological
conquest
conquest is offline
#1
Mar8-12, 01:03 PM
P: 117
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?
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morphism
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#2
Mar8-12, 10:16 PM
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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.
conquest
conquest is offline
#3
Mar9-12, 11:27 AM
P: 117
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?

morphism
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#4
Mar9-12, 05:07 PM
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glueing together normal topological spaces at a closed subset


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.
conquest
conquest is offline
#5
Mar11-12, 07:28 AM
P: 117
aaah thank you!

Suddenly it all makes sense thanks!


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