glueing together normal topological spaces at a closed subsetby conquest Tags: glueing, normal, spaces, subset, topological 

#1
Mar812, 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? 



#2
Mar812, 10:16 PM

Sci Advisor
HW Helper
P: 2,020

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.




#3
Mar912, 11:27 AM

P: 117

Okay I believe this, but I would rather find a way to prove it without using any abstract nonsense so that I have an idea of where it is coming from.
So basically the question is if I have nonintersecting closed sets in this adjunction I look at the preimage in the disjoint union of X and Y (where they are again closed and nonintersecting). 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 nonintersecting? 



#4
Mar912, 05:07 PM

Sci Advisor
HW Helper
P: 2,020

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. 



#5
Mar1112, 07:28 AM

P: 117

aaah thank you!
Suddenly it all makes sense thanks! 


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