Push Out Notes

Pushing out a Topology:

Given a topological space $\QTR{Large}{X}$ , a point set $\QTR{Large}{Y}$ , and a function which is onto, we define a topology on $\QTR{Large}{Y}$ , called the quotient topology under $\QTR{Large}{f}$ , to be the collection of sets is open in $\QTR{Large}{X\}.}$ If is the given topology on $\QTR{Large}{X}$ we write for this "pushed out " topology. That it is a topology, again, follows quickly from the Boolean Algebra above. Moreover,

Lemma

Given a topological space $\QTR{Large}{W}$ , and a function (set map) , then $\QTR{Large}{h}$ is continuous with respect to the topology if and only if $\QTR{Large}{hf}$ is continuous.

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Lemma:

then is closed in the subspace topology if and only if $\QTR{Large}{C}$ is closed in $\QTR{Large}{X}$ .

Proof:

Let $\QTR{Large}{C}$ be closed in $\QTR{Large}{X}$ then is closed in the subspace topology of $\QTR{Large}{A}$ . Let be closed in the subspace topology of $\QTR{Large}{A\ }$ then where $\QTR{Large}{K}$ is closed in $\QTR{Large}{X}$ . But, by hypothesis, $\QTR{Large}{A}$ is closed in $\QTR{Large}{X}$ so is closed in $\QTR{Large}{X.}$

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Theorem (Gluing Lemma)

Let be closed subspaces such that and let , be a function from $\QTR{Large}{T}$ into a topological space $\QTR{Large}{Y}$ . is continuous if and only if for each .

Proof:

Consider the diagram:

Where $\QTR{Large}{p}$ is the union of the $\QTR{Large}{n}$ inclusions and where the left-most $\QTR{Large}{T\ }$ has the push out topology and the right-most $\QTR{Large}{T\ }$ has its given topology. It is a simple argument to check that is a homeomorphism. Use the lemma above and the fact that there is only a finite number of closed sets.

To prove the theorem, now consider the diagram

Since we can consider $\QTR{Large}{T\ }$ as having the push out topology, the glueing lemma is just a restatement of the definition of the push out topology.