Pull-backs and Push-outs

Review of Boolean Algebra and the Definition of a Topology

If $\QTR{Large}{A}$ and $\QTR{Large}{B}$ are sets and $\QTR{Large}{.}$ is a function:

Let $\QTR{Large}{B\ }$ be two subsets of $\QTR{Large}{B}$ then

$\qquad$ 1.

$\qquad$ 2.
More generally let be a set of subsets of $\QTR{Large}{B}$ then

$\qquad$ 1.

$\qquad$ 2.

Given a point set $\QTR{Large}{X}$ , a topology on is a subset of $\QTR{Large}{P(X)}$ , the set of subsets of $\QTR{Large}{X\ }$ that satisfy the 4 topology axioms.

We will sometimes use the notation $\QTR{Large}{\Im }$ for a topology

There are two extreme topologies on any set $\QTR{Large}{X}$ . the indiscrete topology on $\QTR{Large}{X}$ , and the discrete topology on $\QTR{Large}{X}$ .
Given two topologies and , on $\QTR{Large}{X}$ , if then we say is finer than . and is coarser than .
Given two topologies and , (as sets) is also a topology. Note that may not be a topology.

Lemma 1:

Given two topologies and , on $\QTR{Large}{X}$ and ,the identity function, suppose we consider the left-most " $\QTR{Large}{X}$ " with the topology and the right-most with the topology then $\QTR{Large}{1}_{X}$ is continuous if and only if

Proof:

being continuous amounts to saying that every open set is also open

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Pulling Back a Topology

Given a topological space $\QTR{Large}{Y}$ , a point set $\QTR{Large}{X}$ , and a function , the collection of sets , all open $\QTR{Large}{V\}}$ forms a topology on $\QTR{Large}{X}$ , and such that $\QTR{Large}{f}$ is continuous with respect to this topology. If is the given topology on $\QTR{Large}{Y}$ we write for this "pulled back" topology. That it is a topology follows quickly from the Boolean Algebra above. Moreover,

Lemma

Given a topological space $\QTR{Large}{Z}$ , and a function (set map) , then $\QTR{Large}{g}$ is continuous with respect to the topology if and only if $\QTR{Large}{fg}$ is continuous.
If is a second topology on such that $\QTR{Large}{f}$ is continuous then

Proof:

1.Since the composition of continuous functions is continuous, what is left to show is that $\ \QTR{Large}{fg}$ is continuous implies that $\QTR{Large}{g}$ is continuous. In particular, we need to show that

is open for any open set . By definition , for some open

But then an open set.

2. Apply 1. to the case and apply Lemma 1. above.

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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.
If is a second topology on $\QTR{Large}{Y}$ such that $\QTR{Large}{f}$ is continuous then

Proof:

1.Since the composition of continuous functions is continuous, what is left to show is that $\ \QTR{Large}{hf}$ is continuous implies that $\QTR{Large}{h}$ is continuous. In particular, we need to show that

is open for any open set . But then an open set so , by definition is open $\QTR{Large}{.}$

2. Again, apply Lemma 1. to the case