# Classnotes for September 15

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Fall
1 Sep 12 About, Tue, Thu, Std2Disc
2 Sep 19 Tue, Thu, HW1, 14 Sets
3 Sep 26 Tue, Thu, Photo
4 Oct 3 Tue, Thu
5 Oct 10 HW2, Tue, Thu
6 Oct 17 Tue, Thu, HW3
7 Oct 24 Mon, Tue, Thu
8 Oct 31 Tue, Thu, HW4
9 Nov 7 TE1, Tue, Thu
10 Nov 14 Tue, Thu, HW5
11 Nov 21 Tue, Thu
12 Nov 28 Tue, Thu, HW6
13 Dec 5 Tue, Thu
E Dec 12 TE2
Spring
14 Jan 9 Tue, IT83, Thu, HW7
15 Jan 16 Tue, Thu
16 Jan 23 Tue, HW8, Thu
17 Jan 30 Tue, Thu
18 Feb 6 TE3, Tue, Thu
19 Feb 13 Tue, Thu
R Feb 20
20 Feb 27 Tue, Thu, HW9
21 Mar 6 Tue, Thu, HW10
22 Mar 13 Tue, Thu
23 Mar 20 Tue, Thu, HW11
24 Mar 27 Tue, Thu
25 Apr 3 Tue, Thu, HW12
26 Apr 10 Tue, Thu
Study Apr 17 Office Hours
Exams Apr 24 Final, PM

Classnotes for September 15

## Topology Riddle

Theorem: Every two-dimensional orientable surface (with minor conditions) is toplogically equivalent to a sphere with g tunnels and n punctures for some g,n.

See left for an example (g = 4,n = 3).

Riddle:

Oh really? How about the manifold pictured left?

What are the values of g and n, and how can it be deformed into an object of the above format?

## Basic Properties of Open Sets

We begin by reviewing the definitions of an open subset of $\mathbb{R}^n$ and continuous functions from $\mathbb{R}^m$ to $\mathbb{R}^n$ with these hopes of generalizing the notions to more general spaces.

Definition: A set $U \subset \mathbb{R}^n$ is said to be open if $\forall x \in U \ \exists \epsilon > 0$ s.t. $B_\epsilon(x) \subset U$.

Theorem: A function $f \colon \mathbb{R}^m \to \mathbb{R}^n$ is continuous iff. $\forall$ open $U \subset \mathbb{R}^n \ f^{-1}(U)$ is also open.

Open sets as defined above possess three basic properties, which are established by the following theorem:

Theorem:

1. $\emptyset$ and $\mathbb{R}^n$ are open.
2. If I is an index set (of arbitrary cardinality), and if $\forall \alpha \in I \ U_\alpha$ is open, then $\bigcup_{\alpha\in I} U_\alpha$ is open.
3. If I is a finite index set, and if $\forall i \in I \ U_i$ is open, then $\bigcap_{i\in I} U_i$ is open.

Proof: The first part of the theorem is vacuously true, and the proof of the second is a triviality, so we shall focus on the third part. Let I and the Ui be as in the statement of this part of the theorem, and let $x \in U = \bigcap_{i\in I} U_i$. By definition, then, $\forall i \in I \ x \in U_i$. By the openness of each Ui, $\forall i \in I \ \exists \epsilon_i > 0$ such that $B_{\epsilon_i}(x) \subset U_i$. Let ε be the minimum of all the εi, which exists since I is finite, and is thus positive since all the εi are positive. It follows that $\forall i \in I \ B_\epsilon(x) \subset B_{\epsilon_i}(x) \subset U_i$. Thus $B_\epsilon(x) \subset \bigcap_{i\in I} U_i$, and so, $\bigcap_{i\in I} U_i$ is open, q.e.d.

## Topological Spaces

We can now define our more general space: one which has subsets we can rightfully call "open" and which we can therefore use to define continuity in the most general setting possible. Let us remind ourselves that if X is a set, then $\mathcal{P}\left (X \right )$ (alternatively 2X) denotes the power set of X, the set of all subsets of X.

Definition: A topological space is a pair $(X, \mathcal{T})$ where X is a set and $\mathcal{T} \subset \mathcal{P}(X)$ such that:

1. $\emptyset , X \in \mathcal{T}$.
2. If I is an index set (of arbitrary cardinality), and if $\forall \alpha \in I \ U_\alpha \in \mathcal{T}$ then $\bigcup_{\alpha\in I} U_\alpha \in \mathcal{T}$.
3. If I is a finite index set, and if $\forall i \in I \ U_i \in \mathcal{T}$, then $\bigcap_{i\in I} U_i \in \mathcal{T}$.

$\mathcal{T}$ is called a topology on X, and the elements of $\mathcal{T}$ are called open sets.

Some examples of topological spaces as defined above are:

• The so-called standard topology on $\mathbb{R}^n$, given by $\mathcal{T}_{std} = \{U \subset \mathbb{R}^n \colon U$ is open according to the old definition from analysis }.
• The trivial topology on any set X, given by $\mathcal{T}_{tr} = \{\emptyset, X\}$.
• The discrete topology on any set X, given by $\mathcal{T}_{dsc} = \mathcal{P}(X)$. By definition, every subset of X is open in this topology.

From now on we follow the general practice (strictly speaking an abuse of notation) of denoting a topological space $(X, \mathcal{T})$ by X, suppressing the topology (cf. the parallel practice of suppressing the multiplication when referring to a group). Should the need arise to refer explicitly to the topology of a topological space denoted by, say, Y, we can denote it by $\mathcal{T}_Y$.

## Continuity

We now generalize the defintion of continuity to work for topological spaces:

Definition: If X and Y are topological spaces and $f \colon X \to Y$, we say that f is continuous if $f^{-1}(U) \in \mathcal{T}_X$ whenever $U \in \mathcal{T}_Y$.

This new definition of continuity still admits many of the same properties that the old definition from analysis, upon which it was modelled, does.

Theorem:

1. If X is a topological space, then the identity function $id \colon X \to X$ is continuous.
2. If X,Y,Z are topological spaces and $f \colon X \to Y$, $g \colon Y \to Z$ are continuous, it follows that $g \circ f \colon X \to Z$ is continuous.

Proof:

1. Since id = id − 1, it follows that id − 1(U) = id(U) = U. Hence, if $U \subset X$ is open, then I − 1(U) is trivially open too.
2. It is easy to show that $(g\circ f)^{-1}(U) = f^{-1}(g^{-1}(U))$. Hence, if $U \subset X$ is open, then g − 1(U) is open by the continuity of g, and thus $(g\circ f)^{-1}(U) = f^{-1}(g^{-1}(U))$ is open by the continuity of f, q.e.d.

Since every set admits at least two different topologies (i.e. the discrete and the trivial topologies), it is natural to ask, for instance, what happens to the continuity of a function from a set to itself when different topologies are assigned to the domain and codomain.

So, let us consider the case of the real line $\mathbb{R}$, together with the standard topology $\mathcal{T}_{std}$, the trivial topology $\mathcal{T}_{triv}$, and the discrete topology $\mathcal{T}_{dsc}$, and in particular, examine the continuity of the identity function $id \colon \mathbb{R} \to \mathbb{R}$ when various topologies are assigned to the domain and the codomain. We summarise our results in the following table:

 Domain $\left ( \mathbb{R},\mathcal{T}_{triv} \right )$ $\left ( \mathbb{R},\mathcal{T}_{std} \right )$ $\left ( \mathbb{R},\mathcal{T}_{dsc} \right )$ $\left ( \mathbb{R},\mathcal{T}_{triv} \right )$ continuous; continuous; continuous; Codomain $\left ( \mathbb{R},\mathcal{T}_{std} \right )$ not continuous; continuous; continuous; $\left ( \mathbb{R},\mathcal{T}_{dsc} \right )$ not continuous; not continuous; continuous;

We can repeat this analysis for functions in general from $\mathbb{R}$ to itself, identifying which functions are continuous in each case:

 Domain $\left ( \mathbb{R},\mathcal{T}_{triv} \right )$ $\left ( \mathbb{R},\mathcal{T}_{std} \right )$ $\left ( \mathbb{R},\mathcal{T}_{dsc} \right )$ $\left ( \mathbb{R},\mathcal{T}_{triv} \right )$ every function; every function; every function; Codomain $\left ( \mathbb{R},\mathcal{T}_{std} \right )$ constant functions only; continuous (old defintion) functions; every function; $\left ( \mathbb{R},\mathcal{T}_{dsc} \right )$ constant functions only; constant functions only; every function;

These results can be checked with the tools that we already have except for the result that a function $f \colon \left ( \mathbb{R},\mathcal{T}_{std} \right ) \to \left ( \mathbb{R},\mathcal{T}_{dsc} \right )$ is continuous iff. it is a constant function. This particular result will be proved later on.

Some of the results from the above tables can be generalised quite readily:

Proposition: Let X be a set, let $\mathcal{T}_{triv}$ and $\mathcal{T}_{dsc}$ be the trivial and discrete topologies on X, respectively, and let $\mathcal{T}$ be any topology on X. Then:

1. Every function $f \colon \left ( X, \mathcal{T} \right ) \to \left ( X, \mathcal{T}_{triv} \right )$ is continuous.
2. Every function $f \colon \left ( X, \mathcal{T}_{dsc} \right ) \to \left ( X, \mathcal{T} \right )$ is continuous.

Proof:

1. Continuity follows from the fact that $f^{-1} \left ( \empty \right ) = \empty$ and $f^{-1} \left ( X \right ) = X$.
2. Continuity follows from the fact that every subset of X is open in the discrete topology, q.e.d.

Thus, for any given set, the discrete topology can be said to be the "strongest" topology, being stronger than all other possible topologies, and the trivial topology can be said to be the "weakest" topology, being weaker than all other possible topologies.

We end this lesson with some terminology:

Definition: Let X be a set, and let $\mathcal{T}_1$ and $\mathcal{T}_2$ be topologies on X such that $\mathcal{T}_2 \subset \mathcal{T}_1$. Then:

• $\mathcal{T}_1$ is said to be stronger, bigger, or finer than $\mathcal{T}_2$.
• $\mathcal{T}_2$ is said to be weaker, smaller, or coarser, respectively, than $\mathcal{T}_1$.

Using this language we can state a generalisation of the proposition we just proved:

Proposition: Let X be a set, and let $\mathcal{T}_1$ and $\mathcal{T}_2$ be topologies on X, with $\mathcal{T}_1$ stronger than $\mathcal{T}_2$. Then the identity function $id \colon \left ( X, \mathcal{T}_1 \right ) \to \left ( X, \mathcal{T}_2 \right )$ is continuous.

That this last proposition holds follows immediately from the definitions.

## Scanned Notes

### JPG, by Annat Koren

Handwritten notes by Annat Koren
Handwritten notes by Annat Koren
Handwritten notes by Annat Koren
Handwritten notes by Annat Koren

### PDF, by Gabriel Lee

Available here (http://katlas.math.toronto.edu/0506-Topology/index.php?title=Image:1-2-09-15.pdf).

## One draw-based solution for the Topology Riddle.

The Torus with one bite, or the orientable surface with g = 1, n = 1 is homeomorfic to the Seifret Surface of the Trefoil knot.

What does this means? Well, the last picture in the second page.

I hope this is clear for everybody,

Mario 15:09, 23 Sep 2005 (EDT) Mario mario@math.utoronto.ca