Topology

Notation and Conventions

• A topological space is a pair $(X, \mathscr T)$, but henceforth we may omit mentioning $\mathscr T$ and simply say $X$ is a topological space. Of course, $\mathscr T$, a topology on $X$, is still there.
• For a point $p \in X$ we say that $U$ is an open set around $p$ if $p \in U$. This is just to help write more intuitive sentences.
• Saying $A$ is open is equivalent to saying $A \in \mathscr T$. So we might simply use open sets to refer to the (sub)sets in the topology $\mathscr T$. It is important to note that the concept of “open sets” is therefore dependent on the choice of topology $\mathscr T$ on $X$.

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1 - Topological Spaces

Definition (Topological Space): A set $X$ and a collection of subsets of $X$, (called open sets) $\mathscr T$ is a topological space if:

• $\emptyset, X \in \mathscr T$
• if $A_\alpha \in \mathscr T$ is an arbitrary collection of open sets, for each $\alpha \in I$ (where $I$ is any index set), $\bigcup_{\alpha \in I} A_{\alpha} \in \mathscr T$. That is, arbitrary union of open sets, is open.
• Similarly, for a finite collection of open sets $A_i \in \mathscr T$, the intersection $\bigcap_{i=1}^n A_i \in \mathscr T$. That is, finite intersection of open sets are open.

Definition (Neighborhood): Given a topological space $(X,\mathscr T)$, and a point $p \in X$, then $V$ is a neighborhood around $p$ if there is an open set $U$ around $p$ that is contained in $V$. More formally, $V$ is a neighborhood if $p \in U \subset V \subset X$, where $U \in \mathscr T$. The set of all neighborhoods of $p$ is denoted as $N(p)$.

Definition (Metric Space): A set $X$ with a distance function $d: X \times X \rightarrow \mathbb R^+ \cup \{0\}$ is a metric space if:

• $d(x,y) = 0 \iff x = y$ (only identical points have zero distance).
• $d(x,y) = d(y,x)$ (symmetry of distance).
• $d(x,z) + d(z,y) \ge d(x,y)$ (triangle inequality).

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2 - Limit Points, Closed Sets, and Closure

Definition (Limit Point): Let $(X, \mathscr T)$ be a topological space and $S \subset X$. Then a point $p \in X$ is a limit point of $S$ if for all neighborhoods $V$ around $p$, $V$ has a non empty intersection with $S$, and contains points other than $p$. That is, for all $V \in N(p)$, $(V - \{p\}) \cap S \neq \emptyset$. The set of all limit points of $S$ is denoted as $LP(S)$.

Definition (Closed Set): Let $(X, \mathscr T)$ be a topological space and $S \subset X$. Then $S$ is closed if it contains all its limit points. That is, $LP(S) \subset S$.

Closed and Open are NOT Opposites

This is something important to keep in mind. If a set is not closed, that doesn’t mean its open. And if a set is not open, it doesn’t mean that it is closed. It’s a weird case of English names not playing well with their corresponding math definitions.

For example, if $(X, \mathscr T)$ is a topological space, then $X$ is open by definition ($X \in \mathscr T$), However, $X$ also contains all its limit points ($X$ contains all points!) Hence $X$ is also closed. It is both closed and open, or “clopen” (this is an actual term).

Define the standard topology on $\mathbb R$ where the open sets are unions of open intervals $(a,b)$. Consider the set $S =(0,1]$. Clearly this set is not open, as no union of open intervals or finite intersection of open intervals can result in $S$. However, it is also not closed, as $0$ is a limit point of $S$, but $0 \not \in S$. So this is an example of a set that is neither closed nor open in the standard topology on $\mathbb R$.

Definition (Closure of a Set): Let $X$ be a topological space and $S \subset X$. Then the closure of $S$ written $\bar S$ is the union of $S$ with all the limit points of $S$. That is, $\bar S = S \cup LP(S)$.

Theorem: The closure of a set is closed. Let $X$ be a topological space, and $S \subset X$. Then $\bar S = S \cup LP(S)$ is closed.

Proof: Suppose $\overline{S}$ is not closed. Then there exists a point $q \in LP(\overline S)$ with $q \notin \overline S$. In particular, $q \notin LP(S)$, so there is an open set $U$ around $q$ such that $(U-\{q\})\cap S=\emptyset$. Since $q \notin S$, this says $U\cap S=\emptyset$.

Because $q$ is a limit point of $\overline S$, there is some $p\in (U-\{q\})\cap\overline S$. As $U\cap S=\emptyset$, we must have $p\in LP(S)$. But $U$ is also an open set around $p$, so it contains a point of $S$ other than $p$, contradicting $U\cap S=\emptyset$. Thus $\overline S$ is closed. $\square$

Theorem: The closure is the smallest containing closed set. Let $X$ be a topological space and $S \subset X$. Then if $E$ is a closed set such that $S \subset E$, then $\bar S \subset E$. That is, any closed set that contains $S$ must also contain its closure.

Proof: Since $E$ is closed, $E$ contains all its limit points. In particular, $E$ contains all the limit points of $S$ because $E$ contains $S$. Therefore $E$ contains $\bar S$. $\square$

Alternate Definition for Closed Sets

Let $X$ be a topological space. Then $E \subset X$ is closed if and only if $X-E$ is open.

Proof: Well suppose $E$ is closed, then any limit point of $E$ is in $E$. Therefore, no point $p \in X-E$ is a limit point of $E$. Hence, for each $p \in X-E$ there exists a neighborhood $V_p$ around $p$ such that $V_p \cap E = \emptyset$. Equivalently, there exists an open set $U_{p}$ around each $p$ such that $U_p \cap E = \emptyset$. Hence $\bigcup_{p \in {X-E}}U_{p} = X-E$. Hence $X-E$ is open (it is a union of open sets).

Now suppose $X-E$ is open, and (for the sake of contradiction) suppose $q^* \in X-E$ is a limit point of $E$. Since $X-E$ itself is an open set around $q^*$, it must contain a point $e\in E$ other than $q^*$. But then $e\in E\cap(X-E)$, which is a contradiction. Hence no point outside $E$ is a limit point of $E$, so $E$ contains all its limit points and is closed. $\square$

Note: whenever we have a definition of something, A := B iff (“def 1”), and then we also have A := B iff (“some condition”) we can alternatively take that condition itself as a definition, and recover the older definition as a theorem instead. That is to say, if we define closed sets as ones that contain all their limit points, we can show the theorem: E is closed iff its complement in X is open. Dually, we can define E is closed iff its complement in X is open, and recover E is closed iff it contains all its limit points as a theorem.

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3 - Denseness, Subspace Topology, Covers, and Compactness

Definition (Denseness): If $X$ is a topological space, then $A \subset X$ is dense if $\bar A = X$. Denseness of $A$ is a nice property, because any point $x \in X$ is either in $A$, or is a limit point of $A$.

Definition (Subspace Topology): Let $(X,\mathscr T)$ be a topological space and $Y \subset X$. Then inheriting the topology $\mathscr T_Y = \{ U \cap Y: U \in \mathscr T \}$, (basically taking the intersection of all open sets with $Y$) defines a topological space $(Y, \mathscr T_{Y})$. (It is left to the reader to verify that $(Y, \mathscr T_{Y})$ satisfies the axioms of topology.)

Definition (Covers and Subcovers): Let $X$ be a topological space, and $S \subset X$, then an open cover of $S$ is a collection of open sets $C =\{U_{\alpha} : \alpha \in I\}$ such that $S \subset \bigcup_{\alpha \in I}U_{\alpha}$. If $C' \subset C$ is also a cover of $S$, then $C'$ is a subcover of $S$. Note: subcover simply means picking some (maybe all) elements of the cover.

Definition (Compactness): Let $X$ be a topological space and $S \subset X$. Then $S$ is compact if for every open cover $C(S)$ there exists a subcover $C'(S) \subset C(S)$ such that the cardinality of $C'$ is finite.

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4 - Bases, Sub-bases, and Generating Topologies

Definition (Basis for a Topology): Let $(X, \mathscr T)$ be a topological space. Then, a basis of this topology $\mathscr B \subset \mathscr T$ is such that for every point $x \in X$ and open set $U$ around $x$, there exists a “basic open set” $B \in \mathscr B$ such that $x \in B \subset U$. Essentially, the basis is a subcollection of open sets, such that any open set is the union of some basis elements. The elements of the basis are called basic open sets.

• (B1) Since $X$ is itself an open set, for every $x \in X$, there exists a basic open set $B$ around $x$.
• (B2) For any two basic open sets $B_1, B_2$ if $x \in B_1 \cap B_2$ there exists $B_3 \subset B_{1} \cap B_{2}$ around $x$. (this is because the (finite) intersections of open sets are open).

In fact, if there is no topology on $X$, but we have a collection of subsets $\mathscr B$ for which (B1) and (B2) hold, then we can generate a topology from arbitrary unions of elements of $\mathscr B$. By induction on (B2) the generated topology is stable under finite intersection, and by (B1) $X$ is part of the generated topology. And by the process of generation, the generated topology is also stable under arbitrary unions.

Finer and Coarser Topologies

Now, suppose $\mathscr{T}_1$ and $\mathscr{T}_2$ are both topologies on $X$, such that $\mathscr{T}_1 \subseteq \mathscr{T}_2$. Then we say that $\mathscr{T}_2$ is finer than $\mathscr{T}_1$ and that $\mathscr{T}_1$ is coarser than $\mathscr{T}_2$.

The intuition is to take two copies of $X$, then grind it down, one finer than the other. Now, you identify clumps of both, to be the basis open sets. Indeed, I could make the coarser stone, with clumps of finer stones. So the arbitrary unions of coarser stones, can also be made by clumps of finer stone, however, having finer stone lets me make more intricate open sets, so naturally, the topology of the finer stone contains whatever the topology of the coarser stone, but not necessarily the other way around!

Proposition: If $\mathscr{B}_1$ is a basis for $\mathscr{T}_1$ and $\mathscr{B}_2$ is a basis for $\mathscr{T}_2$, such that $\mathscr{T}_2$ is finer than $\mathscr{T}_1$. Then, $\forall x \in X$ and each $B_1 \in \mathscr{B}_1$ containing $x$, there is $B_2 \in \mathscr{B}_2$ such that $x \in B_2 \subseteq B_1$! (this is actually an iff).

Proof: Suppose $\mathscr{T}_2$ is finer than $\mathscr{T}_1$. Let $B_1$ be any basic open set in $\mathscr{T}_1$ and $x$ any point in $B_1$. Since $\mathscr{T}_1 \subseteq \mathscr{T}_2$, $B_1$ is open in $\mathscr{T}_2$. Because $\mathscr{B}_2$ is a basis for $\mathscr{T}_2$, there is some $B_2\in\mathscr B_2$ such that $x\in B_2\subseteq B_1$.

Now, the other direction: let $U \in \mathscr{T}_1$, then, by the basis property, $U = \bigcup_{\alpha \in \Lambda} B_\alpha'$ where $B_\alpha' \in \mathscr{B}_1$. Now for every $x_\alpha \in B_\alpha'$, $\exists B_2^{x_\alpha} \in \mathscr{B}_2$ such that $x_\alpha \in B_2^{x_\alpha} \subseteq B_\alpha'$. Therefore, $\bigcup_{\alpha \in \Lambda} \bigcup_{x_\alpha \in B_\alpha'} B_2^{x_\alpha} = U \in \mathscr{T}_2$! i.e. $\mathscr{T}_1 \subseteq \mathscr{T}_2$. $\square$

Sub-basis

Let $X$ be a set such that $\forall x \in X$, $\exists S \in \mathscr{S}$ for which $x \in S$. Such a collection of subsets of $X$, $\mathscr{S}$ is called the sub-basis for a topology.

Sub-basis $\overset{\text{all finite intersection}}{\longrightarrow}$ basis $\overset{\text{all arbitrary union}}{\longrightarrow}$ Topology.

Lemma: Let $B_i, i = 1$ to $n \in \mathscr{B}$ be any finite collection of basic open sets. Then, for every $x \in \bigcap_{i=1}^{n} B_i$, $\exists B_x^* \in \mathscr{B}$, $x \in B_x^* \subseteq \bigcap_{i=1}^{n} B_i$.

Proof: Indeed, the base case for an inductive argument is in the definition of a basic open set itself. $\forall B_1, B_2 \in \mathscr{B}$, $\forall x \in B_1 \cap B_2$, $\exists B_3$ for which $x \in B_3 \subseteq B_1 \cap B_2$.

Now suppose the inductive hypothesis holds for some $n \ge 2$.

Now, let $x \in \bigcap_{i=1}^{n+1} B_i$. Indeed, $x \in \bigcap_{i=1}^{n} B_i$, so $\exists B_x^n \in \mathscr{B}$ such that $x \in B_x^n \subseteq \bigcap_{i=1}^{n} B_i$. But $x \in B_{n+1}$ too, hence, $x \in B_x^n \cap B_{n+1} \subseteq \bigcap_{i=1}^{n+1} B_i$.

Again due to the definition of basis of topology, $\exists B_x^{n+1} \subseteq B_x^n \cap B_{n+1}$ such that $x \in B_x^{n+1}$. Hence, $x \in B_x^{n+1} \subseteq \bigcap_{i=1}^{n+1} B_i$. $\square$

Theorem: Let $X$ be a set, and $\mathscr{B} = \{B_\alpha : \alpha \in I\}$ a collection of subsets of $X$ satisfying the “basis” properties. Define $\mathscr{T} = \{\bigcup_{\alpha \in \Lambda} B_\alpha : \Lambda \subseteq I\}$, that is, $\mathscr{T}$ consists of unions over all subsets of the index set of $\mathscr{B}$. Then $\mathscr{T}$ forms a topology on $X$.

Proof:

Closed under finite intersection: Let $U_i, i = 1$ to $n$ be a finite collection of elements of $\mathscr{T}$, and hence each $U_i = \bigcup_{\alpha_i \in \Lambda_i} B_{\alpha_i}$ for some $\Lambda_i \subseteq I$.

$$\bigcap_{j=1}^{n} U_j = \bigcap_{j=1}^{n} \left(\bigcup_{\alpha_j \in \Lambda_j} B_{\alpha_j}\right)$$

Repeatedly applying set theory laws (which afaik its okay on uncountable unions too, like I know you can’t give an induction argument but still)

$$\bigcap_{j=1}^{n} U_j = \bigcup_{\alpha_1 \in \Lambda_1} \bigcup_{\alpha_2 \in \Lambda_2} \cdots \bigcup_{\alpha_n \in \Lambda_n} \left(B_{\alpha_1} \cap B_{\alpha_2} \cdots \cap B_{\alpha_n}\right)$$

Indeed, from the previous lemma, $\forall x \in B_{\alpha_1} \cap B_{\alpha_2} \cdots B_{\alpha_n}$, $\exists B_{(\alpha_1, \alpha_2 \cdots \alpha_n)}^x \in \mathscr{B}$, where $x \in B_{(\alpha_1, \alpha_2 \cdots \alpha_n)}^x \subseteq (B_{\alpha_1} \cap B_{\alpha_2} \cdots B_{\alpha_n})$. Hence

$$\bigcup_{x \in \bigcap_{i=1}^{n} B_{\alpha_i}} B_{(\alpha_1, \alpha_2 \cdots \alpha_n)}^x = \bigcap_{j=1}^{n} B_{\alpha_j}$$

If the above argument looks sus to you, just realize the fact that each $B^x_{(\alpha_{1},\alpha_{2},\dots \alpha_{n})}$ for a given $x$ in $\bigcap_{j=1}^{n} B_{\alpha_j}$, contains $x$, and also that each $B^x_{(\alpha_{1},\alpha_{2},\dots \alpha_{n})}$ is a subset of $\bigcap_{j=1}^{n} B_{\alpha_j}$.

Hence, finally

$$\bigcap_{j=1}^{n} U_j = \bigcup_{(\alpha_1,\ldots,\alpha_n)\in \Lambda_1\times\cdots\times\Lambda_n} \left(\bigcup_{x \in \bigcap_{j=1}^{n} B_{\alpha_j}} B_{(\alpha_1, \alpha_2 \cdots \alpha_n)}^x\right)$$

And since its all unions, $\bigcap_{j=1}^{n} U_j \in \mathscr{T}$, so $\mathscr{T}$ is closed under finite intersection.

Closed under arbitrary union: Now, let $U_\beta, \beta \in \Delta$ be an arbitrary subcollection of $\mathscr{T}$. $U_\beta = \bigcup_{\alpha_\beta \in \Lambda_\beta} B_{\alpha_\beta}$. Hence $\bigcup_{\beta \in \Delta} U_\beta = \bigcup_{\beta \in \Delta} \bigcup_{\alpha_\beta \in \Lambda_\beta} B_{\alpha_\beta} \in \mathscr{T}$. So $\mathscr{T}$ is closed under arbitrary union.

Contains $X$ and $\emptyset$: Finally since $\bigcup_{B \in \mathscr{B}} B = X$, $X \in \mathscr{T}$ and similarly have $\emptyset \in \mathscr{T}$. $\square$

Sub-basis Generates a Basis

Let $\mathscr{S}$ be a collection of subsets of $X$ that covers $X$ (sub-basis), then $\mathscr{B} = \{\bigcap_{j=1}^{n} S_j : n \in \mathbb{N}, S_j \in \mathscr{S}\}$ the set of all finite subcollection intersections, forms a basis for a topology on $X$.

Proof: $\forall S \in \mathscr{S}$, $S \in \mathscr{B}$, hence, $\forall x \in X$, $\exists S_x \in \mathscr{B}$ such that $x \in S_x$. Now, let $B_1, B_2 \in \mathscr{B}$. $B_1 = \bigcap_{j=1}^{m} S_j^1$, $B_2 = \bigcap_{j=1}^{n} S_j^2$. Let $x \in B_1 \cap B_2$, $\exists S_x \in \mathscr{S}$ such that $x \in S_x$. Hence $x \in S_x \cap B_1 \cap B_2$ (this is a finite intersection of elements of $\mathscr{S}$) hence indeed $\exists B_x^* = S_x \cap B_1 \cap B_2$ for which $x \in B_x^* \subseteq B_1 \cap B_2$. $\square$

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5 - Continuous Maps

Definition (Continuous Maps on Topological Spaces): Let $(X, \Theta_{X})$ and $(Y, \Theta_{Y})$ be topological spaces. A map $f: X \to Y$ is a homomorphism (continuous) if for any open subset $V \in \Theta_{Y}$ of $Y$, the pre image of $V$, $f^{-1}(V)$ is open in $X$, that is $f^{-1}(V) \in \Theta_{X}$. (Note that since $\emptyset$ is always open in any topology, there is no need for our map to be surjective, if $V$ is not hit by $f$ at all, then its pre-image is empty and so we are okay.)

If $f : X \to Y$ is bijective, and both $f, f^{-1}$ are continuous, then $f$ is a topological isomorphism (called “Homeomorphism”).

• Given three topological spaces $X, Y, Z$, if $f: X \to Y, \ g: Y \to Z$ are both continuous, then $g \circ f : X \to Z$ is continuous. For all $V_{Z} \in \Theta_{Z}$, $g^{-1}(V_Z) \in \Theta_{Y}$. Hence $f^{-1}(g^{-1}(\Theta_{Z})) \in \Theta_{X}$.
• If $f : X \to Y$ is a continuous map, let $S \subset X$. If we put the subspace topology (inheriting from $\Theta_{X}$) on $S$, then the restriction $f|_{S}: S\to Y$ is continuous. Let $V$ be any open set of $Y$. $f^{-1}|_{S}(V) = f^{-1}(V) \cap S$. Since $f^{-1}(V)$ is open in $X$, by the definition of the subspace topology, $f^{-1}(V)\cap S$ is open in $S$.

Continuity via Basis

Proposition: Let $f: X \to Y$ be a map between topological spaces $X, Y$, with $\mathscr{B}_X, \mathscr{B}_Y$ as the respective bases. $f$ is continuous if and only if $\forall b_Y \in \mathscr{B}_Y$, $f^{-1}(b_Y)$ is open in $X$. (not necessarily basic open in $X$).

Proof: The forward direction is trivial. If $f$ is continuous, $b_Y$ is an open set and $f^{-1}(b_Y)$ must be open.

So now, suppose that $\forall b_Y \in \mathscr{B}_Y$, $f^{-1}(b_Y)$ is open in $X$. Let $U$ be any open set in $Y$. $U = \bigcup_{\alpha \in \Lambda} b_Y^\alpha$, hence, $f^{-1}(U) = \bigcup_{\alpha \in \Lambda} f^{-1}(b_Y^\alpha)$ which is a union of open sets in $X$, and hence is open. $\square$

Continuity via Sub-basis

Proposition: Let $f: X \to Y$ be a map between topological spaces $X, Y$. And let $\mathscr{B}_Y$ be a basis for the topology on $Y$, and $\mathscr{S}_Y$ a sub-basis that generates $\mathscr{B}_Y$. $f$ is continuous if and only if for each $s_Y \in \mathscr{S}_Y$, $f^{-1}(s_Y)$ is open in $X$.

Proof: The forward direction is yet again trivial, each $s_Y \in \mathscr{S}_Y$ is indeed an open set in $Y$, so if $f$ is continuous, $f^{-1}(s_Y)$ must be open in $X$. Now, suppose that for every $s_Y \in \mathscr{S}_Y$, $f^{-1}(s_Y)$ is open in $X$. Let $b_Y \in \mathscr{B}_Y$ be any basic open set in $Y$. $b_Y = \bigcap_{i=1}^{n} s_Y^i$. Hence $f^{-1}(b_Y) = \bigcap_{i=1}^{n} f^{-1}(s_Y^i)$ which is a finite intersection of open sets in $X$, and by the previous proposition, $f$ is continuous. $\square$

Some Basic Ideas

$f: X \to Y$, if $f$ is continuous, and we put a coarser topology on $Y$, it will still be continuous. If $f$ is an open map defined: for all open sets $U$ in $X$, $f(U)$ is open in $Y$, if we put a coarser topology on $X$, $f$ is still an open map.

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6 - Disjoint Union and Quotient Topology

Disjoint Union

Let $\Lambda$ be an index set, and $\{A_\alpha : \alpha \in \Lambda\}$ a collection of sets. The disjoint union is defined as

$$\coprod_{\alpha \in \Lambda} A_\alpha = \bigcup_{\alpha \in \Lambda} A_\alpha \times \{\alpha\}$$

Indeed, if $\mathbb{A} = \coprod_{\alpha \in \Lambda} A_\alpha$, we can recover any $A_{\alpha^*}$ as the partition of $\mathbb{A}$ with part $A_{\alpha^*} = \{x : (x, \alpha^*) \in \mathbb{A}\}$. Cool!

Indeed, the canonical injections $\iota_\alpha: A_\alpha \to \mathbb{A}$, $x \mapsto (x, \alpha)$ are very helpful.

Quotient Topology

Let $X$ be a topological space and $\sim$ an equivalence relation on $X$. Then there is a canonical topology (quotient), one where the map $q: X \to X/\sim$, the quotient projection map $x \mapsto [x]$ is continuous. Indeed $q$ is surjective, and $U$ is open in $X/\sim$ if and only if $q^{-1}(U) = \{x \in X : [x] \in U\}$ is open in $X$.

For any set $X$ and any equivalence relation $\sim$, and the quotient set $X/\sim$, we have a natural function, that sends each $x \in X$ to its equivalence class $[x] \in X/\sim$.

In particular, for some index set $\Lambda$, let $X_\alpha, \alpha \in \Lambda$ denote the equivalence classes that partition $X$ under $\sim$.

i.e. $\bigcup_{\alpha \in \Lambda} X_\alpha = X$ and $X_\alpha \cap X_{\alpha'}=\emptyset$ whenever $\alpha \neq \alpha'$. The quotient set is the set of the classes themselves, $X/\sim=\{X_\alpha:\alpha\in\Lambda\}$, and the natural projection sends $x$ to the unique class $[x]$ containing it.

One may choose a representative $x_\alpha^*\in X_\alpha$ for each class, but the representatives are not the elements of the quotient; the equivalence classes are.

If $X$ is a topological space, what is the finest topology we can put on $X/\sim$ for which $q$ is continuous? Okay, put some topology $\mathscr{T}$ on $X/\sim$ such that $q$ is continuous. Then for any $U \in \mathscr{T}$, it is clear that $q^{-1}(U)$ must be open in $X$. This means that any such $\mathscr{T}$ is a subcollection of $\mathscr{M} = \{ U \subset X/\sim : q^{-1}(U) \ \text{is open in }X\}$. Now we only have to show that $\mathscr{M}$ is a topology on $X/\sim$. It is clear that $\emptyset, X/\sim \ \in \mathscr{M}$. Now consider an arbitrary union of elements in $\mathscr{M}$, $V = \bigcup_{a \in A}U_{a}$. Then, $q^{-1}(V) = q^{-1}(\bigcup_{a \in A}U_{a}) = \bigcup_{a \in A}q^{-1}(U_{a})$. But each $q^{-1}(U_{a})$ is open in $X$, so their union is also open in $X$. Thus $q^{-1}(V)$ is open in $X$, and hence $V \in \mathscr{M}$; therefore $\mathscr{M}$ is closed under arbitrary unions. A very similar argument shows that $\mathscr{M}$ is closed under finite intersections. $\square$

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7 - Convergence and the Compact Convergence Topology

On Convergence

Consider a set $E$ as the domain of some set of functions, targeted towards a metric space $(Y,d)$. The sequence of functions $f_1, f_2, \dots$ converges to $f$ pointwise, if for every $\epsilon > 0$ and $x \in E$, there exists $N(x,\epsilon) \in \mathbb N$ such that for $n > N(x,\epsilon)$, $d(f_n(x), f(x)) < \epsilon$. The thing about pointwise convergence on $E$ is that, the tail of the sequence that is epsilon close to $f$ at an input $x$ depends not only on epsilon but $x$ as well.

On the other hand, $f_{1}, f_{2}, \dots$ uniformly converges to $f$ if for any $\epsilon > 0$, there exists $N(\epsilon)$ such that no matter what choice of $x \in E$, whenever $n > N(\epsilon)$, $d(f_{n}(x), f(x)) < \epsilon$.

Definition (Compact Convergence): Let $(X, \mathscr T)$ be a topological space, $(Y, d)$ a metric space, and $^XY$ the set of all functions from $X$ to $Y$. A sequence of functions $f_1, f_2, \dots$ in $^XY$ converges compactly to $f$ in $^XY$ if for every compact subset $K$ of $X$, the restriction of each $f_{n}$ to $K$ uniformly converges to the restriction of $f$ to $K$. That is, for every compact $K\subset X$ and every $\epsilon > 0$, there is $N(K,\epsilon) \in \mathbb N$ such that whenever $n > N(K,\epsilon)$, for all $x \in K$, $d(f_n(x), f(x)) < \epsilon$.

• Since the inequality holds for all $x \in K$, we can say that $\sup_{x \in K} \ d(f_{n}(x),f(x)) < \epsilon$ for all $n > N(K,\epsilon)$.
• Stating the definition compactly (pun intended), for all compact $K \subset X$, $\lim_{ n \to \infty } \sup_{x \in K} d(f_{n}(x), f(x)) = 0$.

In some sense, we want to put a topology on $^X Y$ such that if $f_n \longrightarrow f$ (that is the sequence $f_n$ converges to $f$, both in $^XY$ as a topological space) if and only if $f_n$ compactly converges to $f$.

Take any compact set $C \subset X$, any $\epsilon > 0$ and $f \in {^X}Y$. Define the open ball / basis element given by:

$$B_C(f, \epsilon) = \{ g \in {}^{X}Y : \sup_{x \in C} d(f(x),g(x)) < \epsilon \}$$

Then, using $\kappa X$ to denote all nonempty compact subsets of $X$,

$$\mathscr{B} = \{B_C(f,\epsilon): C \in \kappa X,\ \epsilon > 0,\ f \in {}^{X}Y\}.$$

We now have to verify that $\mathscr B$ is a basis, by (B1) and (B2). So (B1) basically states (in this context) that for each $f \in {}^XY$ there exists $B \in \mathscr B$ such that $f \in B$. Well, pick any compact $C$ and $\epsilon > 0$. Clearly $B_C(f ,\epsilon) \in \mathscr B$, and $f \in B_C(f,\epsilon)$ as $d(f(x), f(x)) = 0 < \epsilon$.

Lemma: If $g \in B_C(f,\epsilon)$ and $\delta_{c} = \epsilon - \sup_{x \in C} \ d(f(x), g(x))$ then $B_C(g, \delta_{c}) \subset B_{C}(f, \epsilon)$.

Proof: Let $h \in B_C(g,\delta_{c} )$, then $\sup_{x \in C} \ d(h(x), g(x)) < \epsilon - \sup_{x \in C} \ d(g(x), f(x))$. Rearrange and use triangle inequality to get $\sup_{x \in C} \ d (h(x), f(x)) < \epsilon$. $\square$

Let $h \in B_C(f, \epsilon) \cap B_{D}(t, \upsilon)$. As defined in the lemma, $B_C(h, \delta_{C}) \subset B_{C}(f, \epsilon)$ and $B_D(h, \delta_{D}) \subset B_{D}(t,\upsilon)$. It follows that $B_{C \cup D}(h,\min(\delta_{C}, \delta_{D})) \subset B_C(f, \epsilon) \cap B_{D}(t, \upsilon)$ and contains $h$. ($C,D$ are compact so $C \cup D$ is compact). Finally showing (B2).

Definition (Convergence of Sequences in Topological Spaces): A sequence $x_n$ converges to $x$ (written $x_n \longrightarrow x$) in a topological space $X$ if for every neighborhood $V$ around $x$, there exists $N(V) \in \mathbb N$ such that for all $n > N(V)$, $x_n \in V$.

Equivalently $x_n \longrightarrow x$ if and only if for each basis element $B$ around $x$ there exists $N(B) \in \mathbb N$ such that for all $n > N(B)$, $x_n \in B$.

Proof: If $V$ is a neighborhood of $x$, then there is an open set $U$ where $x \in U \subset V$. But $U$ is the union of some basis elements, one of them has to contain $x$, call it $B^*$. So it’s sufficient to show that there is a tail of the sequence contained in $B^*$. But all basis elements $B$ around $x$ are themselves neighborhoods around $x$. So it is necessary to show that there is a tail of the sequence contained in any such $B$. $\square$

Definition (Compact Convergence Topology): Let $(X, \mathscr T)$ be a topological space and $(Y,d)$ be a metric space. The compact convergence topology on functions from $X$ to $Y$, $^X Y$ is the one generated by the following basis:

$$\mathscr{B} = \{B_C(f,\epsilon): C \in \kappa X,\ \epsilon > 0,\ f \in {}^{X}Y\}.$$

Where

$$B_C(f, \epsilon) = \{ g \in {}^{X}Y : \sup_{x \in C} d(f(x),g(x)) < \epsilon \}$$

In this topology on $^XY$, $f_n \longrightarrow f$ if and only if $f_n$ converges compactly to $f$.

Indeed, if $f_n\to f$, then for every compact $C\subset X$ and every $\epsilon>0$, the basic neighborhood $B_C(f,\epsilon)$ contains all sufficiently large $f_n$. Thus $\sup_{x\in C}d(f_n(x),f(x))<\epsilon$ eventually.

Conversely, suppose $f_n$ converges compactly to $f$, and let $B_C(g,\delta)$ be any basic neighborhood of $f$. Set $r=\sup_{x\in C}d(f(x),g(x))<\delta$. Compact convergence gives $N$ such that, for $n>N$, $\sup_{x\in C}d(f_n(x),f(x))<\delta-r$. The triangle inequality then gives $\sup_{x\in C}d(f_n(x),g(x))<\delta$, so $f_n\in B_C(g,\delta)$ eventually.

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8 - Injectivity and Surjectivity

Two Views on Injectivity of Functions

1) Assuming $A$ is nonempty, $f: A \to B$ is injective if and only if $\exists g: B \to A$ such that $g \circ f = id_A$ ($f$ has a left inverse). (Without the nonempty assumption, the unique map $\emptyset\to B$ is injective, but it has no left inverse when $B$ is nonempty.)

Proof: Suppose $g$ is the left inverse of $f$. Let $a \neq a' \in A$. Then, $g(f(a)) \neq g(f(a'))$ ($g(f(x)) = x \ \forall x \in A$). This means that $f(a) \neq f(a')$; if they were equal, say $f(a) = f(a') = t$, both $g(f(a))$ and $g(f(a'))$ would equal $g(t)$, a contradiction.

Now suppose $f$ is injective, pick a particular $a^* \in A$ (any one will do). Construct $g: B \to A$, $b \mapsto f^{-1}(\{b\})$ if $b \in f(A)$, $a^*$ if $b \notin f(A)$. $g$ is well defined as $f$ is injective so $f^{-1}(\{b\})$ is always a singleton. Indeed $g \circ f(a) = a \ \forall a \in A$. $\square$

2) $f: A \to B$ is injective iff $\forall$ sets $Z$ and all functions $\alpha, \alpha': Z \longrightarrow A$, $f \circ \alpha = f \circ \alpha' \implies \alpha = \alpha'$.

Proof: Choose $Z = \{p\}$ (any singleton) and make $d_a: \{p\} \to A \ \forall a \in A$, $p \mapsto a$. Suppose $\forall a, a' \in A$, $f \circ d_a = f \circ d_{a'} \implies d_a = d_{a'}$. $\implies f(d_a(p)) = f(d_{a'}(p)) \implies d_a(p) = d_{a'}(p)$ $\implies f(a) = f(a') \implies a = a'$.

Now suppose $f$ is injective, and $\alpha, \alpha': Z \to A$ is any two functions on any set. And suppose $f \circ \alpha = f \circ \alpha'$. Since $f$ has a left inverse $g$, $(g \circ f) \circ \alpha = (g \circ f) \circ \alpha'$. Hence, $id_A \circ \alpha = id_A \circ \alpha'$ hence $\alpha = \alpha'$. $\square$

Two Views on Surjectivity of Functions

3) Assuming the axiom of choice, $f: A \to B$ is surjective iff $\exists g: B \to A$ such that $f \circ g = id_B$ ($f$ has a right inverse).

Proof: Suppose $f$ has a right inverse $g$. Let $b \in B$ be an element of $B$. $f \circ g(b) = b$. That is, $\exists g(b) \in A$ : $f(g(b)) = b$. Hence $f$ is surjective.

Now suppose $f$ is surjective, then $f(A) = B$. Hence $\forall b \in B$, $f^{-1}(\{b\})$ is non empty. So we can choose an element $g(b) \in f^{-1}(\{b\})$ and construct $g$. $\square$

4) $f: A \to B$ is surjective iff $\forall$ sets $Z$ and all functions $\alpha, \alpha': B \to Z$, $\alpha \circ f = \alpha' \circ f \implies \alpha = \alpha'$.

Proof: If $f$ is surjective, and $\alpha \circ f = \alpha' \circ f$, use the right inverse $g$, so $\alpha \circ (f \circ g) = \alpha' \circ (f \circ g)$ $\implies \alpha = \alpha'$. Now suppose whenever $\alpha \circ f = \alpha' \circ f$, $\alpha = \alpha'$. Choose $Z = \{0, 1\}$ and, $\alpha: b \mapsto 0 \ \forall b \in B$ and $\alpha'(b) = 0 \iff b \in f(A)$ else $1$. Indeed $\forall a \in A$, $\alpha(f(a)) = \alpha'(f(a)) = 0$. But this means that $\alpha = \alpha'$. i.e. $\forall b \in B$, $\alpha(b) = 0$, $\alpha'(b) = 0$. Therefore $B = f(A)$. $\square$

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9 - Homotopy Theory

Let $X,Y$ be topological spaces, and $f,g \in Mor(X,Y)$ ($f,g$ are continuous maps from $X$ to $Y$).

Then, $f$ is homotopic to $g$, written $f \sim_H g$ if:

There exists a continuous map $H: X \times [0,1]_{STD} \to Y$ such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for every $x\in X$.

If $H: X \times [0,1] \longrightarrow Y$ is a continuous map, where $H(x,0)=f(x)$ and $H(x,1)=g(x)$ then we call $H$ the (a) homotopy map between $f$ and $g$.

Indeed, $\sim_H$ is an equivalence relation on $Mor(X,Y)$.

Reflexivity

$f \sim_H f$. Choose $H: X \times [0,1] \to Y$ as $H(x,t) = f(x)$. Now, let $V_Y$ be open in $Y$, then, $H^{-1}(V_Y) = f^{-1}(V_Y) \times [0,1]$. But $f$ is continuous. Hence $f^{-1}(V_Y)$ is open in $X$, and $[0,1]$ is trivially open in $[0,1]$. Therefore, by definition of product topology, $f^{-1}(V_Y) \times [0,1]$ is open in $X \times [0,1]$. Indeed, $H(x,0) = f(x)$ and $H(x,1) = f(x)$ too.

We need some lemmas now:

Lemma 1 (Interval Change Map): For $p<q$, define the interval change map $C_{[p,q]}: X \times [p,q] \longrightarrow X \times [0,1]$ by $C_{[p,q]}(x,t) = (x, \frac{t-p}{q-p})$. Then, $C_{[p,q]}$ is a homeomorphism.

Proof:

Bijectivity: $(x, \frac{t_1-p}{q-p}) = (x, \frac{t_2-p}{q-p}) \implies x=x \text{ and } t_1=t_2$. For each $(x,u) \in X \times [0,1]$, $C_{[p,q]}(x, (q-p)u+p) = (x,u)$.

Continuity: It is enough to check a basic open rectangle $\gamma=V_X\times U$, where $V_X$ is open in $X$ and $U=(\bigcup_{\alpha\in\Lambda}(a_\alpha,b_\alpha))\cap[0,1]$ is open in $[0,1]$.

The properties of inverse images gives:

$$C_{[p,q]}^{-1}(\gamma) = V_X \times \left(\left(\bigcup_{\alpha \in \Lambda} ((q-p) a_\alpha+p, (q-p) b_\alpha+p)\right) \cap [p,q]\right)$$

which is indeed an open set in $X \times [p,q]$!

Inverse Continuity: Similarly, it is enough to check a basic open rectangle $\gamma=V\times U$ in $X\times[p,q]$. Its image under $C_{[p,q]}$ is a basic open rectangle in $X\times[0,1]$, since the affine map $t\mapsto (t-p)/(q-p)$ is a homeomorphism. Hence $C_{[p,q]}^{-1}$ is continuous. $\square$

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Lemma 2 (Cut-Paste): Let $H, H^*: X \times [0,1] \longrightarrow Y$ be continuous maps such that $H(x,1) = H^*(x,0) \forall x \in X$. Then, the cut-paste function at $p$, written $H^{[0,p]} \times_{p} H^{*[p,1]}$ or simply $H \times_p H^*$ is the function

$$H \times_p H^*(x,t) = \begin{cases} H(C_{[0,p]}(x,t)) & t \in [0,p] \\ H^*(C_{[p,1]}(x,t)) & t \in [p,1] \end{cases}$$

We claim that $H \times_p H^*$ is a continuous map from $X \times [0,1] \longrightarrow Y$. Firstly, the map $H \times_p H^*$ is well defined, as $H(C_{[0,p]}(x,p)) = H^*(C_{[p,1]}(x,p)) \forall x \in X$. Then, continuity follows by realizing that $X \times [0,p]$ and $X \times [p,1]$ are both closed in $X \times [0,1]$, ($X$ is clopen in $X$, and $[0,p]$ and $[p,1]$ are both closed in $[0,1]$, and product of closed sets are closed in the product topology). And then, we simply invoke the Gluing Lemma, which we will state and prove next. $\square$

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Gluing Lemma

Let $P, Q$ be topological spaces, and suppose $A, B\subseteq P$ are either both closed or both open, with $A \cup B = P$. Let $f: A \to Q, g: B \to Q$ be continuous maps such that whenever $x \in A \cap B$, $f(x) = g(x)$ (the maps agree). Then the glued map $f \cup g: P \to Q$, defined by

$$f \cup g(x) = \begin{cases} f(x) & x \in A \\ g(x) & x \in B \end{cases}$$

is also continuous.

Proof: First suppose $A$ and $B$ are closed. Let $V$ be closed in $Q$. Since the domain is $P=A\cup B$,

$$(f\cup g)^{-1}(V)=((f\cup g)^{-1}(V)\cap A)\cup((f\cup g)^{-1}(V)\cap B)=f^{-1}(V)\cup g^{-1}(V).$$

Now $f^{-1}(V)$ is closed in the subspace $A$, so $f^{-1}(V)=F_1\cap A$ for some closed $F_1\subseteq P$. Similarly, $g^{-1}(V)=F_2\cap B$ for some closed $F_2\subseteq P$. Because $A$ and $B$ are closed in $P$, both preimages are closed in $P$, and their finite union is closed. Hence $f\cup g$ is continuous.

If $A$ and $B$ are open, use the same argument with an open set $V\subseteq Q$: the two preimages are open in the open subspaces $A$ and $B$, hence open in $P$, and their union is open. $\square$

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Lemma 3 (Interval Reversal): Define $\gamma: X \times [p, q] \to X \times [p, q]$ to be the interval reversal map $\gamma(x, t) = (x, (p+q)-t)$. Then $\gamma$ is continuous and bijective (homeomorphism even).

Proof: Proof is very similar to Lemma 1. So we won’t do it here. $\square$

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Transitivity

Let $f,g,h: X \to Y$ be continuous maps. $f \sim_H g$, $g \sim_{H_2} h$. Then, there exists a homotopy $H^*$ between $f$ and $h$.

Indeed, $H^*$ can be chosen to be the cut paste map at $\frac{1}{2}$, $H_1 \times_{\frac{1}{2}} H_2$ from Lemma 2, which also shows continuity of this map.

Indeed $H_1 \times_{\frac{1}{2}} H_2 (x,0) = f(x)$, $H_1 \times_{\frac{1}{2}} H_2 (x,\frac{1}{2}) = g(x)$ and $H_1 \times_{\frac{1}{2}} H_2 (x,1) = h(x)$, for every $x \in X$.

Symmetry

Let $f,g: X \to Y$ be continuous functions. Suppose $f \sim_H g$ then, there exists a homotopy $H'$ such that $g \sim_{H'} f$.

Indeed by Lemma 3, the interval reversal function $\gamma^{[0,1]}: X \times [0,1] \to X \times [0,1]$ given by $\gamma(x,t) = (x, (1-t))$ is continuous. Indeed $H' = H \circ \gamma^{[0,1]}$ is a continuous map $X \times [0,1] \to Y$. $H'(x,0) = H(x,1) = g(x)$ and $H'(x,1) = H(x,0) = f(x)$.

Cool! So indeed the homotopy relation is an equivalence on $Mor(X,Y)$, so we have equivalence classes $[f]$.

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Composing Equivalence Classes of Maps under Homotopy

We need some more lemmas first.

Lemma 4: Let $A, B, C$ be topological spaces, $g_1: A \rightarrow B$, $g_2: A \rightarrow C$. Now construct $g: A \rightarrow B \times C$, $g(a) = (g_1(a), g_2(a))$. (There is a product topology on $B \times C$). $g$ is continuous if and only if $g_1$ and $g_2$ are both continuous.

Proof: Let $U_B \times U_C$ be open in $B \times C$, $g^{-1}(U_B \times U_C) = \{a \in A: (g_1(a), g_2(a)) \in U_B \times U_C \}$. Indeed, $g^{-1}(U_B \times U_C) = \{a \in A: g_1(a) \in U_B \text{ and } g_2(a) \in U_C \}$. Therefore we get: $g^{-1}(U_B \times U_C) = g_1^{-1}(U_B) \cap g_2^{-1}(U_C)$. Indeed, the above equation holds for any $U_B$ open in $B$ and any $U_C$ open in $C$.

Choose $U_B = B$. Then $\forall$ open sets $U_C$ in $C$, $g^{-1}(B \times U_C) = g_1^{-1}(B) \cap g_2^{-1}(U_C)$. i.e. $g^{-1}(B \times U_C) = A \cap g_2^{-1}(U_C) = g_2^{-1}(U_C)$ (pause and justify). Therefore it is necessary that $g_2$ is continuous for $g$ to be continuous.

Now choose $U_C = C$ then $\forall$ open $U_B$ in $B$, $g^{-1}(U_B \times C) = g_1^{-1}(U_B) \cap g_2^{-1}(C) = g_1^{-1}(U_B) \cap A = g_1^{-1}(U_B)$. Therefore it is also necessary that $g_1$ is continuous for $g$ to be continuous.

If $g_1$ and $g_2$ are both continuous, then the preimage of every basic open rectangle is open: $g^{-1}(U_B \times U_C) = g_1^{-1}(U_B) \cap g_2^{-1}(U_C)$. Since such rectangles form a basis for the product topology, $g$ is continuous. $\square$

Lemma 5 (Id Projection Extension): Let $f: A \times B \to C$ be a continuous map. The projection map $\Pi_B: A \times B \to B$, $\Pi_B(a,b) = b$ is also continuous. The map $f \times \Pi_B: A \times B \to C \times B$, $(a,b) \mapsto (f(a,b), \Pi_B(a,b))$ is continuous. (When $A, B, C$ are topological spaces, and product topology on $A \times B$ and $C \times B$.)

Proof: Follows trivially from Lemma 4. $\square$

Theorem (Composing Equivalence Classes of Maps under Homotopy): Let $X, Y, Z$ be topological spaces, put the quotient of homotopy relation on continuous maps between $X$ and $Y$, and $Y$ and $Z$. Now let $[f] \in Mor(X,Y)/\sim_H$ and $[g] \in Mor(Y,Z)/\sim_H$. Then for any $f' \in [f]_{\sim_H}$ and $g' \in [g]_{\sim_H}$, $g' \circ f' \in [g \circ f]_{\sim_H}$.

Proof: Let $H_1: X \times [0,1] \to Y$ be a homotopy from $f$ to $f'$. i.e. $H_1$ is continuous, $H_1(x,0) = f(x)$ and $H_1(x,1) = f'(x) \ \forall x \in X$. Similarly let $H_2: Y \times [0,1] \to Z$ be a homotopy from $g$ to $g'$. $H_2$ is continuous, $H_2(y,0) = g(y)$, $H_2(y,1) = g'(y) \ \forall y \in Y$. Now consider the map $H_2 \circ (H_1 \times \Pi_{[0,1]}): X \times [0,1] \to Z$. This map is continuous, ($H_1 \times \Pi_{[0,1]}$ is continuous by Lemma 5).

Now, $H_2 \circ (H_1 \times \Pi_{[0,1]})(x,0) = H_{2}(H_{1}(x,0),0) = H_{2}(f(x),0) = g(f(x))$, and similarly, $H_2 \circ (H_1 \times \Pi_{[0,1]})(x,1) = g'(f'(x))$. $\square$

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10 - Homotopy Equivalence of Topological Spaces

Definition: Let $X,Y$ be topological spaces, then $X$ is homotopy equivalent to $Y$, written $X \simeq_H Y$, if $\exists$ continuous maps $f: X \to Y$ and $g: Y \to X$ such that $g \circ f \sim_H id_X$, $f \circ g \sim_H id_Y$.

Homotopy Equivalence is an Equivalence Relation

Recall that two topological spaces $X, Y$ are homotopy equivalent if $\exists f: X \to Y$ and $g: Y \to X$ (both continuous) such that $g \circ f \sim_H id_X$, $f \circ g \sim_H id_Y$. In such cases, we write $X \simeq_H Y$. We claim that this relation on topological spaces is indeed an equivalence.

1. Reflexive: $X \simeq_H X$. Indeed, $id_X: X \to X$ is a continuous map, not only homotopic but literally equal to $id_X$.

2. Symmetry: $X \simeq_H Y \iff Y \simeq_H X$ (obvious from the definition itself).

3. Transitivity: Suppose $X \simeq_H Y$ and $Y \simeq_H Z$. We will repeatedly use the theorem on composition of homotopy equivalence classes on maps, which states that if $g, f$ are continuous maps that are composable, and $g \in [g]_{\sim_H}$, and $f \in [f]_{\sim_H}$, then $g \circ f \in [g \circ f]_{\sim_H}$.

(diagram: $X \overset{f}{\longrightarrow} Y \overset{g}{\longrightarrow} Z$ and $Z \overset{h}{\longrightarrow} Y \overset{k}{\longrightarrow} X$)

$h \circ g \in [id_Y]_{\sim_H}$, hence $(h \circ g) \circ f \in [id_Y \circ f]_{\sim_H} = [f]_{\sim_H}$. Therefore, $k \circ (h \circ g) \circ f \in [k \circ f]_{\sim_H}$. But $k \circ f \in [id_X]_{\sim_H}$, and by properties of equivalence classes, $[k \circ f]_{\sim_H} = [id_X]_{\sim_H}$. Finally, $k \circ h \circ g \circ f \in [id_X]_{\sim_H}$.

Similarly, $f \circ k \in [id_Y]_{\sim_H}$, hence $(f \circ k) \circ h \in [id_Y \circ h]_{\sim_H} = [h]_{\sim_H}$. Therefore, $g \circ (f \circ k) \circ h \in [g \circ h]_{\sim_H}$. But $g \circ h \in [id_Z]_{\sim_H}$. Hence, $[g \circ h]_{\sim_H} = [id_Z]_{\sim_H}$. Therefore, $g \circ f \circ k \circ h \in [id_Z]_{\sim_H}$. $\square$