Chapter 3 • Theory & Derivations

Unit 3: Continuity, Homeomorphisms, Weak Topologies & Quotient Spaces

Mappings between topological spaces: continuous functions and their equivalent characterizations (open preimages, closed preimages, closure inclusions $f(\bar{A}) \subseteq \overline{f(A)}$), the Pasting (Gluing) Lemma, homeomorphisms and topological invariants, topological embeddings, initial (weak) and final topologies, function algebras $C(X, \mathbb{R})$, and quotient spaces with 2-manifold identifications (cylinder, Möbius strip, torus, Klein bottle, and real projective plane $\mathbb{RP}^2$).

§3.1 Continuous Functions: Open Preimages & Closure Characterizations

1. Topological Continuity

In elementary calculus, continuity is formulated via $\varepsilon$-$\delta$ bounds. In general topology, continuity is expressed purely in terms of the preimage of open sets.

Definition 3.1 (Continuous Function): Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces. A function $f: X \to Y$ is called continuous if the preimage of every open set in $Y$ is open in $X$:

f^{-1}(V) \in \mathcal{T}_X \qquad \text{for every } V \in \mathcal{T}_Y

We say $f$ is continuous at a point $x_0 \in X$ if for every neighborhood $V$ of $f(x_0)$ in $Y$, the preimage $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$.

2. Fundamental Characterization Theorem

Theorem 3.1 (Equivalent Characterizations of Continuity): Let $f: X \to Y$ be a map between topological spaces. The following are logically equivalent:

$f$ is continuous (preimages of open sets are open).

The preimage of every closed set in $Y$ is closed in $X$:

K \text{ closed in } Y \implies f^{-1}(K) \text{ closed in } X

For every subset $A \subseteq X$, $f(\operatorname{cl}(A)) \subseteq \operatorname{cl}(f(A))$.

For every subset $B \subseteq Y$, $\operatorname{cl}(f^{-1}(B)) \subseteq f^{-1}(\operatorname{cl}(B))$.

For every subset $B \subseteq Y$, $f^{-1}(\operatorname{int}(B)) \subseteq \operatorname{int}(f^{-1}(B))$.

For a basis $\mathcal{B}$ of $Y$, $f^{-1}(B)$ is open in $X$ for all $B \in \mathcal{B}$.

For a subbasis $\mathcal{S}$ of $Y$, $f^{-1}(S)$ is open in $X$ for all $S \in \mathcal{S}$.

Proof of (1) $\iff$ (2): Follows from the set-theoretic identity for complements under preimages:

f^{-1}(Y \setminus B) = X \setminus f^{-1}(B)

If $K \subseteq Y$ is closed, then $Y \setminus K$ is open. By (1), $f^{-1}(Y \setminus K) = X \setminus f^{-1}(K)$ is open in $X$, so $f^{-1}(K)$ is closed in $X$. The converse is identical.

Proof of (1) $\iff$ (3):

(1 $\implies$ 3): Let $A \subseteq X$. The set $\operatorname{cl}(f(A))$ is closed in $Y$.

By (2), $f^{-1}(\operatorname{cl}(f(A)))$ is closed in $X$. Since $f(A) \subseteq \operatorname{cl}(f(A))$, we have $A \subseteq f^{-1}(\operatorname{cl}(f(A)))$. Since $\operatorname{cl}(A)$ is the smallest closed set containing $A$:

\operatorname{cl}(A) \subseteq f^{-1}(\operatorname{cl}(f(A))) \implies f(\operatorname{cl}(A)) \subseteq \operatorname{cl}(f(A))

(3 $\implies$ 2): Let $K \subseteq Y$ be closed. Let $A = f^{-1}(K)$.

By (3), $f(\operatorname{cl}(A)) \subseteq \operatorname{cl}(f(A)) = \operatorname{cl}(f(f^{-1}(K))) \subseteq \operatorname{cl}(K) = K$. Thus $\operatorname{cl}(A) \subseteq f^{-1}(K) = A$. Since $A \subseteq \operatorname{cl}(A)$ always, we have $\operatorname{cl}(A) = A$, so $A = f^{-1}(K)$ is closed. $\blacksquare$

§3.2 The Pasting (Gluing) Lemma & Local Criteria

1. The Pasting (Gluing) Lemma

A crucial tool for constructing continuous maps from piecewise components.

Theorem 3.2 (The Pasting Lemma): Let $X = A \cup B$, where $A$ and $B$ are both closed (or both open) in $X$. Let $f: A \to Y$ and $g: B \to Y$ be continuous functions that agree on the overlap:

f(x) = g(x) \qquad \forall x \in A \cap B

Then the combined function $h: X \to Y$ defined by:

h(x) = \begin{cases} f(x) & \text{if } x \in A \\ g(x) & \text{if } x \in B \end{cases}

is continuous on $X$.

Proof (Closed Case): Let $K \subseteq Y$ be any closed set in $Y$. The preimage under $h$ is:

h^{-1}(K) = \{x \in X : h(x) \in K\} = \{x \in A : f(x) \in K\} \cup \{x \in B : g(x) \in K\} = f^{-1}(K) \cup g^{-1}(K)

Since $f: A \to Y$ is continuous, $f^{-1}(K)$ is closed in the subspace $A$.

Since $A$ is closed in $X$, by Proposition 2.1, $f^{-1}(K)$ is closed in $X$.

Similarly, $g^{-1}(K)$ is closed in $B$, and since $B$ is closed in $X$, $g^{-1}(K)$ is closed in $X$.

The union of two closed sets in $X$ is closed in $X$:

h^{-1}(K) = f^{-1}(K) \cup g^{-1}(K) \text{ is closed in } X

By Theorem 3.1, $h$ is continuous on $X$. $\blacksquare$

Cautionary Remark: The Pasting Lemma fails if one set is open and the other is closed, or if neither is closed/open! For example, in $\mathbb{R} = (-\infty, 0) \cup [0, \infty)$, $f(x) = -1$ on $(-\infty, 0)$ and $g(x) = 1$ on $[0, \infty)$ are continuous on their domains, but the combined Heaviside step function is discontinuous at $x = 0$!

§3.3 Homeomorphisms, Topological Invariants & Embeddings

1. Homeomorphisms

A homeomorphism is the topological analogue of an isomorphism: two spaces connected by a homeomorphism are structurally indistinguishable from the standpoint of topology.

Definition 3.2 (Homeomorphism): A function $f: X \to Y$ between topological spaces is called a homeomorphism if:

$f$ is a bijection (one-to-one and onto).

$f$ is continuous.

The inverse function $f^{-1}: Y \to X$ is continuous (equivalently, $f$ is an open map).

If a homeomorphism exists between $X$ and $Y$, we say that $X$ and $Y$ are homeomorphic (denoted $X \cong Y$).

Warning: A continuous bijection is NOT necessarily a homeomorphism! Example: Let $X = [0, 2\pi)$ with the subspace topology of $\mathbb{R}$, and $Y = S^1 = \{e^{i\theta}\} \subset \mathbb{C}$. The map $f(\theta) = e^{i\theta}$ is a continuous bijection, but its inverse $f^{-1}$ is discontinuous at $(1, 0)$! (The half-open interval $[0, \pi)$ is open in $X$, but its image $f([0, \pi))$ is not open in $S^1$).

2. Topological Invariants

Definition 3.3 (Topological Invariant): A property $P$ of a topological space is a topological invariant (or topological property) if whenever $X \cong Y$ and $X$ possesses property $P$, then $Y$ must also possess property $P$.

Canonical Topological Invariants:

Compactness, Sequential Compactness, Local Compactness

Connectedness, Path-Connectedness, Number of Connected Components

Separation Axioms ($T_0, T_1, T_2, T_3, T_4$)

Countability Axioms (First-countability, Second-countability, Separability)

Fundamental Group $\pi_1(X)$, Homology Groups $H_k(X)$

Topological Dimension (Invariance of Domain: $\mathbb{R}^m \not\cong \mathbb{R}^n$ if $m \ne n$).

3. Topological Embeddings

Definition 3.4 (Embedding): An injective continuous map $f: X \to Y$ is called a topological embedding if the corestriction:

f: X \to f(X)

is a homeomorphism onto its image $f(X)$ (equipped with the subspace topology inherited from $Y$).

§3.4 Initial Topologies, Weak Topologies & Function Algebras

1. Initial (Weak) Topology

Given a set $X$ and a family of maps $f_\alpha: X \to Y_\alpha$ into topological spaces $(Y_\alpha, \mathcal{T}_\alpha)$, what is the most economical topology on $X$ that makes all $f_\alpha$ continuous?

Definition 3.5 (Initial / Weak Topology): The initial topology (or weak topology) on $X$ induced by the family of maps $\{f_\alpha: X \to Y_\alpha\}_{\alpha \in I}$ is the coarsest (smallest) topology on $X$ with respect to which every map $f_\alpha$ is continuous. A subbasis for the initial topology is:

\mathcal{S} = \{f_\alpha^{-1}(V_\alpha) : \alpha \in I, \; V_\alpha \in \mathcal{T}_\alpha\}

Theorem 3.3 (Universal Property of Initial Topologies): A map $g: Z \to X$ from an arbitrary topological space $Z$ into $(X, \mathcal{T}_{\text{initial}})$ is continuous if and only if each composition:

f_\alpha \circ g: Z \to Y_\alpha

is continuous for all $\alpha \in I$.

2. The Product Topology as an Initial Topology

Definition 3.6 (Product Topology): Let $\{X_\alpha\}_{\alpha \in I}$ be a family of topological spaces. The product topology on the Cartesian product $X = \prod_{\alpha \in I} X_\alpha$ is precisely the initial topology induced by the canonical projection maps:

\pi_\beta: \prod_{\alpha \in I} X_\alpha \to X_\beta, \qquad \pi_\beta((x_\alpha)_{\alpha \in I}) = x_\beta

A basis for the product topology consists of cylinders:

B = \prod_{\alpha \in I} U_\alpha, \qquad \text{where } U_\alpha \text{ is open in } X_\alpha \text{ and } U_\alpha = X_\alpha \text{ for all but finitely many } \alpha \in I

3. Function Algebras $C(X, \mathbb{R})$

For any topological space $X$, the set of continuous real-valued functions $C(X, \mathbb{R})$ forms an associative, commutative $\mathbb{R}$-algebra under pointwise addition, scalar multiplication, and pointwise multiplication:

(f + g)(x) = f(x) + g(x), \qquad (\lambda f)(x) = \lambda f(x), \qquad (f \cdot g)(x) = f(x) g(x)

The weak topology on $X$ induced by $C(X, \mathbb{R})$ plays a central role in Tychonoff spaces and Gelfand duality.

§3.5 Quotient Spaces, Identification Maps & Surface Topologies

1. The Quotient Topology

Definition 3.7 (Quotient Space & Quotient Map): Let $(X, \mathcal{T}_X)$ be a topological space and let $\sim$ be an equivalence relation on $X$. Let $Y = X / \sim$ be the set of equivalence classes, and let $q: X \to Y$ be the canonical projection $q(x) = [x]$. The quotient topology on $Y$ is the finest (largest) topology that makes $q$ continuous:

\mathcal{T}_Y = \{V \subseteq Y : q^{-1}(V) \in \mathcal{T}_X\}

More generally, a surjective map $p: X \to Y$ is called a quotient map (or identification map) if a subset $V \subseteq Y$ is open in $Y$ if and only if $p^{-1}(V)$ is open in $X$.

Theorem 3.4 (Universal Mapping Property of Quotient Spaces): Let $p: X \to Y$ be a quotient map. A function $g: Y \to Z$ is continuous if and only if the composite map $g \circ p: X \to Z$ is continuous.

2. Classical Surface Topologies from the Unit Square $I^2$

Let $I^2 = [0, 1] \times [0, 1] \subset \mathbb{R}^2$ with the Euclidean subspace topology.

Identification 1: The Cylinder $S^1 \times [0, 1]$ Identify left and right edges with the same orientation:

(0, y) \sim (1, y), \qquad \forall y \in [0, 1]

The resulting quotient space is homeomorphic to the standard cylinder $S^1 \times [0, 1]$. It is an orientable 2-manifold with boundary.

Identification 2: The Möbius Strip Identify left and right edges with a half-twist (reversed orientation):

(0, y) \sim (1, 1 - y), \qquad \forall y \in [0, 1]

The resulting quotient space is the Möbius strip. It is a non-orientable 2-manifold with boundary (a single closed boundary curve homeomorphic to $S^1$).

Identification 3: The Torus $T^2 = S^1 \times S^1$ Identify both pairs of opposite edges with standard orientation:

(0, y) \sim (1, y) \quad \text{and} \quad (x, 0) \sim (x, 1)

The quotient is a compact, connected, orientable 2-manifold without boundary, with Euler characteristic $\chi(T^2) = 0$.

Identification 4: The Klein Bottle $K^2$ Identify one pair of edges directly and the other with a twist:

(0, y) \sim (1, 1 - y) \quad \text{and} \quad (x, 0) \sim (x, 1)

The Klein bottle is a compact, connected, non-orientable 2-manifold without boundary, with $\chi(K^2) = 0$. It cannot be embedded in $\mathbb{R}^3$ without self-intersection, but embeds smoothly in $\mathbb{R}^4$.

Identification 5: The Real Projective Plane $\mathbb{RP}^2$ Identify antipodal points on the boundary:

(x, 0) \sim (1 - x, 1) \quad \text{and} \quad (0, y) \sim (1, 1 - y)

Equivalently, $\mathbb{RP}^2 \cong S^2 / \{\pm x\}$. It is a closed non-orientable surface with $\chi(\mathbb{RP}^2) = 1$.

Tiered Solved Examination Problems & Rigorous Derivations

Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.

Foundational Example 3.1: Rigorous Application of the Pasting Lemma

Consider the piecewise function $f: \mathbb{R} \to \mathbb{R}$ defined by:

f(x) = \begin{cases} x^2 + 2x & \text{if } x \le 1 \\ 4 - x & \text{if } x > 1 \end{cases}

Decompose the domain $\mathbb{R}$ into two closed subsets $A$ and $B$.
State the continuity of the restricted functions $f|_A$ and $f|_B$.
Check the gluing compatibility condition on $A \cap B$.
Formally invoke the Pasting Lemma (Theorem 3.2) to conclude that $f$ is continuous on the entire real line $\mathbb{R}$.

1. Closed Domain Decomposition

Let:

A = (-\infty, 1] \quad \text{and} \quad B = [1, \infty)

Notice that:

$A$ is closed in $\mathbb{R}$ because its complement $(1, \infty)$ is open.
$B$ is closed in $\mathbb{R}$ because its complement $(-\infty, 1)$ is open.
$A \cup B = (-\infty, 1] \cup [1, \infty) = \mathbb{R}$.

2. Continuity of Domain Restrictions

On $A$, $f|_A(x) = x^2 + 2x$ is a polynomial function, which is continuous on $\mathbb{R}$, hence continuous on the closed subspace $A$.
On $B$, $f|_B(x) = 4 - x$ is an affine polynomial function, which is continuous on $\mathbb{R}$, hence continuous on the closed subspace $B$.

3. Compatibility Condition on Overlap

The intersection of the two closed domains is the singleton:

A \cap B = (-\infty, 1] \cap [1, \infty) = \{1\}

Evaluate both functions at the single intersection point $x = 1$:

$$f|_A(1) = 1^2 + 2(1) = 1 + 2 = 3$$

$$f|_B(1) = 4 - 1 = 3$$

Since $f|_A(1) = f|_B(1) = 3$, the functions agree on $A \cap B$.

4. Conclusion via Pasting Lemma

By the Pasting Lemma (Theorem 3.2): Since $\mathbb{R} = A \cup B$ where both $A$ and $B$ are closed in $\mathbb{R}$, and $f|_A, f|_B$ are continuous on $A, B$ respectively with $f|_A|_{A \cap B} = f|_B|_{A \cap B}$, the combined function $f: \mathbb{R} \to \mathbb{R}$ is continuous on all of $\mathbb{R}$. $\blacksquare$

Advanced Example 3.2: Stereographic Projection as a Homeomorphism

Let $S^n = \{x \in \mathbb{R}^{n+1} : \|x\|_2 = 1\}$ be the $n$-dimensional sphere, and let $N = (0, \dots, 0, 1)$ be the North Pole. The stereographic projection map $\sigma: S^n \setminus \{N\} \to \mathbb{R}^n$ is defined by projecting from $N$ onto the equatorial hyperplane $x_{n+1} = 0$:

\sigma(x_1, \dots, x_n, x_{n+1}) = \left( \frac{x_1}{1 - x_{n+1}}, \dots, \frac{x_n}{1 - x_{n+1}} \right)

Derive the explicit formula for the inverse map $\sigma^{-1}: \mathbb{R}^n \to S^n \setminus \{N\}$.
Prove rigorously that $\sigma$ and $\sigma^{-1}$ are both continuous.
Conclude that $S^n \setminus \{N\} \cong \mathbb{R}^n$ are homeomorphic.

1. Inversion Formula

Let $y = (y_1, \dots, y_n) \in \mathbb{R}^n$. Let $\|y\|^2 = \sum_{i=1}^n y_i^2$. The line connecting $N = (0, \dots, 0, 1)$ to $(y_1, \dots, y_n, 0)$ is parametrized by:

L(t) = (1 - t)N + t(y, 0) = (t y_1, \dots, t y_n, 1 - t)

We seek the non-trivial intersection of $L(t)$ with $S^n$:

\|L(t)\|^2 = t^2 \|y\|^2 + (1 - t)^2 = 1

t^2 \|y\|^2 + 1 - 2t + t^2 = 1 \implies t [ t(\|y\|^2 + 1) - 2 ] = 0

The root $t = 0$ corresponds to $N$. The intersection point on $S^n \setminus \{N\}$ corresponds to:

t = \frac{2}{\|y\|^2 + 1}

Substituting this $t$ into $L(t)$:

\sigma^{-1}(y) = \left( \frac{2 y_1}{\|y\|^2 + 1}, \dots, \frac{2 y_n}{\|y\|^2 + 1}, \frac{\|y\|^2 - 1}{\|y\|^2 + 1} \right)

2. Continuity of $\sigma$ and $\sigma^{-1}$

Continuity of $\sigma$:

Each component function $\sigma_i(x) = \frac{x_i}{1 - x_{n+1}}$ is a rational function of the coordinates $(x_1, \dots, x_{n+1})$. On the domain $S^n \setminus \{N\}$, $x_{n+1} < 1$, so the denominator $1 - x_{n+1} > 0$ never vanishes! Since rational functions with non-zero denominators are continuous, each component $\sigma_i$ is continuous. Hence $\sigma$ is continuous.

Continuity of $\sigma^{-1}$:

Each component function of $\sigma^{-1}(y)$ has denominator $\|y\|^2 + 1 \ge 1 > 0$, which is non-zero everywhere on $\mathbb{R}^n$. Hence all component functions are rational functions with non-vanishing denominators, proving $\sigma^{-1}$ is continuous on $\mathbb{R}^n$.

3. Conclusion

Since $\sigma \circ \sigma^{-1} = \operatorname{id}_{\mathbb{R}^n}$ and $\sigma^{-1} \circ \sigma = \operatorname{id}_{S^n \setminus \{N\}}$, $\sigma$ is a continuous bijection with a continuous inverse. Therefore, $\sigma: S^n \setminus \{N\} \to \mathbb{R}^n$ is a homeomorphism:

S^n \setminus \{N\} \cong \mathbb{R}^n \quad \blacksquare

Honors / Proof Challenge Example 3.3: Universal Property & Quotient Homeomorphism $I^2 / \sim \; \cong S^1 imes S^1$

Let $I^2 = [0, 1] \times [0, 1]$ equipped with the standard topology. Let $\sim$ be the equivalence relation on $I^2$ identifying opposite sides:

(0, y) \sim (1, y) \quad \text{and} \quad (x, 0) \sim (x, 1), \qquad \forall x, y \in [0, 1]

Let $q: I^2 \to I^2 / \sim$ be the canonical quotient projection. Consider the map $F: I^2 \to S^1 \times S^1 \subset \mathbb{C}^2$ defined by:

F(x, y) = \left( e^{2\pi i x}, e^{2\pi i y} \right)

Prove that $F$ is continuous and constant on the equivalence classes of $\sim$.
By the Universal Mapping Property of quotient spaces, show there exists a unique continuous bijection $\bar{F}: I^2 / \sim \; \to S^1 \times S^1$.
Prove that $\bar{F}$ is a homeomorphism, using the fact that $I^2 / \sim$ is compact and $S^1 \times S^1$ is Hausdorff.

1. Continuity and Invariance of $F$

The map $F: I^2 \to S^1 \times S^1$ has coordinate functions:

F_1(x, y) = e^{2\pi i x} = \cos(2\pi x) + i \sin(2\pi x)

F_2(x, y) = e^{2\pi i y} = \cos(2\pi y) + i \sin(2\pi y)

Both component functions are standard trigonometric continuous maps, so $F$ is continuous.

Now verify invariance on equivalence classes:

For $(0, y)$ and $(1, y)$:

F(0, y) = (e^0, e^{2\pi i y}) = (1, e^{2\pi i y})

F(1, y) = (e^{2\pi i}, e^{2\pi i y}) = (1, e^{2\pi i y})

Thus $F(0, y) = F(1, y)$.

For $(x, 0)$ and $(x, 1)$:

F(x, 0) = (e^{2\pi i x}, 1) = F(x, 1)

For the four corner points $(0,0), (1,0), (0,1), (1,1)$, all evaluate to $(1, 1)$.

Thus, $(x_1, y_1) \sim (x_2, y_2) \implies F(x_1, y_1) = F(x_2, y_2)$. $F$ is constant on all equivalence classes!

2. Induced Continuous Map $\bar{F}$

By the Universal Property of Quotient Topologies (Theorem 3.4): Since $q: I^2 \to I^2 / \sim$ is a quotient map and $F: I^2 \to S^1 \times S^1$ is continuous with $F$ constant on fibres of $q$, there exists a unique continuous map:

\bar{F}: I^2 / \sim \; \to S^1 \times S^1

satisfying $\bar{F} \circ q = F$, defined by $\bar{F}([x, y]) = F(x, y)$.

Injectivity of $\bar{F}$: Suppose $\bar{F}([x_1, y_1]) = \bar{F}([x_2, y_2])$. Then $e^{2\pi i x_1} = e^{2\pi i x_2} \implies x_1 - x_2 \in \mathbb{Z}$. Since $x_1, x_2 \in [0, 1]$, this implies either $x_1 = x_2$, or $\{x_1, x_2\} = \{0, 1\}$. Similarly, $e^{2\pi i y_1} = e^{2\pi i y_2} \implies y_1 = y_2$ or $\{y_1, y_2\} = \{0, 1\}$. In all cases, $(x_1, y_1) \sim (x_2, y_2)$, so $[x_1, y_1] = [x_2, y_2]$. Thus $\bar{F}$ is injective.

Surjectivity of $\bar{F}$: For any $(e^{i\theta}, e^{i\phi}) \in S^1 \times S^1$, choosing $x = \frac{\theta \pmod{2\pi}}{2\pi} \in [0, 1)$ and $y = \frac{\phi \pmod{2\pi}}{2\pi} \in [0, 1)$, we have $F(x, y) = (e^{i\theta}, e^{i\phi})$. Thus $\bar{F}$ is surjective.

Hence $\bar{F}$ is a continuous bijection.

3. Proof that $\bar{F}$ is a Homeomorphism

The unit square $I^2 = [0, 1] \times [0, 1]$ is closed and bounded in $\mathbb{R}^2$, hence compact by the Heine-Borel theorem.
The canonical quotient map $q: I^2 \to I^2 / \sim$ is surjective and continuous.

The continuous image of a compact space is compact; therefore, the quotient space $I^2 / \sim$ is compact.

The torus $S^1 \times S^1 \subset \mathbb{C}^2 \cong \mathbb{R}^4$ is a subspace of Euclidean space, hence it is Hausdorff ($T_2$).

Now we apply the classical topological theorem:

Theorem: Any continuous bijection from a compact space to a Hausdorff space is a homeomorphism!

Since $\bar{F}: I^2 / \sim \; \to S^1 \times S^1$ is a continuous bijection from a compact space to a Hausdorff space, its inverse $\bar{F}^{-1}$ is automatically continuous. Therefore, $\bar{F}$ is a homeomorphism:

I^2 / \sim \; \cong S^1 \times S^1 = T^2 \quad \blacksquare