Continuity and Homeomorphisms

Definition. Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be topological spaces. A function $f : X \to Y$ is continuous if the preimage of every open set is open: for all $U \in \tau_Y$, $f^{-1}(U) \in \tau_X$.

Equivalently, $f$ is continuous if the preimage of every closed set is closed.

Proposition 4.1. The following are equivalent for $f : X \to Y$:

1. $f$ is continuous.
2. $f^{-1}(V)$ is closed in $X$ for every closed $V \subseteq Y$.
3. $\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})$ for every $B \subseteq Y$.
4. $f(\overline{A}) \subseteq \overline{f(A)}$ for every $A \subseteq X$.

Proof. (1) $\Leftrightarrow$ (2): for any $B \subseteq Y$, $f^{-1}(Y \setminus B) = X \setminus f^{-1}(B)$, so preimages of open sets being open is exactly preimages of closed sets being closed.

(2) $\Rightarrow$ (3): the closure of $B$ is the intersection of all closed supersets of $B$, and preimages preserve intersections, so

By (2) each $f^{-1}(C)$ is closed in $X$, and each contains $f^{-1}(B)$. The closure $\overline{f^{-1}(B)}$ is the smallest closed set containing $f^{-1}(B)$, so it is contained in every member of the family on the right; in particular in their intersection, $f^{-1}(\overline{B})$.

(3) $\Rightarrow$ (2): let $V$ be closed in $Y$ and set $B = V$. Then $\overline{f^{-1}(V)} \subseteq f^{-1}(\overline{V}) = f^{-1}(V) \subseteq \overline{f^{-1}(V)}$, where the last inclusion is extensivity of closure. So $f^{-1}(V)$ equals its closure and is closed.

(1) $\Rightarrow$ (4): let $x \in \overline{A}$ and let $V$ be any open neighbourhood of $f(x)$. By (1), $f^{-1}(V)$ is an open neighbourhood of $x$, and $x \in \overline{A}$, so $f^{-1}(V) \cap A \neq \emptyset$. Pick $a$ in the intersection: then $f(a) \in V \cap f(A)$, so every open neighbourhood of $f(x)$ meets $f(A)$, i.e. $f(x) \in \overline{f(A)}$.

(4) $\Rightarrow$ (1): let $U$ be open in $Y$; we show $f^{-1}(U)$ is open. Set $A = X \setminus f^{-1}(U)$. We claim $f^{-1}(U) \subseteq X \setminus \overline{A}$: if $x \in f^{-1}(U)$ were in $\overline{A}$, then by (4) $f(x) \in \overline{f(A)}$, so the open neighbourhood $U$ of $f(x)$ meets $f(A)$; say $f(a) \in U$ for some $a \in A$. But then $a \in f^{-1}(U) \cap A = \emptyset$, a contradiction. Hence $f^{-1}(U) \subseteq X \setminus \overline{A}$. Conversely $X \setminus \overline{A} \subseteq X \setminus A = f^{-1}(U)$. So $f^{-1}(U) = X \setminus \overline{A}$, which is open. $\blacksquare$

Direction (3) is the one to be careful with: the inclusion runs from the closure of the preimage into the preimage of the closure, and it can be strict. For the constant map $f : \mathbb{R} \to \mathbb{R}$, $f \equiv 0$, with $B = (0, 1)$: $\overline{f^{-1}(B)} = \overline{\emptyset} = \emptyset$ while $f^{-1}(\overline{B}) = f^{-1}([0,1]) = \mathbb{R}$.

Example 4.1. Every constant function $f : X \to Y$ is continuous.

Example 4.2. The identity map $\operatorname{id} : X \to X$ is always continuous.

Example 4.3. If $\tau_1$ and $\tau_2$ are topologies on $X$ with $\tau_1 \subseteq \tau_2$, then $\operatorname{id} : (X, \tau_2) \to (X, \tau_1)$ is continuous, but $\operatorname{id} : (X, \tau_1) \to (X, \tau_2)$ need not be.

Proposition 4.2. The composition of continuous functions is continuous: if $f : X \to Y$ and $g : Y \to Z$ are continuous, then $g \circ f : X \to Z$ is continuous.

Proof. For $W$ open in $Z$, $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$. The set $g^{-1}(W)$ is open in $Y$ by continuity of $g$; its preimage under $f$ is open in $X$ by continuity of $f$. $\blacksquare$

Continuity at a point. A function $f : X \to Y$ is continuous at $x_0 \in X$ if for every neighbourhood $V$ of $f(x_0)$, there exists a neighbourhood $U$ of $x_0$ such that $f(U) \subseteq V$. This local definition coincides with the global definition: $f$ is continuous iff it is continuous at every point.

Proposition 4.2.1. $f$ is continuous if and only if it is continuous at every point of $X$.

Proof. ($\Rightarrow$) Let $V$ be a neighbourhood of $f(x_0)$. It contains an open set $V^\circ \ni f(x_0)$. Then $U = f^{-1}(V^\circ)$ is an open neighbourhood of $x_0$ with $f(U) \subseteq V$.

($\Leftarrow$) Let $W$ be open in $Y$; show $f^{-1}(W)$ is open. If $x \in f^{-1}(W)$, then $W$ is a neighbourhood of $f(x)$, so continuity at $x$ gives a neighbourhood $U_x$ of $x$ with $f(U_x) \subseteq W$. Shrinking $U_x$ to its open interior, $x$ is an interior point of $f^{-1}(W)$. Every point of $f^{-1}(W)$ is interior, so $f^{-1}(W)$ is open. $\blacksquare$

Proposition 4.3 (Pasting Lemma). If $X = A \cup B$ where $A$ and $B$ are closed (or open) in $X$, and $f : A \to Y$, $g : B \to Y$ are continuous functions that agree on $A \cap B$, then the function $h : X \to Y$ defined by $h|_A = f$ and $h|_B = g$ is continuous.

Proof. Closed case. Let $V$ be closed in $Y$. Then $h^{-1}(V) = (f^{-1}(V) \cup g^{-1}(V))$: indeed $x \in h^{-1}(V)$ means $h(x) \in V$, and $h(x)$ is either $f(x)$ or $g(x)$. Now $f^{-1}(V)$ is closed in $A$ (continuity of $f$) and $A$ is closed in $X$, so $f^{-1}(V)$ is closed in $X$; likewise $g^{-1}(V)$. A finite union of closed sets is closed, so $h^{-1}(V)$ is closed, and $h$ is continuous by Proposition 4.1(2).

Open case. Dually, let $V$ be open in $Y$. Then $h^{-1}(V) = f^{-1}(V) \cup g^{-1}(V)$ as above, each of $f^{-1}(V)$, $g^{-1}(V)$ is open in the open (in $X$) subspace $A$, resp. $B$, hence open in $X$, and their union is open. $\blacksquare$

Why the hypotheses matter: if one piece is closed and the other open, the argument breaks down in both cases at once; a set closed in a closed subspace is closed in $X$, but a set open in a closed subspace need not be. The classic counterexample: $f(x) = 0$ on $(-\infty, 0]$ and $f(x) = 1$ on $(0, \infty)$ is continuous on each piece and the pieces agree on their (empty) intersection, but one piece is closed while the other is merely open; and $f$ is discontinuous at $0$.

Example 4.4 (Piecewise functions). The absolute value function $f(x) = |x|$ is continuous on $\mathbb{R}$ by the pasting lemma: on $(-\infty, 0]$ it equals $-x$ (continuous), on $[0, \infty)$ it equals $x$ (continuous), and the two agree at $x = 0$.

Example 4.5. The function $f(x) = \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ is not continuous at $0$, even though it satisfies the intermediate value property on every neighbourhood.

Definition. A function $f : X \to Y$ is open if the image of every open set is open; it is closed if the image of every closed set is closed. Continuity does not imply openness: the constant function $f(x) = c$ is continuous but not open (the image $\{c\}$ is not open in $\mathbb{R}$). The projection $\pi : X \times Y \to X$ is open but not necessarily closed.

Proposition 4.4. A bijective continuous function is a homeomorphism iff it is open (equivalently, closed).

Proof. Let $f : X \to Y$ be a continuous bijection. If $f$ is a homeomorphism, then $f^{-1}$ is continuous, so for open $U \subseteq X$ the set $(f^{-1})^{-1}(U) = f(U)$ is open in $Y$: $f$ is open. Conversely, if $f$ is open, then for open $U \subseteq X$, $(f^{-1})^{-1}(U) = f(U)$ is open in $Y$, which is exactly continuity of $f^{-1}$; so $f$ is a homeomorphism.

For the equivalence with closed: given that $f$ is a bijection, $f$ maps complements to complements, $f(X \setminus U) = Y \setminus f(U)$, so $f$ is open iff $f$ is closed. $\blacksquare$

Definition. A function $f : X \to Y$ is a homeomorphism if $f$ is bijective and both $f$ and $f^{-1}$ are continuous. We write $X \cong Y$ and say $X$ and $Y$ are homeomorphic.

A topological property (or topological invariant) is a property preserved by homeomorphisms.

Example 4.6. $(0, 1)$ is homeomorphic to $\mathbb{R}$ via $f(x) = \tan\left(\pi x - \frac{\pi}{2}\right)$.

Example 4.7. Any two open intervals $(a, b)$ and $(c, d)$ in $\mathbb{R}$ are homeomorphic via an affine map.

Example 4.8 (Coffee cup and donut). A coffee cup (with a handle) is homeomorphic to a torus $S^1 \times S^1$ via a continuous deformation. This illustrates that homeomorphisms permit stretching and bending but not tearing or gluing.

Proposition 4.5. Homeomorphism is an equivalence relation: reflexive, symmetric, and transitive.

Proof. Reflexive: $\operatorname{id}_X$ is a bijection and its own continuous inverse. Symmetric: if $f$ is a homeomorphism then $f^{-1}$ is a bijection with $(f^{-1})^{-1} = f$ continuous, so $f^{-1}$ is a homeomorphism. Transitive: if $f : X \to Y$ and $g : Y \to Z$ are homeomorphisms, $g \circ f$ is a bijection with inverse $f^{-1} \circ g^{-1}$, continuous by Proposition 4.2 applied to the continuous maps $g^{-1}$ and $f^{-1}$. $\blacksquare$

The following are topological invariants (preserved by homeomorphisms):

• Compactness
• Connectedness
• Separation axioms ($T_0$, $T_1$, $T_2$, etc.)
• Countability axioms (first-countable, second-countable)
• The fundamental group $\pi_1(X)$

Proposition 4.6. “Boundedness” is not a topological property: $(0, 1)$ is bounded but is homeomorphic to the unbounded $\mathbb{R}$.

Proof. The map $f : (0, 1) \to \mathbb{R}$ of Example 4.6, $f(x) = \tan(\pi x - \pi/2)$, is a bijection with inverse $f^{-1}(t) = \frac{1}{2} + \frac{1}{\pi}\arctan(t)$; both are continuous as compositions of continuous functions ($\tan$ and $\arctan$ are continuous on their domains, which the maps avoid leaving). So $(0, 1) \cong \mathbb{R}$. If boundedness were a topological property, $\mathbb{R}$ would be bounded; but $\mathbb{R}$ is not bounded in its standard metric. Hence boundedness is a property of the metric, not of the topology it generates. $\blacksquare$

Definition. An embedding is a homeomorphism onto its image: $f : X \to Y$ such that $f : X \to f(X)$ (with the subspace topology) is a homeomorphism.

Definition. A surjective continuous map $q : X \to Y$ is a quotient map if $U \subseteq Y$ is open iff $q^{-1}(U)$ is open in $X$. Quotient maps are used to construct spaces by gluing: if $\sim$ is an equivalence relation on $X$, then $X/{\sim}$ with the quotient topology is the space of equivalence classes with the finest topology making the projection continuous.

Example 4.9. The unit interval $[0, 1]$ with endpoints identified yields the circle $S^1$: $[0, 1] / \{0 \sim 1\} \cong S^1$. The unit square with opposite edges identified yields the torus.

Problem. Show that $(0, 1)$ and $[0, 1]$ are not homeomorphic.

A homeomorphism is the topological version of “the same shape.” Two spaces are homeomorphic if there exists a continuous bijection between them whose inverse is also continuous. Informally, this means you can stretch, bend, and deform one space into the other, but you cannot tear or glue. A coffee cup and a donut are homeomorphic because they both have exactly one hole; a sphere and a torus are not homeomorphic because they have different numbers of holes.

The key insight is that homeomorphisms preserve topological properties such as compactness, connectedness, and the fundamental group. Properties that are not preserved by homeomorphisms, like “boundedness” or “being a subset of $\mathbb{R}^3$,” are geometric rather than topological. The pasting lemma captures a practical criterion for constructing continuous functions: if you define a function separately on two closed (or open) pieces that agree on the overlap, the result is continuous. This is why absolute value, piecewise linear functions, and most “glued together” constructions in analysis turn out to be continuous.

Problem. Determine whether $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x\sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ is continuous at $x = 0$.

Mistake 1: Assuming that a continuous bijection is automatically a homeomorphism. A continuous bijection is not necessarily a homeomorphism. The inverse function must also be continuous. For example, the map $f: [0, 2\pi) \to S^1$ defined by $f(t) = (\cos t, \sin t)$ is a continuous bijection, but its inverse is not continuous at $(1, 0)$. A bijective continuous map is a homeomorphism if and only if it is open (or equivalently, closed).

Mistake 2: Confusing open maps with continuous maps. A function can be continuous without being open, and vice versa. The constant function $f(x) = c$ is continuous but not open (the image of any open set is $\{c\}$, which is not open). The projection $\pi: \mathbb{R}^2 \to \mathbb{R}$ is open but not closed (the image of the closed hyperbola $\{(x, y) : xy = 1\}$ is $\mathbb{R} \setminus \{0\}$, which is not closed).

Mistake 3: Forgetting that homeomorphism is an equivalence relation. Homeomorphism is reflexive (the identity map is a homeomorphism), symmetric (the inverse of a homeomorphism is a homeomorphism), and transitive (the composition of homeomorphisms is a homeomorphism). Failing to check any of these can lead to errors when classifying spaces up to homeomorphism.

Mistake 4: Assuming that topological properties are preserved by continuous maps. Only homeomorphisms preserve all topological properties. A continuous map need not preserve compactness, connectedness, or separation axioms in both directions. For example, a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, but this is a special case, not a general rule.

Mistake 5: Misapplying the pasting lemma. The pasting lemma requires that the two pieces agree on their intersection and that both pieces are either both closed or both open. If one piece is closed and the other is open, or if they disagree on the intersection, the resulting function may not be continuous.

• Topological Spaces: Topological spaces provide the foundation for defining continuity and homeomorphisms.
• Compactness: Compactness is preserved by continuous functions and characterizes spaces where extreme value theorem holds.
• Connectedness: Connectedness is a topological invariant preserved by homeomorphisms and continuous images.

• Closed Sets, Closure, Interior, and Boundary
• Common Pitfalls
• Compactness
• Connectedness