Continuous Maps Between Topological Spaces

Continuity is the central concept linking topological spaces. In point-set topology, continuity is defined purely in terms of open sets, generalizing the $\epsilon$ - $\delta$ definition from analysis. This abstraction reveals that continuity is fundamentally about the preservation of topological structure.

Definition of Continuity

Definition2.1Continuous Map

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

This is the most general and most useful definition. It makes no reference to metrics or distances.

ExampleBasic Examples of Continuous Maps

Constant maps: If $f: X \to Y$ is defined by $f(x) = y_0$ for all $x$ , then $f$ is continuous. For any open $V \subseteq Y$ , we have $f^{-1}(V) = X$ if $y_0 \in V$ and $f^{-1}(V) = \emptyset$ if $y_0 \notin V$ .
Identity map: The identity $\operatorname{id}: (X, \tau) \to (X, \tau)$ is continuous. More generally, $\operatorname{id}: (X, \tau_1) \to (X, \tau_2)$ is continuous if and only if $\tau_1 \supseteq \tau_2$ (i.e., $\tau_1$ is finer than $\tau_2$ ).
Inclusion map: If $A \subseteq X$ carries the subspace topology, then the inclusion $\iota: A \hookrightarrow X$ is continuous, since $\iota^{-1}(U) = U \cap A$ is open in $A$ for every open $U$ in $X$ .
Projection maps: The projections $\pi_1: X \times Y \to X$ and $\pi_2: X \times Y \to Y$ (with the product topology) are continuous.

Equivalent Formulations

Theorem2.1Equivalent Conditions for Continuity

Let $f: X \to Y$ be a function between topological spaces. The following are equivalent:

$f$ is continuous (preimages of open sets are open).
For every closed set $C \subseteq Y$ , $f^{-1}(C)$ is closed in $X$ .
For every $A \subseteq X$ , $f(\overline{A}) \subseteq \overline{f(A)}$ .
For every $B \subseteq Y$ , $\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})$ .
For every $x \in X$ and every neighborhood $V$ of $f(x)$ in $Y$ , $f^{-1}(V)$ is a neighborhood of $x$ in $X$ .

Proof

(1 $\Leftrightarrow$ 2): $f^{-1}(Y \setminus C) = X \setminus f^{-1}(C)$ , so $f^{-1}$ of complements commutes. Thus preimages of open sets are open if and only if preimages of closed sets are closed.

(1 $\Rightarrow$ 3): Let $y \in f(\overline{A})$ , say $y = f(x)$ with $x \in \overline{A}$ . Let $V$ be any open set containing $y$ . Then $f^{-1}(V)$ is open and contains $x$ , so $f^{-1}(V) \cap A \neq \emptyset$ . Hence $V \cap f(A) \neq \emptyset$ , giving $y \in \overline{f(A)}$ .

(3 $\Rightarrow$ 2): Let $C$ be closed in $Y$ . Setting $A = f^{-1}(C)$ , condition (3) gives $f(\overline{A}) \subseteq \overline{f(A)} \subseteq \overline{C} = C$ . Thus $\overline{A} \subseteq f^{-1}(C) = A$ , so $A = \overline{A}$ is closed.

(1 $\Rightarrow$ 4): We have $f^{-1}(\overline{B})$ is closed (by (2)) and contains $f^{-1}(B)$ , so $\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})$ .

(4 $\Rightarrow$ 2): If $C$ is closed in $Y$ , set $B = C$ . Then $\overline{f^{-1}(C)} \subseteq f^{-1}(\overline{C}) = f^{-1}(C)$ , so $f^{-1}(C)$ is closed.

(1 $\Leftrightarrow$ 5): Direct from definitions.

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Continuity via Bases and Subbases

Theorem2.2Basis Criterion for Continuity

Let $f: X \to Y$ be a function and let $\mathcal{B}$ be a basis (resp. $\mathcal{S}$ a subbasis) for the topology on $Y$ . Then $f$ is continuous if and only if $f^{-1}(B) \in \tau_X$ for every $B \in \mathcal{B}$ (resp. $f^{-1}(S) \in \tau_X$ for every $S \in \mathcal{S}$ ).

Proof

The forward direction is immediate. For the converse (basis case): any open set $V$ in $Y$ is a union $V = \bigcup_\alpha B_\alpha$ of basis elements. Then $f^{-1}(V) = \bigcup_\alpha f^{-1}(B_\alpha)$ , which is a union of open sets, hence open.

For the subbasis case: any basis element is a finite intersection of subbasis elements, and $f^{-1}$ preserves finite intersections. So preimages of basis elements are open, reducing to the basis case.

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RemarkPractical Importance

The subbasis criterion is especially useful for the product topology: a map $f: X \to \prod_\alpha Y_\alpha$ is continuous if and only if $\pi_\alpha \circ f$ is continuous for each $\alpha$ . This is because the subbasis for the product topology consists of sets $\pi_\alpha^{-1}(U_\alpha)$ , and $f^{-1}(\pi_\alpha^{-1}(U_\alpha)) = (\pi_\alpha \circ f)^{-1}(U_\alpha)$ .

Continuity at a Point

Definition2.2Continuity at a Point

A function $f: X \to Y$ is continuous at the point $x_0 \in X$ if for every open set $V \subseteq Y$ containing $f(x_0)$ , there exists an open set $U \subseteq X$ containing $x_0$ such that $f(U) \subseteq V$ . Equivalently, every neighborhood of $f(x_0)$ has a preimage that is a neighborhood of $x_0$ .

A function is continuous (globally) if and only if it is continuous at every point of $X$ .

ExamplePointwise vs Global Continuity

Consider $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ . This is continuous at every point. At $x_0$ , given $\epsilon > 0$ , the preimage $f^{-1}((x_0^2 - \epsilon, x_0^2 + \epsilon))$ contains an open interval around $x_0$ .

In contrast, the characteristic function $\chi_{\mathbb{Q}}: \mathbb{R} \to \mathbb{R}$ (which takes value 1 on rationals, 0 on irrationals) is continuous at no point of $\mathbb{R}$ with the standard topology: every open interval contains both rationals and irrationals.