In [metric spaces](/page/Continuity%20(Metric%20Spaces)), continuity is defined by the $\varepsilon$-$\delta$ condition: small perturbations of the input produce small perturbations of the output. But many of the spaces that arise in modern mathematics — weak topologies on Banach spaces, quotient spaces, spaces of distributions, the Zariski topology in algebraic geometry — carry no natural metric. The product space $\mathbb{R}^\mathbb{R}$ (all [functions](/page/Function) $\mathbb{R} \to \mathbb{R}$ with the topology of pointwise convergence) is not metrisable. Neither is the space $\mathcal{D}'(\Omega)$ of distributions with its standard topology. Yet continuity in these spaces is essential: [weak convergence](/page/Weak%20Convergence), [distributional](/page/Distribution) convergence, and algebraic continuity all require a definition that works without distance. The [topological](/page/Topology) definition eliminates metrics entirely. A function $f: X \to Y$ between topological spaces is continuous if and only if the preimage of every [open set](/page/Open%20Set) in $Y$ is open in $X$. This single condition — expressible in terms of the open-set structure alone — captures every metric-space notion of continuity as a special case, while extending to spaces where $\varepsilon$ and $\delta$ are meaningless. It also reveals that continuity is the *structure-preserving map* of topology: continuous functions are to topological spaces what homomorphisms are to [groups](/page/Group) and [linear maps](/page/Linear%20Map) are to vector spaces. [example:Continuity Without A Metric] Consider the space $\mathbb{R}^\mathbb{N}$ of all real [sequences](/page/Sequence), equipped with the product topology (the topology of pointwise convergence). This space is not normable: there is no single norm that generates the product topology. Yet we can verify continuity of maps into this space using only open sets. Define $T: C([0, 1]) \to \mathbb{R}^\mathbb{N}$ by $T(f) = (f(1/n))_{n=1}^\infty$ — the map that samples a continuous function at the points $1, 1/2, 1/3, \ldots$. To check continuity, we need only verify that preimages of subbasic open sets are open. A subbasic open set in $\mathbb{R}^\mathbb{N}$ has the form $\pi_k^{-1}(U) = \{(x_n) : x_k \in U\}$ for some $k \in \mathbb{N}$ and open $U \subseteq \mathbb{R}$. The preimage is: \begin{align*} T^{-1}(\pi_k^{-1}(U)) = \{f \in C([0,1]) : f(1/k) \in U\} = \operatorname{ev}_{1/k}^{-1}(U), \end{align*} which is open in the sup-norm topology on $C([0,1])$ because point evaluation $\operatorname{ev}_{1/k}$ is continuous. No $\varepsilon$-$\delta$ argument involving a metric on $\mathbb{R}^\mathbb{N}$ was needed — the topological definition handled a non-metrisable codomain directly. [/example] This page develops continuity at the level of general topological spaces. For the metric-space theory (the $\varepsilon$-$\delta$ definition, sequential characterisation, uniform and Lipschitz continuity) see [Continuity (Metric Spaces)](/page/Continuity%20(Metric%20Spaces)). For concrete examples on $\mathbb{R}$ see [Continuity (Real Analysis)](/page/Continuity%20(Real%20Analysis)). ## Definition [definition:Continuous 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 $f^{-1}(V) = \{x \in X : f(x) \in V\}$ is open in $X$: $f^{-1}(V) \in \tau_X$. [/definition] The definition asks nothing about the *image* of open sets — forward images of open sets under continuous functions need not be open. (The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ sends the open set $(-1, 1)$ to $[0, 1)$, which is not open.) It is the *preimage* that must respect the topology. There is an equivalent formulation in terms of closed sets, obtained by taking complements: $f$ is continuous if and only if $f^{-1}(F)$ is [closed](/page/Closed%20Set) in $X$ for every closed $F \subseteq Y$. This dual characterisation is often more convenient for real-valued functions, since many natural [sets](/page/Set) in $\mathbb{R}$ are closed: $\{0\}$, $[a, b]$, $(-\infty, c]$. [example:Closed Sets As Continuous Preimages] If $f: X \to \mathbb{R}$ is continuous, then: the zero set $Z(f) = f^{-1}(\{0\})$ is closed (preimage of the closed set $\{0\}$); the sublevel set $\{x : f(x) \leq c\}$ is closed (preimage of $(-\infty, c]$); and the level set $f^{-1}(\{c\})$ is closed for every $c$. These facts allow us to verify that specific sets are closed by expressing them as preimages: the unit sphere $S^{n-1} = \{x \in \mathbb{R}^n : \|x\| = 1\}$ is closed because $\|{\cdot}\|$ is continuous and $\{1\}$ is closed. [/example] ### Why Preimages, Not Images? The asymmetry in the definition — preimages of open sets must be open, but nothing is required of forward images — is a deliberate and essential feature, not an arbitrary convention. Forward images of open sets under continuous functions can fail badly: the squaring map $x \mapsto x^2$ sends the open interval $(-1, 1)$ to $[0, 1)$, which is not open. The projection $\pi_1: \mathbb{R}^2 \to \mathbb{R}$ sends the open set $\{(x, y) : xy = 1,\, x > 0\}$ to $(0, \infty)$, but the closed set $\{(x, y) : xy = 1\}$ to $\mathbb{R} \setminus \{0\}$, which is not closed. Forward images simply do not respect the topology in a predictable way. Preimages, by contrast, commute with all set-theoretic operations: $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$, $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$, and $f^{-1}(Y \setminus A) = X \setminus f^{-1}(A)$. This algebraic compatibility is what makes the preimage condition equivalent to the closed-set formulation, and it is why continuous functions automatically preserve convergence of nets, compactness, and connectedness. The few situations where forward images *are* well-behaved (open maps, closed maps, proper maps) are important enough to merit separate study later in this page. ### Continuity at a Point The global definition asks for *every* open preimage to be open — a condition on all of $X$ at once. But many arguments in analysis are local: we care about behaviour near a specific point $a$, and we want to express the idea that $f$ does not "tear" the space at $a$. The pointwise definition captures this by requiring only that neighbourhoods of $f(a)$ pull back to neighbourhoods of $a$, without imposing any condition at other points. [definition:Continuous At A Point Topological] A function $f: X \to Y$ is **continuous at** $a \in X$ if for every open set $V \subseteq Y$ containing $f(a)$, there exists an open set $U \subseteq X$ containing $a$ with $f(U) \subseteq V$. Equivalently, for every neighbourhood $V$ of $f(a)$, the preimage $f^{-1}(V)$ is a neighbourhood of $a$. [/definition] A function is continuous (globally) if and only if it is continuous at every point — this follows immediately from the definitions, since a set is open if and only if it is a neighbourhood of each of its points. ### The Closure Operator Characterisation A function $f: X \to Y$ is continuous if and only if $f(\overline{A}) \subseteq \overline{f(A)}$ for every $A \subseteq X$: continuous functions cannot "escape" the closure by mapping [limit](/page/Limit) points outside the closure of the image. The forward direction follows because $\overline{f(A)}$ is closed, so $f^{-1}(\overline{f(A)})$ is closed and contains $A$, hence contains $\overline{A}$. The reverse direction: if $F \subseteq Y$ is closed and $A = f^{-1}(F)$, then $f(\overline{A}) \subseteq \overline{f(A)} \subseteq \overline{F} = F$, so $\overline{A} \subseteq f^{-1}(F) = A$, proving $A$ is closed. ## Composition and the Category of Topological Spaces The most basic structural property of continuous functions is that they compose: continuity is preserved under sequential application. This makes topological spaces and continuous maps into a category, and it is the reason that continuity — rather than, say, openness of forward images — is the "correct" notion of morphism in topology. Continuous functions compose: if $f: X \to Y$ and $g: Y \to Z$ are continuous, then $g \circ f: X \to Z$ is continuous. The proof is one line: $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$, which is open by two applications of continuity. The identity map $\operatorname{id}_X: X \to X$ is always continuous. These two facts make topological spaces and continuous maps into a **category** — the category $\mathbf{Top}$. Continuous functions also interact well with topological constructions: **Restriction.** If $f: X \to Y$ is continuous and $A \subseteq X$ carries the subspace topology, then $f|_A: A \to Y$ is continuous. **Corestriction.** If $f: X \to Y$ is continuous and $f(X) \subseteq B \subseteq Y$, then $f: X \to B$ (with the subspace topology on $B$) is continuous. **Products.** A function $f: Z \to X \times Y$ (with the product topology) is continuous if and only if both component functions $\pi_X \circ f$ and $\pi_Y \circ f$ are continuous. More generally, $f: Z \to \prod_{\alpha} X_\alpha$ is continuous if and only if $\pi_\alpha \circ f$ is continuous for every $\alpha$. ### The Pasting Lemma A common technique for constructing continuous functions is to define them piecewise on overlapping or adjacent closed (or open) subsets and "glue" the pieces together. The Pasting Lemma guarantees that the result is continuous, provided the pieces agree on overlaps. If $X = A \cup B$ where $A$ and $B$ are both closed (or both open), and $f: A \to Y$ and $g: B \to Y$ are continuous with $f|_{A \cap B} = g|_{A \cap B}$, then the function $h: X \to Y$ defined by $h(x) = f(x)$ for $x \in A$ and $h(x) = g(x)$ for $x \in B$ is continuous. For the closed case: if $C \subseteq Y$ is closed, then $h^{-1}(C) = f^{-1}(C) \cup g^{-1}(C)$, which is a union of two sets that are closed in $A$ and $B$ respectively; since $A$ and $B$ are closed in $X$, these sets are closed in $X$, so $h^{-1}(C)$ is closed. The result extends to finite unions of closed sets but fails for infinite unions. For open covers, the result extends to arbitrary unions — reflecting the fact that arbitrary unions of open sets are open. [example:Constructing A Loop By Pasting] Define $\gamma: [0, 1] \to \mathbb{R}^2$ by: \begin{align*} \gamma(t) = \begin{cases} (2t, 0) & \text{if } 0 \leq t \leq 1/2, \\ (\cos(\pi(2t-1)), \sin(\pi(2t-1))) & \text{if } 1/2 \leq t \leq 1. \end{cases} \end{align*} The two pieces are continuous on the closed sets $[0, 1/2]$ and $[1/2, 1]$, and they agree at $t = 1/2$ (both give $(1, 0)$). By the Pasting Lemma, $\gamma$ is continuous on $[0, 1]$. This construction — building paths and loops by concatenating pieces — is fundamental in algebraic topology. [/example] ## Preservation Theorems The deepest consequences of continuity come from its interaction with topological properties of the domain. Two properties are preserved by continuous images: compactness and connectedness. These preservation results are the abstract engines behind the Extreme Value Theorem and the Intermediate Value Theorem. ### Continuous Images of Compact Sets [quotetheorem:305] The proof is a single manipulation of open covers: if $\{V_i\}$ is an open cover of $f(K)$, then $\{f^{-1}(V_i)\}$ is an open cover of $K$ (by continuity); extract a finite subcover $\{f^{-1}(V_{i_1}), \ldots, f^{-1}(V_{i_m})\}$ (by compactness of $K$); then $\{V_{i_1}, \ldots, V_{i_m}\}$ is a finite subcover of $f(K)$. This result is the abstract engine behind many concrete facts. Continuous real-valued functions on compact spaces are bounded and attain their bounds: [quotetheorem:304] The Extreme Value Theorem follows from the chain: $f(X)$ is compact in $\mathbb{R}$ (by preservation of compactness), hence closed and bounded (by [Heine-Borel](/theorems/271)); a closed bounded subset of $\mathbb{R}$ contains its [supremum and infimum](/page/Supremum%20and%20Infimum), so $f$ attains its maximum and minimum. The key insight is that *none* of the individual steps require a metric on $X$ — compactness and continuity are purely topological, and the metric structure of $\mathbb{R}$ enters only at the very end, through Heine-Borel. This is why the Extreme Value Theorem generalises effortlessly from $\mathbb{R}^n$ to any compact topological space: the topology does all the work. The theorem fails without compactness ($f(x) = 1/x$ on $(0,1)$ is continuous but unbounded) and without continuity ($\mathbb{1}_{(0,1)}$ on $[0,1]$ is bounded but does not attain its supremum on any compact set containing the [boundary](/page/Boundary)). ### Continuous Images of Connected Sets [quotetheorem:296] The proof is by contradiction: if $f(X) = U \cup V$ were a disconnection (with $U, V$ nonempty, open in $f(X)$, and disjoint), then $f^{-1}(U)$ and $f^{-1}(V)$ would be nonempty, open in $X$, disjoint, and covering $X$ — disconnecting $X$, contradicting connectedness. The most important consequence is the Intermediate Value Theorem: if $f: X \to \mathbb{R}$ is continuous and $X$ is connected, then $f(X)$ is connected in $\mathbb{R}$, hence an interval. Every value between $\inf f$ and $\sup f$ is attained. [quotetheorem:629] [example:Fixed Points From Connectedness] Every continuous $f: [0, 1] \to [0, 1]$ has a fixed point. The function $g(x) = f(x) - x$ is continuous with $g(0) = f(0) \geq 0$ and $g(1) = f(1) - 1 \leq 0$. By the [Intermediate Value Theorem](/theorems/629), $g(c) = 0$ for some $c$, giving $f(c) = c$. This is the one-dimensional Brouwer fixed-point theorem — in higher dimensions, the proof requires algebraic topology. [/example] ## The Closed Map Lemma and Homeomorphisms A continuous bijection need not have a continuous inverse: the map $f: [0, 2\pi) \to S^1$ defined by $f(t) = (\cos t, \sin t)$ is a continuous bijection, but $f^{-1}$ is discontinuous at $(1, 0)$ because $[0, 2\pi)$ is not compact. The following result gives the decisive sufficient condition. [quotetheorem:317] The proof uses the interplay between compactness and the Hausdorff property. If $C \subseteq X$ is closed, then $C$ is compact (closed subset of a compact space). By the [Continuous Image of Compact Sets](/theorems/305), $f(C)$ is compact in $Y$. Since $Y$ is Hausdorff, compact subsets are closed, so $f(C)$ is closed. [quotetheorem:318] This follows immediately from the [Closed Map Lemma](/theorems/317): a continuous closed bijection has a continuous inverse (the inverse sends closed sets to closed sets, hence open sets to open sets). The theorem explains why the map $[0, 2\pi) \to S^1$ fails to be a homeomorphism: $[0, 2\pi)$ is not compact. If we replace it with $[0, 2\pi] / \{0 \sim 2\pi\}$ (the quotient identifying the endpoints), the resulting space is compact, and the induced map to $S^1$ is indeed a homeomorphism. ### Homeomorphisms and Embeddings The Closed Map Lemma and its corollary raise a fundamental question: when are two topological spaces "the same"? Group theory has isomorphisms, linear algebra has invertible linear maps, and topology needs an analogue — a bijection that preserves the entire open-set structure in both directions. A continuous bijection is not enough (as $[0, 2\pi) \to S^1$ shows), because the inverse may fail to be continuous. The correct notion requires continuity in *both* directions. [definition:Homeomorphism] A **homeomorphism** is a continuous bijection $f: X \to Y$ whose inverse $f^{-1}: Y \to X$ is also continuous. Equivalently, $f$ is a bijection such that $U \subseteq X$ is open if and only if $f(U) \subseteq Y$ is open. If a homeomorphism exists, $X$ and $Y$ are **homeomorphic**, written $X \cong Y$. [/definition] Homeomorphic spaces are topologically identical — every topological property (compactness, connectedness, Hausdorffness, number of connected components, fundamental group, ...) is the same for both. Homeomorphisms are the isomorphisms of the category $\mathbf{Top}$. Often we do not need a bijection between two entire spaces, but rather a way to realise one space as a "copy" sitting inside another. A subspace $A \subseteq Y$ with the subspace topology is an obvious example, but more generally we want to recognise when a continuous injection $f: X \to Y$ faithfully represents $X$ as a subspace of $Y$ — that is, when $f$ is not just injective and continuous but also identifies $X$ homeomorphically with $f(X)$. [definition:Topological Embedding] A **topological embedding** is a continuous injection $f: X \to Y$ that is a homeomorphism onto its image $f(X)$ (with the subspace topology). Not every continuous injection is an embedding. [/definition] [example:Continuous Bijection That Is Not A Homeomorphism] The identity map $\operatorname{id}: (X, \tau_{\text{disc}}) \to (X, \tau_{\text{coarse}})$ from the discrete topology to the coarsest topology (only $\varnothing$ and $X$ are open) is a continuous bijection — every function from a discrete space is continuous. But the inverse $\operatorname{id}: (X, \tau_{\text{coarse}}) \to (X, \tau_{\text{disc}})$ is not continuous whenever $|X| \geq 2$: singletons are open in the discrete topology but not in the coarse topology. This is the simplest example showing that a continuous bijection need not be a homeomorphism — the two topologies on $X$ may have different open sets despite the identity being a bijection. [/example] ## Semicontinuity Many functions that arise naturally in analysis — suprema, infima, indicator functions of open sets, norms — fail to be continuous but satisfy a one-sided condition. Semicontinuity captures this asymmetry in purely topological terms. [definition:Lower Semicontinuous Function] A function $f: X \to \mathbb{R} \cup \{+\infty\}$ on a topological space $X$ is **lower semicontinuous** (l.s.c.) if for every $c \in \mathbb{R}$, the sublevel set $\{x \in X : f(x) \leq c\}$ is closed. Equivalently, $\{x : f(x) > c\}$ is open for every $c$, or $\liminf_{x \to a} f(x) \geq f(a)$ for every $a$. [/definition] [definition:Upper Semicontinuous Function] A function $f: X \to \mathbb{R} \cup \{-\infty\}$ is **upper semicontinuous** (u.s.c.) if $-f$ is lower semicontinuous. Equivalently, the superlevel set $\{x : f(x) \geq c\}$ is closed for every $c$. [/definition] A function is continuous if and only if it is both lower and upper semicontinuous. Semicontinuity appears in two key settings: **Variational problems.** The direct method of the [calculus of variations](/page/Calculus%20of%20Variations) seeks to minimise a functional $I(u)$ over admissible functions. The key hypothesis is that $I$ is l.s.c. with respect to the appropriate topology — typically the [weak topology](/page/Weak*%20Topology) on a [Sobolev space](/page/Sobolev%20Space). Lower semicontinuity ensures that $I(u) \leq \liminf I(u_n)$ along a minimising sequence $u_n$, so the infimum is attained at the weak limit. **Indicator functions.** The indicator $\mathbf{1}_U$ of an [open set](/page/Open%20Set) $U$ is lower semicontinuous: $\{\mathbf{1}_U \leq c\}$ equals $X$ for $c \geq 1$, $X \setminus U$ for $0 \leq c < 1$, and $\varnothing$ for $c < 0$ — all closed. Dually, $\mathbf{1}_F$ for a [closed set](/page/Closed%20Set) $F$ is upper semicontinuous. [example:The Supremum Of A Family Of Continuous Functions] If $\{f_\alpha\}_{\alpha \in A}$ is a family of continuous functions $f_\alpha: X \to \mathbb{R}$, then $g(x) = \sup_\alpha f_\alpha(x)$ is lower semicontinuous (when the supremum is finite or $+\infty$). The superlevel set $\{x : g(x) > c\} = \bigcup_\alpha \{x : f_\alpha(x) > c\}$ is a union of open sets, hence open. Similarly, $\inf_\alpha f_\alpha$ is upper semicontinuous. This explains a common phenomenon: the supremum of continuous functions need not be continuous (it may "jump up" at limit points), but it can only jump in one direction. [/example] ## Initial and Final Topologies Many topologies are defined by the requirement that certain maps be continuous. The initial and final topologies are the universal solutions to this problem. ### Initial Topologies In practice, we often *construct* topologies by specifying which maps should be continuous. The subspace topology on $A \subseteq Y$ is the coarsest topology making the inclusion $\iota: A \hookrightarrow Y$ continuous; the product topology on $\prod X_\alpha$ is the coarsest topology making every projection $\pi_\alpha$ continuous. These are both instances of a single construction: given a set $X$ and maps $f_\alpha: X \to Y_\alpha$ into known topological spaces, there is a unique coarsest topology on $X$ that makes all the $f_\alpha$ continuous. [definition:Initial Topology] Let $X$ be a set and let $\{f_\alpha: X \to Y_\alpha\}_{\alpha \in A}$ be a family of functions into topological spaces $Y_\alpha$. The **initial topology** on $X$ (with respect to $\{f_\alpha\}$) is the coarsest topology on $X$ making every $f_\alpha$ continuous. It is generated by the sets $f_\alpha^{-1}(V_\alpha)$ for $V_\alpha$ open in $Y_\alpha$. [/definition] The initial topology has a **universal property**: a function $g: Z \to X$ (from any topological space $Z$) is continuous if and only if $f_\alpha \circ g: Z \to Y_\alpha$ is continuous for every $\alpha$. The most important instances are: the **subspace topology** on $A \subseteq Y$ (the initial topology for the inclusion $\iota: A \hookrightarrow Y$) and the **product topology** on $\prod_\alpha X_\alpha$ (the initial topology for the projections $\pi_\alpha: \prod X_\alpha \to X_\alpha$). The weak topology on a [Banach space](/page/Banach%20Space) $X$ is the initial topology for the family of functionals $\{f \in X^*\}$; the weak-* topology on $X^*$ is the initial topology for the evaluation maps $\{\operatorname{ev}_x : x \in X\}$. ### Final Topologies The dual situation arises when we have maps *into* a set $Y$ from known topological spaces, and we want the finest topology on $Y$ making all these maps continuous. Where the initial topology asks "what is the least we must declare open to make incoming maps continuous?", the final topology asks "what is the most we can declare open while keeping outgoing maps continuous?" The quotient topology is the central example: we identify points in a space and want the resulting quotient to carry as much topological structure as possible. [definition:Final Topology] Let $Y$ be a set and let $\{g_\alpha: X_\alpha \to Y\}_{\alpha \in A}$ be a family of functions from topological spaces $X_\alpha$. The **final topology** on $Y$ is the finest topology making every $g_\alpha$ continuous. A set $V \subseteq Y$ is open if and only if $g_\alpha^{-1}(V)$ is open in $X_\alpha$ for every $\alpha$. [/definition] The final topology has the dual universal property: $h: Y \to Z$ is continuous if and only if $h \circ g_\alpha: X_\alpha \to Z$ is continuous for every $\alpha$. The most important instance is the **quotient topology** on $Y = X / {\sim}$ (the final topology for the quotient map $q: X \to X/{\sim}$). A set $V \subseteq X/{\sim}$ is open if and only if $q^{-1}(V)$ is open in $X$. The **coproduct (disjoint union) topology** is another instance: $V \subseteq \bigsqcup_\alpha X_\alpha$ is open if and only if $V \cap X_\alpha$ is open for every $\alpha$. [example:The Quotient Map $\mathbb{R} \to S^1$] The map $q: \mathbb{R} \to S^1$ defined by $q(t) = (\cos 2\pi t, \sin 2\pi t)$ is continuous and surjective. The [equivalence relation](/page/Equivalence%20Relation) $t \sim s \iff t - s \in \mathbb{Z}$ gives a bijection $\mathbb{R}/\mathbb{Z} \cong S^1$, and the quotient topology on $\mathbb{R}/\mathbb{Z}$ is exactly the topology inherited from $S^1$. A function $f: S^1 \to Z$ is continuous if and only if $f \circ q: \mathbb{R} \to Z$ is continuous and periodic — this is the universal property of the quotient in action. By contrast, restricting $q$ to $[0, 1)$ gives a continuous bijection $[0, 1) \to S^1$ that is *not* a homeomorphism ($[0, 1)$ is not compact, so the [Topological Inverse Function Theorem](/theorems/318) does not apply). The quotient construction $[0, 1]/\{0 \sim 1\}$, which *is* compact, gives the correct homeomorphism. [/example] ## Open and Closed Maps Continuity controls *preimages*: open sets pull back to open sets. But in many situations — quotient constructions, projections, covering maps — we need to understand the *forward* behaviour of a map. When does a map send open sets to open sets, or closed sets to closed sets? These are genuinely independent properties: a map can be continuous without being open or closed, open without being continuous, or any other combination. The maps that do preserve open or closed sets forward have special structural significance, particularly in the theory of quotient spaces and in constructing homeomorphisms. [definition:Open Map] A function $f: X \to Y$ is an **open map** if $f(U)$ is open in $Y$ for every open $U \subseteq X$. [/definition] [definition:Closed Map] A function $f: X \to Y$ is a **closed map** if $f(C)$ is closed in $Y$ for every closed $C \subseteq X$. [/definition] A bijection $f$ is a homeomorphism if and only if it is continuous and open, or equivalently, continuous and closed. Projections $\pi: X \times Y \to X$ are always open maps (a basic open set $U \times V$ in the product projects to $U$, which is open) but need not be closed (the hyperbola $\{(x, y) : xy = 1\}$ is closed in $\mathbb{R}^2$ but its projection $\mathbb{R} \setminus \{0\}$ onto the $x$-axis is not closed). The [Closed Map Lemma](/theorems/317) says that continuous maps from compact to Hausdorff spaces are always closed — this is the key to the [Topological Inverse Function Theorem](/theorems/318). Quotient maps $q: X \to X/{\sim}$ are always continuous and surjective, but they are open only if the equivalence relation is "well-behaved" (e.g., if the equivalence classes are orbits of a topological group action by homeomorphisms). ## Proper Maps Closed maps ensure that closed sets push forward to closed sets, but they impose no control on the "size" of fibres. A proper map adds a compactness condition on fibres that prevents mass from escaping to infinity — making it the topological analogue of the coercivity conditions that appear throughout analysis and PDE theory. Properness is the key hypothesis in degree theory, in the global analysis of fibre bundles, and in many compactness arguments where one needs preimages of bounded sets to be bounded. [definition:Proper Map] A continuous map $f: X \to Y$ is **proper** if $f^{-1}(K)$ is compact for every compact $K \subseteq Y$. [/definition] Every continuous map from a compact space is proper (since every preimage is a closed subset of a compact space, hence compact). For non-compact domains, properness is a strong condition: the inclusion $\iota: (0, 1) \hookrightarrow \mathbb{R}$ is continuous but not proper, because $\iota^{-1}([0, 1]) = (0, 1)$ is not compact. Proper maps to locally compact Hausdorff spaces are closed. This gives an alternative route to the Closed Map Lemma: a continuous map from a compact space to a Hausdorff space is proper (preimages of compact sets are closed subsets of the compact domain, hence compact), hence closed. Properness is the topological analogue of a "coercivity" condition in analysis: it ensures that sequences cannot escape to infinity without their images also escaping. This makes proper maps the natural setting for degree theory and for the global analysis of fibre bundles. ## Extension of Continuous Functions A recurring problem in topology and analysis is the *extension problem*: given a continuous function defined on a subspace, can it be extended to a continuous function on the whole space? This question is not merely technical — it underlies the construction of partitions of unity, the theory of retracts, and the very definition of normal spaces. On [metric spaces](/page/Metric%20Space), extensions are often constructed explicitly (e.g., by the Tietze formula or McShane's extension for Lipschitz functions). In general topology, existence depends on separation axioms: the space must have enough open sets to "interpolate" between the values prescribed on the subspace. [quotetheorem:888] The Tietze-Urysohn theorem is the definitive positive result: on normal spaces (which include all metric spaces and all compact Hausdorff spaces), bounded continuous real-valued functions on closed subsets always extend, and the extension preserves the sup-norm. The hypothesis that $C$ be closed is essential — on an open subset, the function might oscillate near the boundary in a way that prevents continuous extension. The normality hypothesis cannot be dropped either: Tietze extension *characterises* normality, in the sense that a $T_1$ space is normal if and only if every bounded continuous real-valued function on a closed subset extends continuously. The extension problem becomes far more subtle for maps into spaces other than $\mathbb{R}$. The Borsuk-Dugundji theorem extends the result to maps into locally convex [topological vector spaces](/page/Topological%20Vector%20Space), but for maps into spheres or other non-contractible targets, extension is obstructed by algebraic topology — the obstructions are measured by cohomology classes. ## Bibliography - Munkres, *Topology* (2000). - Willard, *General Topology* (1970). - Engelking, *General Topology* (1989). - Kelley, *General Topology* (1955).