Continuous map ($f:X\to Y$)

A map $f:X\to Y$ between topological spaces is continuous when every open set $V\subseteq Y$ has an inverse image $f^{-1}(V)=\{x\in X:f(x)\in V\}$ that is an open set in $X$. The notation denotes a subset, so $f$ need not have an inverse function.

For the continuous map $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, with the usual metric topology,The two open intervals form an open set; an inverse image can split into separate pieces without violating continuity. The defining condition concerns inverse images, rather than images: here $f((-1,1))=[0,1)$ is not an open set in the real line. Checking this one inverse image illustrates the definition; it alone does not prove continuity for all target open sets.

For a continuous map into a metric space, $f(\overline{A})\subseteq \overline{f(A)}$: an open neighborhood of an image point pulls back to a neighborhood meeting $A$. A bounded set lies in a closed ball, and its closure lies in the same ball because distance from the centre is a continuous function. The displayed boundedness conclusion follows. Neither compactness of the domain nor uniform continuity of the map is needed.

A function of several variables is separately continuous when it is continuous in each variable with the others fixed. This is weaker than joint continuity in the product topology. Functions with narrow rectangular supports provide useful examples distinguishing convergence in topologies generated by separate continuity from Euclidean convergence.

Let $(u_n,v_n)\to (0,0)$ with both coordinates nonzero, $r_n=|u_n|/4$, $s_n=|v_n|/4$, and $\varphi (t)=\max (1-|t|,0)$. The displayed function equals one at every $(u_n,v_n)$ and zero on both coordinate axes. Each coordinate line off the corresponding axis meets only finitely many supports, so its restriction is a finite maximum of continuous functions. On the axes it is zero. Hence $B$ is separately continuous and disproves convergence of this sequence in any topology making every separately continuous function continuous. Isotropic radial bumps need not have this property when one coordinate approaches zero much faster than the other.

A function $f:X\to Y$ between topological spaces is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$ for every subset $A\subseteq X$, where the bars denote closure. For a continuous $f$, pull back an open neighbourhood of $f(x)$ for $x\in \overline{A}$; its preimage meets $A$, so the neighbourhood meets $f(A)$. Conversely, apply the inclusion to $A=f^{-1}(C)$ with $C$ closed. Then $f(\overline{A})\subseteq C$ implies $\overline{A}\subseteq A$, so the preimage of every closed subset is closed and $f$ is continuous.

A function $f:X\to Y$ between topological spaces is continuous exactly when the inverse image of every closed set in $Y$ is closed in $X$. The defining open set criterion is equivalent, because $f^{-1}(Y\setminus A)=X\setminus f^{-1}(A)$. This criterion concerns inverse images; a continuous map need not be a closed map.

A continuous map that is also an injective function. A continuous injection need not be a topological embedding: its inverse on its image need not be continuous. If its source is a compact space and its target is a Hausdorff space, however, the compact-to-Hausdorff continuous bijection theorem makes that inverse continuous.

A continuous bijection is a continuous map that is a bijection. It need not be a homeomorphism, since its inverse function need not be continuous. The compact-to-Hausdorff continuous bijection theorem supplies a useful sufficient condition for the stronger conclusion.

For continuous maps $p:E\to B$ and $f:X\to B$, a continuous lift of $f$ through $p$ is a continuous map $\tilde{f}:X\to E$ with $p\tilde{f}=f$. Existence and uniqueness require additional hypotheses; the lifting criterion for a covering space states a useful existence criterion when $p$ is a covering map and $X$ is path-connected and locally path-connected.

The continuous map projecting a real number onto the closed interval $[a,b]$, with $a\le b$. It equals $a$ below the interval, equals the input inside it, and equals $b$ above it. It is nondecreasing and has Lipschitz constant one.

A continuous map injective map $f:X\to Y$ that is a homeomorphism onto its image with the subspace topology. A continuous map injective map from a compact space into a Hausdorff space is a topological embedding.

Continuous functions $X\to [0,1]$ on a normal Hausdorff space separate points and closed sets by Urysohn's lemma. The evaluation map into the product of their codomains is injective and continuous. For each point in an open set, a coordinate equal to one there and zero outside the open set detects a smaller coordinate neighborhood, proving the inverse topology.

A subset $Y\subseteq X$ is C-embedded if every continuous real-valued function on its subspace topology extends continuously to $X$. Closed subsets of normal spaces have this property by the Tietze extension theorem. The condition also makes sense for nonclosed subsets.

A subset is Cb-embedded if every bounded continuous real-valued function on it extends to a bounded continuous real-valued function on the ambient space. A bounded extension can be clipped to preserve the original bound. This is weaker than C-embeddedness: the natural numbers in their Stone-Čech compactification have the bounded extension property, but their unbounded identity function cannot extend to a compact space.

A family of maps with a common domain determines the displayed map into the Cartesian product of their codomains. It is injective exactly when the family separates points. For continuous coordinate maps it is continuous in the product topology. If inverse images of open sets under the coordinates generate the original topology, the inverse from the image is also continuous; thus the map is a topological embedding. A basis of such inverse images, as opposed to a subbasis, is a sufficient stronger hypothesis.

Every topological space, including one without the T0 space property, embeds into a power of this three-point space. For each open set $U$, use a coordinate equal to $2$ on $U$ and $0$ elsewhere. Its inverse image of the only proper nonempty open set is $U$. For each point $a$, add a coordinate equal to $1$ at $a$ and $0$ elsewhere. These coordinates are continuous because the subspace $\{0,1\}$ has the indiscrete topology, and they separate all points. The combined diagonal map is injective and induces precisely the original topology, hence is a topological embedding.