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If $f:X\to Y$ is a continuous map and $X$ is a connected space, then its image $f(X)$, with the subspace topology from $Y$, is a connected subset of $Y$. The whole codomain $Y$ need not be connected; the assertion concerns the values actually attained. No injectivity is required.^[ref]

For example, the continuous function $f(x)=x^2$ maps the connected closed interval $[-2,2]$ onto the connected closed interval $[0,4]$, even though opposite inputs have the same output. The same continuous function maps the disconnected union $[-2,-1]\cup [1,2]$ onto $[1,4]$. Thus a connected image does not imply a connected source: a continuous map can merge different connected components.

For a continuous function $r:[0,1]\to \mathbb{R}$ representing a continuous reading, $r(0)=0$ and $r(1)=1$ force some input with $r(t)=1/2$. Its image is a connected subset of the real line, hence a real interval containing both endpoint readings. For the synthetic model $r(t)=t^2$, the missing-value test is explicit: $r(1/\sqrt{2})=1/2$. A step reading equal to $0$ for $t<1/2$ and $1$ for $t\ge 1/2$ skips $1/2$, but fails continuity at $t=1/2$.

Proof outline: a separation of a topological space of $f(X)$ into nonempty open subsets pulls back under $f$ to nonempty open subsets of $X$. Their intersection is empty and their union is $X$, contradicting connectedness. This also shows why a continuous map sends each connected component of its source into a single connected component of its target.

A connected component of a topological space $X$ is a nonempty connected subset that is maximal under subset inclusion. If $C$ is a connected component and $C\subseteq D\subseteq X$ with $D$ a connected subset, then $D=C$. Maximality means that no larger connected subset contains $C$; it does not mean largest cardinality among all connected subsets.

For example, the connected components of $[-2,2]\setminus \{0\}$ in the usual subspace topology of the real line are $[-2,0)$ and $(0,2]$. Each is a real interval, and no connected subset of the real line can meet both without containing the missing point $0$. For the finite set $\{-2,0,2\}$ in the same subspace topology, the three connected components are its singleton sets.

Every point $x\in X$ lies in exactly one connected component, constructed by taking the unionAll members contain $x$, so their union is connected by connected union with a connected hub. The union includes every connected subset through $x$, so it is maximal. Equivalently, defineThis is an equivalence relation: singleton sets give reflexivity, symmetry is immediate, and two connected subsets sharing the middle point have a connected union, giving transitivity. The equivalence classes are precisely the connected components. They form a partition of a set: they cover $X$ and any two are either equal or disjoint.^[ref]

Every nonempty connected subset lies within one connected component. Moreover, every connected component is closed in $X$: closure of a connected set makes its closure connected, and maximality forces that closure to equal the original connected component. A connected component need not be open. For example, in $S=\{0\}\cup \{1/n:n=1,2,\dots \}$ with its usual subspace topology from the real line, all connected components are singleton sets, but $\{0\}$ is not open: every neighbourhood of $0$ in $S$ contains some $1/n$.^[ref]

The map $z\mapsto \cosh (\pi z)$ is a conformal bijection fromonto the upper half-plane. Its three boundary intervals map respectively to $(1,\infty )$, $(-1,1)$ and $(-\infty ,-1)$.

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