= Continuous map
{title2=$f:X\to Y$}
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>,
$$
f^{-1}((1,4))=(-2,-1)\cup(1,2).
$$
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>.
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