Two continuous maps into a Hausdorff space that agree on a dense subset are equal

Let $Z$ be a topological space, let $D \subseteq Z$ be dense (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets), let $Y$ be Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and let $f, g : Z \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally) with

$f(d) = g(d) \qquad \text{for every } d \in D .$

Then $f = g$.

So a continuous map into a Hausdorff space is determined by its restriction to any dense subset of its domain. Nothing is asserted about which functions on $D$ extend: the statement is about uniqueness of an extension, not existence.

• The Hausdorff hypothesis is spent exactly once, inside [L1], and the density hypothesis exactly once, at step 2.1. Neither is used anywhere else, and neither can be weakened to the other: a dense agreement set alone does not force equality without a separation hypothesis on the codomain, and a Hausdorff codomain alone plainly does not.

• Density is a hypothesis about $Z$, not about $Y$. In particular the statement is about one domain and one dense subset of it; it says nothing about restrictions to subsets that are merely large in some other sense, and there is no cardinality condition anywhere in it.

• The uniqueness/existence split matters. A continuous $f : D \to Y$ need not extend continuously to $Z$ at all. What this corollary rules out is two different extensions, and that is exactly what makes an extension, when it exists, worth naming.