Proposition 4.8.2. Let $X$ be a topological space, $Y$ be a Hausdorff space, $A \subset X$ be a dense subset, and $F, G \in C(X; Y)$. If $F|_{A} = G|_{A}$, then $F = G$.
Proof. Let $x \in X$. By (4) Definition 4.5.2, there exists a filter base $\fB \subset 2^{A}$ that converges to $x$. By (2) of local continuity, $F(\fB)$ converges to $F(x)$ and $G(\fB)$ converges to $G(x)$. Since $F(\fB) = G(\fB)$, $F(x) = G(x)$ by (4) of Definition 4.8.1.$\square$