--- > [!proposition]+ Proposition. ([[basis characterization of continuity]]) > Let $(X, \tau_{X})$ and $(Y, \tau_{Y})$ be two [[topological space|topological spaces]], and $\mathscr{B}_{Y} \subset \tau_{Y}$ be a [[basis for a topology|basis]] [[topology generated by a basis|generating]] $\tau_{Y}$. A map $f: X \to Y$ is [[continuous]] if and only if $f^{-1}(B) \in \tau_{X}$ for any $B \in \mathscr{B}_{Y}$. ^proposition > [!proof]+ Proof. ([[basis characterization of continuity]]) > $\to.$ Suppose $f^{-1}(B) \in \tau_{X}$ for all $B \in \mathscr{B}_{Y}$. Let $U \in \tau_{Y}$. Using [[open sets are unions of basis elements]], write $U$ as union of elements in $\mathscr{B}_{Y}$ thus: $U= \bigcup_{y}^{} B_{y}.$ Denote by $O_{y}$ the preimage of $B_{y}$ under $f$, $O_{y}:=f^{-1}(B_{y})$. Since $f$ is [[continuous]], $O_{y} \in \tau_{X}$. [[preimages and unions commute]] we obtain $f^{-1}(U)= f^{-1} (\bigcup_{y}^{}B_{y})=\bigcup_{y}^{}f^{-1}(B_{y})= \bigcup_{y}^{}O_{y} \in \tau_{X}$ where the last step follows from the fact that $\tau_{X}$ is stable under arbitrary unions. > $\leftarrow.$ Suppose $f$ is continuous. Then for any open set $U$ in $Y$, we have $f^{-1}(U)=\tau_{X}$. In particular, each basis element $B \in \mathscr{B}_{Y}$ is open in $Y$, so the preimage $f^{-1}(B)$ is in $\tau_{X}$ as required. ^proof --- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag #reformatrevisebbatch2