Let \(X=\{1,2,3\}\) with \(\mathcal{T}=\{\varnothing,\{2\},\{1,2\},\{2,3\},\{1,2,3\}\}\) and \(Y=\{a,b,c\}\) with \(\mathcal{T}=\{\varnothing,\{b\},\{a,b\},\{b,c\},\{a,b,c\}\}\). Which function \(f:X\to Y\) is continuous?
It suffices that the preimage of each proper open set of \(Y\) be open in \(X\). For the correct one: \(f^{-1}(\{b\})=\{2\}\); \(f^{-1}(\{a,b\})=\{1,2\}\); \(f^{-1}(\{b,c\})=\{2,3\}\), all open. By contrast, for \(f(1)=a,\ f(2)=a,\ f(3)=b\) we get \(f^{-1}(\{b\})=\{3\}\), which is not open.
A function \(f:X\to Y\) between topological spaces \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) is continuous if the preimage of every open set \(V\) of \(Y\) is an open set of \(X\).