Suppose you have an open map \(p\) between topological spaces, and if you have a subet \(A\) of \(p\)’s domain such that \(p(A)\) is open. Can you then conclude that \(A\) is open? Nope! Consider the following spaces \(X=\{x_1,x_2\}\) and \(Y=\{y_1,y_2\}\) with topologies \(\tau_X=\{\varnothing, X, \{x_1\}\}\) and \(\tau_Y=\{\varnothing,Y,\{y_1\}\}\), respectively and let \(p: X\times Y\to X\) be the projection onto its first fator. This is an open map. If we consider \(A=X\times\{y_2\}\) we see that \(A\) is not open in \(X\times Y\), but we have that \(p(A)=p(X\times\{y_2\})= X\) which is trivially open in \(X\).

# Pitfalls of open maps

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