Munkers Section 16 Exercise 6 Proof
This is question 4 on page 92 in Chapter 16 of James R. Munkers’ Topology 2nd Edition textbook.
Question
A map \(f: X \rightarrow Y\) is said to be an open map if for every open set \(U\) of \(X\), the set \(f(U)\) is open in Y. Show that $\pi_1 : X \times Y \rightarrow X$ and $\pi_2 : X \times Y \rightarrow Y$ are open maps.
\(X\) and \(Y\) are both topologies.
Proof
For \(\pi_1 : X \times Y \rightarrow X\):
Let \(U\) be an open set of \(X\). Because \(\pi_1\) maps to the entirety of \(X\), discarding \(Y\) in the process, it follows that \(U\) must be in the resulting topological space \(X\).
Similar logic holds for \(\pi_2 : X \times Y \rightarrow Y\):
Let \(V\) be an open set of \(Y\). Because \(\pi_2\) maps to the entirety of \(Y\), discarding \(X\) in the process, it follows that \(V\) must be in the resulting topological space \(Y\).
\[\tag*{$\blacksquare$}\]Extras
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