Exercise · ZW-TOPOLOGICAL-HOMEOMORPHISM-005 · ★☆☆ Basic
Let \(X=\{1,2,3\}\) with open sets \(\varnothing,\{1\},\{1,2\},X\), and \(Y=\{a,b,c\}\) with open sets \(\varnothing,\{c\},\{b,c\},Y\). If \(f:X\to Y\) is a homeomorphism , what is \(f(3)\)?
Choose one option.
A \(b\) B \(a\) C \(c\) D It can be any of the three Show answer ← Back to the question
Explanation The open sets of each space form a chain. \(f\) must send \(\{1\}\) to \(\{c\}\) and \(\{1,2\}\) to \(\{b,c\}\), so \(f(1)=c\), \(f(2)=b\) and hence \(f(3)=a\). It is the only homeomorphism .
A bijection \(f\) between topological spaces \(X\) and \(Y\) that is continuous and whose inverse \(f^{-1}\) is also continuous.
\[f:X\to Y\ \text{bijective};\quad f,\ f^{-1}\ \text{continuous}\]
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