Activity 19.2.
In this activity we prove Theorem 19.3.
Theorem 19.3.
Let \(X\) and \(Y\) be topological spaces and let \(f : X \to Y\) be a continuous function. If \(A\) is a path connected subspace of \(X\text{,}\) then \(f(A)\) is a path connected subspace of \(Y\text{.}\)
Assume that \(X\) and \(Y\) are topological spaces, \(f : X \to Y\) is a continuous function, and \(A \subseteq X\) is path connected. To prove that \(f(A)\) is path connected, we choose two elements \(u\) and \(v\) in \(f(A)\text{.}\) It follows that there exist elements \(a\) and \(b\) in \(A\) such that \(f(a) = u\) and \(f(b) = v\text{.}\)
(a)
Explain why there is a continuous function \(p: [0,1] \to A\) such that \(p(0) = a\) and \(p(1) = b\text{.}\)
(b)
Determine how \(p\) and \(f\) can be used to define a path \(q: [0,1] \to f(A)\) from \(u\) to \(v\text{.}\) Be sure to explain why \(q\) is a path. Conclude that \(f(A)\) is path connected.
