(Purdue ’10) Prove or disprove: If \(X\) is path connected, and \(f:X\to Y\) is continuous, then \(f(x)\) is path connected.
Solution
As printed, \(f(x)\) denotes a single point, so the intended assertion is that the image \(f(X)\) is path connected. That assertion is true.
1. Let \(y_0,y_1\in f(X)\).Choose \(x_0,x_1\in X\) with\[ f(x_0)=y_0, \qquad f(x_1)=y_1. \]
This is the definition of membership in the image \(f(X)\).
2. There exists a path\[ \gamma:[0,1]\to X \]with \(\gamma(0)=x_0\) and \(\gamma(1)=x_1\).
The space \(X\) is path connected.
3. The composite\[ f\circ\gamma:[0,1]\to f(X) \]is a path from \(y_0\) to \(y_1\).
The composite is continuous because both \(\gamma\) and \(f\) are continuous, and its image lies in \(f(X)\). Moreover, \[ (f\circ\gamma)(0)=f(x_0)=y_0, \qquad (f\circ\gamma)(1)=f(x_1)=y_1. \]
4. Therefore \(f(X)\) is path connected.
The points \(y_0,y_1\in f(X)\) in 1 were arbitrary, and 3 constructs a path between them.