Given two spaces \(X\) and \(Y\), let \(\pi_X:X\times Y\to X\) and \(\pi_Y:X\times Y\to Y\) be the two standard projection maps.
Show that, for any other space \(Z\), a function \(f:Z\to X\times Y\) is continuous if and only if \(\pi_X\circ f\) and \(\pi_Y\circ f\) are continuous.
Use this to show that the cartesian product of two path-connected topological spaces is path-connected.
Solution
1. If \(f:Z\to X\times Y\) is continuous, then both coordinate maps \(\pi_X\circ f\) and \(\pi_Y\circ f\) are continuous.
The projection maps \[ \pi_X:X\times Y\to X, \qquad \pi_Y:X\times Y\to Y \] are continuous for the product topology. Therefore each composite of a projection with the continuous map \(f\) is continuous.
2. Conversely, if both coordinate maps are continuous, then \(f\) is continuous.
Assume \[ f_X=\pi_X\circ f:Z\to X, \qquad f_Y=\pi_Y\circ f:Z\to Y \] are continuous. A basis for the product topology on \(X\times Y\) consists of sets \(U\times V\) with \(U\subseteq X\) open and \(V\subseteq Y\) open. For every such basic open set, \[ f^{-1}(U\times V) =f_X^{-1}(U)\cap f_Y^{-1}(V). \] Both sets on the right are open in \(Z\), so their intersection is open. Thus the inverse image under \(f\) of every basic open subset of \(X\times Y\) is open, which proves that \(f\) is continuous.
3. Therefore a map into \(X\times Y\) is continuous if and only if both of its coordinate maps are continuous.
Combine 1 and 2.
4. If \(X\) and \(Y\) are path-connected, then \(X\times Y\) is path-connected.
Take arbitrary points \[ (x_0,y_0),(x_1,y_1)\in X\times Y. \] Since \(X\) is path-connected, there is a continuous path \[ \alpha:[0,1]\to X \] with \(\alpha(0)=x_0\) and \(\alpha(1)=x_1\). Likewise, path-connectedness of \(Y\) gives a continuous path \[ \beta:[0,1]\to Y \] with \(\beta(0)=y_0\) and \(\beta(1)=y_1\). Define \[ \gamma:[0,1]\longrightarrow X\times Y, \qquad \gamma(t)=(\alpha(t),\beta(t)). \] Its coordinate maps are \[ \pi_X\circ\gamma=\alpha, \qquad \pi_Y\circ\gamma=\beta, \] so 3 shows that \(\gamma\) is continuous. Moreover, \[ \gamma(0)=(x_0,y_0), \qquad \gamma(1)=(x_1,y_1). \] Thus every two points of \(X\times Y\) can be joined by a path, and \(X\times Y\) is path-connected.